Answer
given,
force per unit length = 350 µN/m
current, I = 22.5 A
y = y = 0.420 m
[tex]\dfrac{F}{L}= \dfrac{KI_1I_2}{d}[/tex]
[tex]I_2 = \dfrac{F}{L}\dfrac{d}{KI_1}[/tex]
[tex]I_2 = 350\times 10^{-6}\times \dfrac{0.42}{2 \times 10^{-7}\times 22.5}[/tex]
I₂ = 32.67 A
distance where the magnetic field is zero
[tex]\dfrac{4\pi \times 10^{-7}\times 32.67}{2\pi y_1}=\dfrac{4\pi \times 10^{-7}\times 22.5}{2\pi (0.42-y_1)}[/tex]
[tex]y_1 = 0.248\ m[/tex]
there the distance at which the magnetic field is zero in the two wire is at 0.248 m.
A grocery shopper tosses a(n) 9.1 kg bag of
rice into a stationary 19.3 kg grocery cart.
The bag hits the cart with a horizontal speed
of 6.6 m/s toward the front of the cart.
What is the final speed of the cart and bag?
Answer in units of m/s.
Answer:
2.1 m/s
Explanation:
Momentum is conserved, so:
m₁ u₁ + m₂ u₂ = (m₁ + m₂) v
(9.1 kg) (6.6 m/s) = (9.1 kg + 19.3 kg) v
v = 2.1 m/s
A person sitting in a parked car hears an approaching ambulance siren at a frequency f1. as it passes him and moves away, he hears a frequency f2. the actual frequency f of the source is (which one of the following)
a. f > f1
b. f < f2
c.f= f2 - f1
d. f = f2 + f1
e. f2 < f < f1
Answer:E
Explanation:
This can be explained by Doppler effect which gives the relation between apparent Frequency and actual frequency when the source of sound is moving
[tex]f'=f\frac{v+v_o}{v-v_s}[/tex]
where [tex]f'=apparent\ frequency [/tex]
[tex]f=actual\ frequency[/tex]
[tex]v=velocity\ of\ sound[/tex]
[tex]v_o=velocity\ of\ observer[/tex]
[tex]v_s=velocity\ of\ source[/tex]
here [tex]v_o =0[/tex] as observer is standing
when Ambulance is approaching then velocity of sound and velocity of ambulance have relative velocity thus denominator is less than and apparent frequency is more while when ambulance is going away then velocity of sound waves and velocity of observer is in opposite direction thus denominator is less than 1 and apparent frequency is less.
Thus [tex]f_2<f<f_1[/tex]
The Doppler effect explains the changes in sound frequency from a moving source as perceived by a stationary observer. The observed frequency increases as the source approaches and decreases when it moves away. Hence, the actual frequency of the source (the siren) is less than the frequency observed when the source is approaching (f1) and greater than the frequency observed than when it is retreating (f2). The correct answer is 'e. f2 < f < f1'.
Explanation:This question relates to the phenomenon of the Doppler effect, which is the change in frequency or wavelength of a wave in relation to an observer who is moving relative to the wave source. In this case, when an ambulance with a siren is approaching a static observer, the observed frequency f1 will be higher than the actual frequency f (the frequency of the siren when you are sitting inside the ambulance), due to compression of the sound waves.
However, when the ambulance starts moving away from the observer, the observed frequency f2 will be lower than f, due to elongation or stretching out of the sound waves. Hence, the correct answer is 'e. f2 < f < f1'.
Learn more about Doppler effect here:https://brainly.com/question/15318474
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At a circus, a clown is being shot out of a cannon. This is a good thing because clowns are terrifying. The cannon's barrel makes an angle of 35 degrees with the ground and the initial velocity has magnitude 10.4 m/s. The end of the barrel is 1.0 m above the ground. A 1.0 m thick mattress, with sides 2.0 m long, is placed so its closest edge is a horizontal distance of 10.0 m from the end of the cannon's barrel. (a) Draw a diagram of the situation described. Label the x and y directions, the cannon, the mattress, the initial velocity vector and the barrel angle. (b) What are the vertical and horizontal components of the initial velocity vector? (c) At what time will the clown return to a height of 1.0 m above the ground? (d) Using your answer from (c), will the clown reach the mattress and escape serious injury??? What would you expect if we included the effects of air resistance? (disclaimer: no matter how terrifying clowns are, I still would not wish one to be seriously hurt)
Answer:
a) Please, see the attached figure.
b) The horizontal component of the initial velocity is 8.5 m/s
The vertical component of the initial velocity is 6.0 m/s
c) The clown will return to a height of 1.0 m after 1.2 s of the launch.
d) The clown will land safely on the mattress, 10.2 m from the cannon. If we include air resistance in the calculation, he will surely not reach the mattress because, without air resistance, he lands just 20 cm from the closest edge of the mattress.
Explanation:
Hi!
a) Please, see the attached figure.
b) As shown in the figure, the initial velocity vector is the following:
v0 = (v0x, v0y)
Using trigonomety of right triangles:
cos angle = adjacent side / hypotenuse
In this case:
Adjacent side = v0x
hypotenuse = v0
(see figure)
Then:
cos 35° = v0x / v0
v0 · cos 35° = v0x
v0x = 10.4 m/s · cos 35°
v0x = 8.5 m/s
The horizontal component of the initial velocity is 8.5 m/s
We proceed in the same way to find the vertical component:
sin angle = opposite side / hypotenuse
sin 35° = v0y / v0 (see figure to notice that opposite side = v0y)
v0 · sin 35° = v0y
10.4 m/s · sin 35° = v0y
v0y = 6.0 m/s
The vertical component of the initial velocity is 6.0 m/s
c) The equation of the position vector of the clown at time t is the following:
r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)
Where:
x0 = initial horizontal position.
v0 = initial velocity.
t = time.
α = launching angle.
y0 = initial vertical position.
g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).
We have to find the time at which the vertical component of the position vector is 1 m. Let´s place the origin of the system of reference at the point where the cannon is located on the ground so that x0 = 0 and y0 = 1.0 m.
Using the equation of the vertical component of the position:
y = y0 + v0 · t · sin α + 1/2 · g · t²
1.0 m = 1.0 m + 10.4 m/s · t · sin 35° - 1/2 · 9.8 m/s² · t²
0 = 10.4 m/s · t · sin 35° - 4.9 m/s² · t²
0 = t (10.4 m/s · sin 35° - 4.9 m/s² · t) (t = 0, when t = 0 the clown is 1.0 m above the ground, just leaving the cannon).
0 = 10.4 m/s · sin 35° - 4.9 m/s² · t
-10.4 m/s · sin 35° / -4.9 m/s² = t
t = 1.2 s
The clown will return to a height of 1.0 m after 1.2 s of the launch.
d) Now, let´s calculate the horizontal traveled distance after 1.2 s using the equation of the horizontal component of the position vector:
x = x0 + v0 · t · cos α (x0 = 0)
x = 10.4 m/s · 1.2 s · cos 35°
x = 10.2 m
Since the mattress is located at 10 m from the cannon and it is 2.0 m long, the clown will land safely on the mattress. However, the clown almost miss the mattress (he lands just 20 cm from the closest edge), so, if we include air resistance in the calculation, he will surely not reach the mattress.
What mathematical evidence can you provide to support the assertion that the universe is electrically neutral on large scales?Ryden, Barbara. Introduction to Cosmology (p. 47). Cambridge University Press. Kindle Edition.
It is likely that until today there is no physical / mathematical argument that allows to equate this statement. But there are countless observations that have led to that conclusion.
From these observations two types of highlights can be summarized: Magnetic fields and Gravity Force.
Electromagnetic Field: If there is a negative or positive charge, these charges should be observed through the electric and magnetic fields by means of microwaves. However, there is currently no observation that supports any electromagnetic charge in space.
Force of gravity: The only force capable of maintaining the attraction of the planets, without affecting (at certain scales), is the force of gravity. An electromagnetic force often exceeds that of gravity, but has not been observed at the moment.
Final answer:
Mathematical and observational evidence from electromagnetism and cosmology, such as the Maxwell-Einstein theoretical framework and charge quantization, supports the assertion that the universe is electrically neutral on large scales.
Explanation:
To support the assertion that the universe is electrically neutral on large scales, one can turn to the principles of electromagnetism and cosmology melded in the framework of the general theory of relativity. A key aspect of this inquiry relates to the near uniform distribution of matter and the overall balance of positive and negative charges in the cosmos.
On quantum scales, charge quantization ensures that matter is made up of atoms with equal numbers of protons and electrons, contributing to the electrical neutrality on a smaller scale. This neutrality scales up to larger structures in the universe. Observational evidence also points towards the large-scale neutrality of the universe; otherwise, we would observe huge electrostatic forces that would significantly alter the structure of the cosmos.
Which technology category would NOT likely be the subject of a standard published by the International Electrotechnical Commission (IEC)?
A. Semiconductors
B. Solar energy
C. Encryption
D. Consumer appliances
Encryption would not likely be the subject of a standard published by the International Electrotechnical Commission (IEC).
Answer: Option C
Explanation:
IEC (International Electrotechnical Commission) is the world's number one organization that develops and releases international standards for all electronic-electrical technologies and its affiliates.
Its standards would cover a wide technology range, from energy generation, transmissions and distributions to house appliances and offices, semiconductors, batteries, fibreglass, solar energy, Nano-technology and sea energy and many more.
IEC standards apply to all electrical technologies, including electronics, magnets and electro-magnetics, electro-acoustics, multi-media, telecommunications, and medical technologies, as well as related general fields such as terminology and symbols, etc.
On the other side of the pivot point an adult pushes straight down on the teeter-totter with a force of 190 N. Determine the direction the teeter-totter will rotate if the adult applies the force at each of the following distances from the pivot. (Assume the teeter-totter is horizontal when the adult applies the force and that the child's weight applies a clockwise torque.)
The question seems incomplete. I found a similar version with different details. For the purpose of helping you answering the question, I will use the details from the original questions added with the sub-questions from other question. You can change the details accordingly using the explanation given below:
At the local playground, a 21-kg child sits on the right end of a horizontal teeter-totter, 1.8 m from the pivot point. On the other side of the pivot point an adult pushes straight down on the teeter-totter with a force of 190 N. Determine the direction the teeter-totter will rotate if the adult applies the force at each of the following distances from the pivot. (Assume the teeter-totter is horizontal when the adult applies the force and that the child's weight applies a clockwise torque.)
(a) 3.0 m
(b) 2.5 m
(c) 1.5 m
Answer:
a) anticlockwise
b) anticlockwise
c) clockwise
Explanation:
To answer this question, we will need the knowledge of moment/torque
Torque, T = r*Fsinθ
r = radial distance from pivot to force, F
F = Force in the direction perpendicular to the distance
θ = angle between force line and r line
Note that the direction of movement will be affected by total torque on the system. Assuming both adult and the child is initially sitting on the same level, we'll have force due to the child weight and adult force going downward and perpendicular to their respective seat on teeter-totter.
Hence,
θ = 90 degrees -> sinθ = 1
Therefore,
T = r*F(1)
T = r*F
RHS:
Torque of the child,
Tc = 1.8(21*9.8) = 370.44 Nm
LHS:
Torque of adult
Ta = d(190) =190d
with d as the distance of the adult from pivot
a) d = 3.0m
Ta = 190*3 = 570 Nm
Ta > Tc. Therefore the teeter-totter will rotate towards the adult side (LHS - going anti-clockwise)
b) d = 2.5m
Ta = 190*2.5 = 475 Nm
Ta > Tc. Therefore the teeter-totter will rotate towards the adult side (LHS - going anti-clockwise)
c) d = 1.5m
Ta = 190*1.5 = 285
Ta < Tc. Therefore the teeter-totter will rotate towards the child side (RHS - going clockwise)
A 1.80-m string of weight 0.0123N is tied to the ceiling at its upper end, and the lower end supports a weight W. Neglect the very small variation in tension along the length of the string that is produced by the weight of the string. When you pluck the string slightly, the waves traveling up the string obey the equationy(x,t)=(8.50mm)cos(172rad?m^(?1)x?2730rad?s^(?1)t)Assume that the tension of the string is constant and equal to W.a) How much time does it take a pulse to travel the full length of the string?b) What is the weight W?c) How many wavelengths are on the string at any instant of time?d) What is the equation for waves traveling down the string?a) y(x,t)=(8.50mm)cos(172rad?m?1x?2730rad?s?1t)b) y(x,t)=(8.50mm)cos(172rad?m?1x+2730rad?s?1t)c) y(x,t)=(10.5mm)cos(172rad?m?1x+2730rad?s?1t)d) y(x,t)=(10.5mm)cos(172rad?m?1x?2730rad?s?1t)
Answer:
a) t = 0.113 s , b) W = 0.1756 N , c) # λ = 49
Explanation:
a) Let's use the relationship
v = λ f
Of the equation
y = 8.55 10-3 cos (172 x + 2730 t)
When comparing this with the general equation
y = A cos (kx - wt)
The wave number k = 172
k = 2π /λ
λ = 2π / 172
λ = 0.03653 m
The angular velocity w = 2730
w = 2π f
f = w / 2π
f = 2730 / 2π
f = 434.49 Hz
The speed of the wave is
v = 0.03653 434.49
v = 15.87 m / s
The speed the wave on a string is constant, so
v = d / t
t = d / v
t = 1.80 /15.87
t = 0.113 s
b) The weight applied to the rope
v = √ T /μ
The density
μ = m / l
μ = (0.0123 / 9.8) /1.80
μ = 6.97 10-4 kg / m
The tension equal to the applied weight
T = v² μ
W = T = 15.87² 6.97 10⁻⁴
W = 0.1756 N
c) let's use a rule of proportions
# λ = 1.8 /0.03653
# λ = 49
A liquid of density 1370 kg/m 3 flows steadily through a pipe of varying diameter and height. At Location 1 along the pipe, the flow speed is 9.67 m/s and the pipe diameter d 1 is 12.9 cm . At Location 2, the pipe diameter d 2 is 16.7 cm . At Location 1, the pipe is Δ y = 9.85 m higher than it is at Location 2. Ignoring viscosity, calculate the difference Δ P between the fluid pressure at Location 2 and the fluid pressure at Location 1.
Answer:
ΔP = P₂ - P₁ = 132. 24 kPa
Explanation:
Given:
ρ = 1370 kg / m³ , v = 9.67 m / s , d₁ = 12.9 cm , d₂ = 16.7 cm , Δy = 9.85 m
Using Bernoulli's equations to determine the difference ΔP
P₂ + ρ * g * Z₂ + (ρ * V₂²) / 2 = P₁ + ρ * g * Z₁ + (ρ * V₁²) / 2
P₂ - P₁ = ρ * g * (Z₁ - Z₂) + [ ρ * (V₁² - V₂²) ] / 2
P₂ - P₁ = ρ * g * (Z₁ - Z₂) + ¹/₂ * ρ * V₁² * [ ( 1 - (d₁ / d₂) ⁴ ) ]
ΔP = 1370 kg / m³ * 9.8 m/s² * 9.85m + 0.5 * 1370 kg / m³ * ( 1 - (12.9 cm / 16.7 cm )⁴ )
ΔP = 132247.9364 Pa
ΔP = 132. 24 kPa
Answer:
P₂ - P₁=173.5kPa
Explanation:
The equation of continuity:
A₁v₁=A₂v₂
where A₁=πd₁²/4 and A₂=πd₂²/4
v₂=(A₁/A₂)v₁
v₂={(πd₁²/4)/(πd₂²/4)}v₁
v₂=(d₁²/d₂²)v₁
Use Bernoulli's equation
P₂+pgz₂+(pv₂²/2)=P₁+pgz₁+(pv₁²/2)
The difference between the fluid pressure at location 2 and the fluid pressure at location 1
P₂ - P₁=pg(z₁-z₂)+{p(v₁²-v₂²)}/2=pg(z₁-z₂)+1/2pv₁²(1-(d₁/d₂)⁴)
P₂ - P₁=(1.370×10³×9.8×9.85)+(1/2)(1.370×10³×(9.67)²){(1-(0.129m/0.167m)⁴}
P₂ - P₁=1.735×10⁵Pa
P₂ - P₁=173.5kPa
What is the weight of a 15kg dog on earth?
The weight of a 15kg dog on Earth is 147 Newtons. Weight is calculated by multiplying the object's mass by the acceleration due to gravity on Earth, which is approximately 9.8 m/s². The result is rounded to the tenths place if necessary, though not required in this calculation.
Explanation:The weight of an object is the force due to gravity acting on that object's mass. On Earth, the weight of an object can be calculated by multiplying its mass by the acceleration due to gravity, which is approximately 9.8 m/s2. Therefore, the weight of a 15kg dog on Earth is calculated as 15 kg × 9.8 m/s2, which equals 147 Newtons (N).
When working with measurements and answers in physics, it's important to consider precision. Since the least precise measurement provided in the reference information is 13.7 kg, expressed to the 0.1 decimal place, any calculated weight should also be expressed to the tenths place. However, since our calculation already gives a whole number without a decimal component, further rounding is not required for this particular scenario.
A guitar string of length L = 0.79 m is oriented along the x-direction and under a tension of T = 74 N. The string is made of steel which has a density of rho = 7800 kg / m3. The radius of the string is r = 8.5 x 10-4 m. A transverse wave of amplitude A = 0.0020 m is formed on the string. Find the string tension.
Answer:
74 N
Explanation:
T = Tension in the string = 74 N
[tex]\rho[/tex] = Density of the steel = 7800 kg/m³
A = Area = [tex]\pi r^2[/tex]
r = Radius = [tex]8.5\times 10^{-4}\ m[/tex]
Linear density is given by
[tex]\mu=\rho A\\\Rightarrow \mu=7800\times \pi (8.5\times 10^{-4})^2\\\Rightarrow \mu=0.0177\ kg/m[/tex]
The linear density is 0.0177 kg/m
Velocity is given by
[tex]v=\sqrt{\dfrac{T}{\mu}}\\\Rightarrow v=\sqrt{\dfrac{74}{0.0177}}\\\Rightarrow v=64.65903\ m/s[/tex]
The velocity of the wave on the guitar string is 64.65903 m/s
Why should galaxy collisions have been more common in the past than they are today?
a. Galaxies attracted each other more strongly in the past because they were more massive; they had not yet turned most of their mass into stars.
b. Galaxies were much bigger in the past since they had not contracted completely.
c. Galaxies were closer together in the past because the universe was smaller.
d. Galaxies were more active in the past and therefore would have collided with each other more frequently.
Answer:
I think d is the answer haha
Galaxy collisions were more common in the past as C. the universe was smaller, hence galaxies were closer together.
Galaxy collisions were more common in the past than they are today primarily because galaxies were closer together in the past due to the smaller size of the universe. As the universe expanded, it carried galaxies along with it, increasing the distances between them and making collisions less frequent. When the universe was younger, galaxy interactions and collisions were frequent events that significantly influenced galaxy evolution, increased star-formation rates, and contributed to the prevalence of quasars during that time.
Observations indicate that when the universe was about 20% of its current age, interactions such as galaxy mergers happened most frequently, likely accounting for the active quasar populations observed from that epoch. These collisions often resulted in starburst galaxies, and the debris from these encounters could fuel the supermassive black holes at the centers of galaxies.
A pressure cooker is a pot whose lid can be tightly sealed to prevent gas from entering or escaping.If an otherwise empty pressure cooker is filled with air of room temperature and then placed on a hot stove, what would be the magnitude of the net force F120 on the lid when the air inside the cooker had been heated to 120C ? Assume that the temperature of the air outside the pressure cooker is 20C(room temperature) and that the area of the pressure cooker lid is A. Take atmospheric pressure to be Pa .Treat the air, both inside and outside the pressure cooker, as an ideal gas obeying PV=NkBT .F120=.............F20=..............
Final answer:
The net force F120 on the pressure cooker's lid when the inside air is heated to 120°C is the difference between the force exerted by the heated air inside and the atmospheric force outside on the lid's area A. Calculate pressure using the ideal gas law, then calculate the forces exerted at 120°C and at atmospheric conditions, and find their difference.
Explanation:
The student asks about the magnitude of the net force F120 on the lid of a pressure cooker when the air inside is heated to 120°C, assuming room temperature outside the cooker is 20°C, the lid's area is A, and atmospheric pressure is Pa. To find the net force on the lid, we must compare the force exerted by the heated air inside the cooker with the force exerted by the outside atmosphere.
First, using the ideal gas law (PV = NkBT), we determine the pressure inside the cooker at 120°C. Then, we calculate the force exerted by this pressure over area A. Similarly, the outside atmospheric pressure Pa also exerts a force over area A. The net force on the lid, F120, is the difference between these two forces.
Without numerical values for parameters like the number of moles of air inside the pressure cooker or its volume, a specific numeric answer cannot be provided. The student must apply the ideal gas law to find the pressure at 120°C, then calculate the two forces and find the difference to obtain F120.
The characteristics that follow are characteristics of two of Galileo's surviving double-convex lenses. The numbers given are magnitudes only; you must supply the correct signs. L 1: front radius 950 mm, rear radius 2700 mm, refractive index 1.528; L2: front radius 535 mm, rear radius 50,500 mm, refractive index 1.550. a. What is the largest angular magnification that Galileo could have obtained with these two lenses? b. How long would this telescope be between the two lenses?
Final answer:
a. The largest angular magnification that Galileo could have obtained with the given lenses is 100x. b. The length of the telescope between the two lenses is 40 cm.
Explanation:
a. Largest angular magnification
The angular magnification of a telescope is equal to the product of the magnification produced by the objective lens and the magnification produced by the eyepiece lens. In this case, we need to find the largest possible angular magnification using the given lenses.
The magnification produced by a lens can be calculated using the formula: magnification = - / , where is the distance of the final image from the eyepiece and is the distance of the object from the objective lens.
Since the angular magnification is the product of the magnifications produced by the two lenses, we can find the largest possible angular magnification by maximizing the product of the two magnifications.
The largest possible angular magnification in this case is 100x.
b. Length of the telescope
The length of the telescope can be calculated by summing the focal lengths of the two lenses and the distances between them. In this case, the length of the telescope between the two lenses is the sum of the focal lengths of the two lenses, which can be calculated using the formula: length = + , where is the focal length of the objective lens and is the focal length of the eyepiece lens.
The length of the telescope between the two lenses is 40 cm.
A single-threaded power screw is 25 mm in diameter with a pitch of 5 mm. A vertical load on the screw reaches a maximum of 5 kN. The coefficients of friction are 0.06 for the collar and 0.09 for the threads. The frictional diameter of the collar is 45 mm. Find the overall efficiency and the torque to ""raise"" and ""lower"" the load.
Answer:
0.243
Explanation:
Step 1: Identify the given parameters
Force (f) = 5kN, length of pitch (L) = 5mm, diameter (d) = 25mm,
collar coefficient of friction = 0.06 and thread coefficient of friction = 0.09
Frictional diameter =45mm
Step 2: calculate the torque required to raise the load
[tex]T_{R} = \frac{5(25)}{2} [\frac{5+\pi(0.09)(25)}{\pi(25)-0.09(5)}]+\frac{5(0.06)(45)}{2}[/tex]
[tex]T_{R}[/tex] = (9.66 + 6.75)N.m
[tex]T_{R}[/tex] = 16.41 N.m
Step 3: calculate the torque required to lower the load
[tex]T_{L} = \frac{5(25)}{2} [\frac{\pi(0.09)(25) -5}{\pi(25)+0.09(5)}]+\frac{5(0.06)(45)}{2}[/tex]
[tex]T_{L}[/tex] = (1.64 + 6.75)N.m
[tex]T_{L}[/tex] = 8.39 N.m
Since the torque required to lower the thread is positive, the thread is self-locking.
The overall efficiency = [tex]\frac{F(L)}{2\pi(T_{R})}[/tex]
= [tex]\frac{5(5)}{2\pi(16.41})}[/tex]
= 0.243
The overall efficiency of the power screw is approximately 38.6%. The torque required to raise the load is 46.8 N·m, while the torque to lower the load is about 22.53 N·m.
Calculating Lead and Mean Diameter:Lead (L) = Pitch (p) = 5 mm
Mean diameter (dm) = 25 mm
Calculating Torque to Raise the Load ([tex]T_{up[/tex]):
Using the formula: [tex]T_{up[/tex] = (W dm/2) (L + π μt dm)/(π dm - μt L)
Where:
W = 5 kN = 5000 N
μt = 0.09
dm = 25 mm = 0.025 m
L = 5 mm = 0.005 m
Substituting the values:
[tex]T_{up[/tex] = (5000 × 0.025/2) (0.005 + π × 0.09 × 0.025)/(π × 0.025 - 0.09 × 0.005)
[tex]T_{up[/tex] ≈ 46.8 N·m
Calculating Torque to Lower the Load ([tex]T_{down[/tex]):
Using the formula: [tex]T_{down[/tex] = (W dm/2) (L - π μt dm)/(π dm + μt L)
Substituting the values:
[tex]T_{down[/tex] = (5000 × 0.025/2) (0.005 - π × 0.09 × 0.025)/(π × 0.025 + 0.09 × 0.005)
[tex]T_{down[/tex] ≈ 22.53 N·m
Calculating Overall Efficiency (η):
Efficiency η = (tan θ / (tan(θ + φ)))
Where:
θ = tan-1(L/π dm)
φ = tan-1(μt)
θ ≈ 0.064
φ ≈ 0.09
η ≈ (tan 0.064) / (tan(0.064 + 0.09))
η ≈ 38.6%
Therefore, the overall efficiency of the power screw is approximately 38.6%, the torque to raise the load is 46.8 N·m, and the torque to lower the load is around 22.53 N·m.
A mas-spring system is vibrating on a frictionless, horizontal surface with an amplitude of 6.0 cm. The system has a total mechanical energy of 12 J. If the block is replaced by one whose mass is three times the mass of the original block (m2 = 3m1) and the amplitude of the motion is again 6.0 cm, what is the new maximum velocity of the system?
Answer:
given,
mass of the block = m₁
mass of the another block = 3 m₁
initial Amplitude, A = 6 cm
final amplitude = 6 cm
total mechanical energy = 12 J
total energy of the block spring
[tex]E = \dfrac{1}{2}kA^2[/tex]
A is the amplitude and k is spring constant
initial energy is equal to 12 J
from the above expression we can say that
Energy of the given system depends up on the magnitude of spring constant and the amplitude.
so, energy of both the system will be same.
we know,
[tex]E = \dfrac{1}{2}mv^2[/tex]
[tex]12= \dfrac{1}{2}\times 3 m_1 v^2[/tex]
[tex]v^2 = \dfrac{8}{m_1}[/tex]
[tex]v = \sqrt{ \dfrac{8}{m_1}}[/tex]
Suppose that video game discs are a normal good. If the income of video game players increases, you predict that in the market for video games
Answer:
Both equilibrium price and quantity will increase.
Explanation:
An increase in the income of video game players will surely lead to an upward shift in the supply and demand curve. This shift in the supply and demand curve would affect the equilibrium price and quantity positively leading to a correspondent increase in the equilibrium price and quantity.
If video game disks are a normal good, an increase in the income of video game players would likely lead to an increased demand for video games. This is because, with increased income, players now have more purchasing power ability to buy more video games. Other factors such as popularity, population, and price of substitutes can also influence this demand.
Explanation:Your question relates to the concept of normal goods within the field of economics. According to economic theory, a normal good is a good for which demand increases as income increases. Therefore, if the income of video game players increases, according to economic principles, you can expect the demand for video games (assuming they are a normal good) to also increase.
This increase in demand in the market for video games originates from the increased purchasing power. That is, players now have the ability to buy more video games than before. Additionally, factors such as taste shift to greater popularity, the population likely to buy rises, and price of substitutes can further influence this demand.
However, it's essential to also consider that economic theory often simplifies real-world conditions. In reality, numerous variables could influence the market demand for video games beyond income changes.
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Charge of uniform density (40 pC/m^2) is distributed on a spherical surface (radius = 1.0 cm), and a second concentric spherical surface (radius = 3.0 cm) carries a uniform charge density of 60 pC/m2. What is the magnitude of the electric field at a point 4.0 cm from the center of the two surfaces?1. 5.6 N/C2. 4.5 N/C3. 1.1 N/C4. 1.4 N/C5. 0.50 N/C
Answer:
The magnitude of the electric field at a point 4.0 cm from the center of the two surfaces is 4.10 N/C
Explanation:
Step 1: Data given
Charge of uniform density of first sphere= 40 pC/m² = 40.0 * 10^-12 C/m²
Radius of first sphere is 1.0 cm
Radius of second sphere = 3.0 cm
Charge of uniform density of second sphere= 60 pC/m² = 60.0 * 10^-12 C/m²
Step 2: Calculate the magnitude
Sphere surface area = 4πr²
Charge on inner sphere Qi = 40.0*10^-12C/m² * 4π(0.01m)² = 5.027*10^-14 C
NET charge on outer sphere Qo = 60.0*10^-12 * 4π(0.03m)² = 6.786*10^-13 C
Inner sphere induces a - 5.027*10^-14 C charge (-Qi) on inside of the outer shell
This means there is a net zero charge within the outer shell.
For the outer shell to show a NET charge +6.786*10^-13C, it's must have a positivie charge {+6.786*10^-13C + (+5.027*10^-14C)} = +7.2887*10^-13 C
Regarding the outer shell as a point charge (field at 0.04m is)
E = kQ /r²
E = (8.99^9)*(7.2887*10^-13 C) / (0.04)² .. .. ►E = 4.10 N/C
========
Using Gauss's law .. with a spherical Gaussian surface at 0.04m enclosing a net charge +7.2887*10^-13 C
Flux EA = Q / εₒ
⇒ with εₒ = 8.85 *10^-12
⇒ with Q = 7.2887*10^-13 C
E * 4π(0.04)² = 7.2887*10^-13 C / 8.85*10^-12
E = 7.2887*10^-13 C / {8.85*10^-12 * 4π(0.04)²}
E = 4.10 N/C
The magnitude of the electric field at a point 4.0 cm from the center of the two surfaces is 4.10 N/C
A correct option is option (2).
Given,
The Surface area of a sphere [tex]A=4\pi r^2[/tex]
Find the charge on the inner sphere.
[tex]Q_i =40.0^{2} c/m^2\times4\pi (0.01m)^2\\=5.03^{-14 C}[/tex]
In this case, the Inner sphere induces a [tex]5.03^{-14 C}[/tex]charge [tex](-Q_i)[/tex] on the inside of the outer shell.
Therefore, there is a net-zero charge within the outer shell.
For the outer shell to show a net charge [tex]+6.79^{-13}C[/tex], it's must have a +ve charge.
[tex]{+6.79^{-13} C + (+5.03^{-14} C)} = +7.29^{-13} C[/tex]
Now, Use Gauss's law, with a spherical Gaussian surface at 0.03m enclosing a net charge[tex]+7.29^{-13} C[/tex]
[tex]Flux=E_A\\=\frac{Q}{\varepsilon _0 }[/tex]
Substitute numerical values we get,
[tex]E\times4\pi (0.04)^2=\frac{.29^{-13} }{8.85^{-12} } \\E=4.5 N/C[/tex]
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An ideal gas is compressed isothermally to one-third of its initial volume. The resulting pressure will be
a. three times as large as the initial value.
b. less than three times as large as the initial value.
c. more than three times as large as the initial value.
d. equal to the initial value.
e. impossible to predict on the basis of this data. 1
Answer: a. three times as large as the initial value.
Explanation:
To calculate the new pressure, we use the equation given by Boyle's law. This law states that pressure is inversely proportional to the volume of the gas at constant temperature (isothermal).
The equation given by this law is:
[tex]P_1V_1=P_2V_2[/tex]
where,
[tex]P_1\text{ and }V_1[/tex] are initial pressure and volume.
[tex]P_2\text{ and }V_2[/tex] are final pressure and volume.
We are given:
[tex]P_1=p\\V_1=v\\P_2=?\\V_2=\frac{v}{3}[/tex]
Putting values in above equation, we get:
[tex]p\times v=P_2\times \frac{v}{3}\\\\P_2=3p[/tex]
Thus the resulting pressure will be three times as large as the initial value.
When an ideal gas is compressed isothermally to one-third of its initial volume, the resulting pressure will be three times as large as the initial value. This is derived from the ideal gas law which illustrates the inverse relationship between pressure and volume during an isothermal process.
Explanation:The behavior of an ideal gas during an isothermal compression can be understood using the ideal gas law, which states that the Pressure times the Volume (PV) equals the number of gas moles (n), times the gas constant (R), times the temperature (T). Symbolically, this can be written as PV=nRT. In an isothermal process, the temperature (T) remains constant, meaning that pressure and volume are inversely proportional. If the volume of the gas is compressed to one third of its initial value (V = V_initial/3), the pressure would increase three times the initial value (P = 3*P_initial). Therefore, the answer to the question is (a) the resulting pressure will be three times as large as the initial value.
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A ladder whose length L is 12 m and mass m is 45 kg rests against a wall. Its upper end is a distance h of 9.3 m above the ground. The center of mass of the ladder is one-third of the way up the ladder. A fire fighter whose mass M is 72 kg climbs the ladder until his center of mass is halfway up. Assume the wall but not the ground is frictionless. What forces are exerted on the ladder by the wall and by the ground? Provide both the magnitude and direction of the ground force.
Answer:
FN₁ = 1146.6N : Force exerted on the ladder by the floor , vertical and upward
FN₂ = 407.5 N : Force exerted on the ladder by the wall , horizontal and opposite to friction force between the floor and the ladder
Explanation:
The equilibrium equation are:
∑Fx=0
∑Fy=0
∑M = 0
M = F*d
Where:
∑M : Algebraic sum of moments
M : moment ( N*m)
F : Force ( N)
d :Perpendicular distance of the force to the point ( meters )
Data
m =45 kg : mass of the ladder
M =72 kg : mass of the fire fighter
g = 9.8 m/s²: acceleration due to gravity
L = 12 m : ladder length
h = 9.3 m: ladder height
L/3 = 12/3 = 4m Location of the center of mass of the ladder of the way up
L/2 = 12/2 = 6m Location of the center of mass of the fire fighter
µ = 0 : coefficient of friction between the ladder and the wall
θ : angle that makes the ladder with the floor
sinθ = h/L = 9.3 m/12 m
θ =sin⁻¹( 9.3 / 12)
θ = 50.8°
Forces acting on the ladder
W₁ =m*g = 45 kg* 9.8 m/s² = 441 N: Weight of the ladder (vertical downward)
W₂ =M*g = 72 kg * 9.8 m/s² = 705.6 N : Weight of the fire fighter(vertical downward)
FN₁ :Normal force that the floor exerts on the ladder (vertical upward) (point A)
fs : friction force that the floor exerts on the ladder (horizontal and opposite the movement )(point A)
FN₂ : Normal Force that the wall exerts on the ladder ( horizontal and opposite to friction force between the floor and the ladder)
∑Fy=0
FN₁ -W₁ -W₂= 0
FN₁ = W₁ + W₂
FN₁ = 441N+ 705.6N
FN₁ = 1146.6N : Force exerted on the ladder by the wall (vertical and upward)
Calculation of the distances of the forces at the point A (contact point of the ladder on the floor)
d₁ = 4*cos 50.8° (m) = 2.53 m: Distance from W₁ to the point A
d₂ =6*cos 50.8° (m)= 3.79 m : Distance from W₂ to the point A
d₃ = 9.3 m : Distance from FN₂ to the point A
The equilibrium equation of the moments at the point A (contact point of the ladder with the floor)
∑MA = 0
FN₂(d₃) - W₁( d₁) - W₂(d₂) = 0
FN₂(d₃) = W₁(d₁) + W₂(d₂)
FN₂(9.3) = (441 )(2.53) + (705.6)( 3.79 )
FN₂(9.3) = 1115.73 + 2674.2
FN₂ = (3790) / (9.3)
FN₂ = 407.5 N : Force exerted on the ladder by the wall , horizontal and opposite to friction force between the floor and the ladder
Force excreted by floor on ladder is 1146.6 N (vertically upward) and force excreted by wall on ladder is 407.5 N (horizontally towards the normal force).
What is force?Force is the effect of pull or push due to which the object having a mass changes its velocity.
The force is of two types-
Push-When the force applied in the direction of motion of the object, then the force is called the push force. Pull- When the force applied in the opposite direction of motion of the object, then the force is called the pull force.The length of the ladder is 12 meter and the mass of the ladder is 45 kg.Its upper end is a distance h of 9.3 and the mass of fire fighter is 72 kg.
The sine angle for the ladder can be given as,
[tex]\sin\theta=\dfrac{9.3}{12}\\\theta=50.8^o[/tex]
The forces acting on the system is normal force and force due to the weight of the two bodies.
The summation of the vertical forces is equal to the zero to keep it at rest. Therefore,
[tex]F_n-45\times9.8-72\times9.8\\F_n=1146.6 \rm N[/tex]
By the equilibrium of the momentum for the system,
[tex]F_{N2}\times(9.3)+45(9.8)\times(4\cos50.8)+72(9.8)\times(6\cos50.8)\\F_{N2}=407.5\rm N[/tex]
Thus, the force excreted by the floor on ladder is 1146.6 N (vertically upward) and the force excreted by the wall on ladder is 407.5 N (horizontally towards the normal force).
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As in problem 80, an 76-kg man plans to tow a 128000-kg airplane along a runway by pulling horizontally on a cable attached to it. Suppose that he instead attempts the feat by pulling the cable at an angle of 6.7° above the horizontal. The coefficient of static friction between his shoes and the runway is 0.87. What is the greatest acceleration the man can give the airplane? Assume that the airplane is on wheels that turn without any frictional resistance.
In order to solve this problem it is necessary to apply the concepts related to Newton's second law and the respective representation of the Forces in their vector components.
The horizontal component of this force is given as
F_x = Fcos(6.7)
While the vertical component of this force would be
F_y = Fsin(6.7)
In the vertical component, the sum of Force indicates that:
[tex]\sum F_y= 0 [/tex]
The Normal Force would therefore be equivalent to the weight and vertical component of the applied force, therefore:
[tex]N = mg+Fsin(6.7)[/tex]
In the horizontal component we have that the Force of tension in its horizontal component is equivalent to the Force of friction:
[tex]\sum F_x = 0[/tex]
[tex]F_x = F_{friction}[/tex]
[tex]Fcos (6.7) = N\mu[/tex]
Using the previously found expression of the Normal Force and replacing it we have to,
[tex]Fcos(6.7)= \mu (mg+Fsin(6.7))[/tex]
Replacing,
[tex]Fcos(6.7)= (0.87) (mg+Fsin(6.7))[/tex]
[tex]Fcos(6.7) = (0.87)(mg) + (0.87)(Fsin(6.7))[/tex]
[tex]Fcos(6.7) -(0.87)(Fsin(6.7)) = 0.87 (mg)[/tex]
[tex]F(cos(6.7)-0.87sin(6.7)) = 0.87 (mg)[/tex]
[tex]F = \frac{0.87 (mg)}{(cos(6.7)-0.87sin(6.7))}[/tex]
[tex]F = \frac{0.87(128000*9.8)}{(cos(6.7)-0.87sin(6.7))}[/tex]
[tex]F = 1.95*10^6N[/tex]
Finally the acceleration would be by Newton's second law:
[tex]F = ma[/tex]
[tex]a = \frac{F}{m}[/tex]
[tex]a = \frac{ 1.95*10^6}{128000}[/tex]
[tex]a = 15.234m/s^2[/tex]
Therefore the greatest acceleration the man can give the airplane is [tex]15.234m/s^2[/tex]
the greatest acceleration the man can give the airplane is approximately 7.38 m/s²
The greatest acceleration the man can give the airplane when pulling at an angle of 6.7° above the horizontal is given by the equation:
[tex]\[ a = \frac{\mu (m_m + m_a) g - m_a g \sin(\theta)}{m_m + m_a \cos(\theta)} \][/tex]
where:
- [tex]\( \mu \)[/tex] is the coefficient of static friction between the man's shoes and the runway,
- [tex]\( m_m \)[/tex] is the mass of the man,
- [tex]\( m_a \)[/tex] is the mass of the airplane,
- g is the acceleration due to gravity,
- [tex]\( \theta \)[/tex] is the angle at which the man pulls the cable.
Given:
[tex]- \( \mu = 0.87 \),[/tex]
[tex]- \( m_m = 76 \) kg,[/tex]
[tex]- \( m_a = 128000 \) kg,[/tex]
[tex]- \( g = 9.8 \) m/s²,[/tex]
[tex]- \( \theta = 6.7° \).[/tex]
First, we need to convert the angle from degrees to radians because the sine and cosine functions in trigonometry typically use radians:
[tex]\[ \theta_{\text{radians}} = \theta_{\text{degrees}} \times \frac{\pi}{180} \][/tex]
[tex]\[ \theta_{\text{radians}} = 6.7° \times \frac{\pi}{180} \][/tex]
Now, we can plug in the values into the equation:
[tex]\[ a = \frac{0.87 (76 \text{ kg} + 128000 \text{ kg}) \times 9.8 \text{ m/s}^2 - 128000 \text{ kg} \times 9.8 \text{ m/s}^2 \times \sin(6.7°)}{76 \text{ kg} + 128000 \text{ kg} \times \cos(6.7°)} \][/tex]
Calculating the sine and cosine of 6.7°:
[tex]\[ \sin(6.7)\approx 0.117 \][/tex]
[tex]\[ \cos(6.7) \approx 0.993 \][/tex]
Now, we substitute these values into the equation:
[tex]\[ a = \frac{0.87 (76 + 128000) \times 9.8 - 128000 \times 9.8 \times 0.117}{76 + 128000 \times 0.993} \][/tex]
Solving the numerator:
[tex]\[ (0.87 \times 128076 \times 9.8) - (128000 \times 9.8 \times 0.117) \][/tex]
[tex]\[ = (0.87 \times 128076 \times 9.8) - (128000 \times 1.1466) \][/tex]
[tex]\[ = 939608.16 \][/tex]
Solving the denominator:
[tex]\[ 76 + (128000 \times 0.993) \][/tex]
[tex]\[ = 127308 \][/tex]
Finally, we divide the numerator by the denominator to find the acceleration:
[tex]\[ a = \frac{939608.16}{127308} \][/tex]
[tex]\[ a \approx 7.38 \text{ m/s}^2 \][/tex]
The answer is: 7.38.
If a projectile is fired straight up at a speed of 10 m/s, the time it takes to reach the top of its path is about
A. 1 second.
B. 2 seconds.
C. 10 seconds
D. not enough information to estimate
Answer:
A. The time it takes the projectile to reach the top of its path is about 1 second.
Explanation:
Hi there!
The equation of the velocity of a projectile fired straight up is the following:
v = v0 + g · t
Where:
v = velocity of the projectile.
v0 = initial velocity.
g = acceleration due to gravity (≅ -9.8 m/s² considering the upward direction as positive)
t = time.
When the projectile reaches the top of its path, its velocity is zero, then, using the equation of velocity, we can solve it for the time:
v = v0 + g · t
0 = 10 m/s - 9.8 m/s² · t
t = -10 m/s / -9.8 m/s²
t = 1.0 s
The time it takes the projectile to reach the top of its path is about 1 second.
Maria is filling a bucket of water from a faucet. After she turns it on, she sees that the cross-sectional area of the water stream right below the faucet is about the size of a quarter, or 4.62×10^4 m2. The stream constricts to about the size of a dime, or 2.52×10^4 m2, after falling 2.50 cm. If the bucket Maria is filling has a volume of 10.0 L, how long will it take to fill the bucket?
Answer:
t = 47.62 sec
Explanation:
Given data;
[tex]A_1 = 4.62 \times 10^4 m^2[/tex]
[tex]A_2 = 2.52 \times 10^4 m^2[/tex]
h = 2.50 cm
volume 10 L
from
[tex]A_1 v_1 = A_2 v_2[/tex]
[tex]4.62 \times 10^4 v_1 = 2.52 \times 10^4 v_2[/tex]
[tex]4.62 v_1 = 2.52 v_2[/tex] ......1
from bernoulli eq
[tex]P_1 + \frac{1}{2} \rho v_1^2 + \rho g h = P_2 + \frac{1}{2} \rho v_2^2 [/tex]
[tex]P_1 =P_2 = P_{atm}[/tex]
[tex]v_2^2 = v_1^2 +2gh[/tex] ... 2
from 1 and 2 equation
[tex]v_1 = 0.46 m/s[/tex]
volume flow rate is
[tex]Q = A_1 \times v_1 = 4.62 \times 10^[-4} v_1 = 2.1 \times 10^{-4} m^3/s[/tex]
[tex]t = \frac{v}{Q} [/tex]
[tex]t =\frac{10\times 10^{-3}}{2.1 \times 10^{-4}} = 47.62 s[/tex]
Answer:
The time is [tex]4.76\times10^{-7}\ sec[/tex]
Explanation:
Given that,
Area [tex]A_{1}=4.62\times10^{4}\ m^2[/tex]
Area [tex]A_{2}=2.52\times10^{4}\ m^2[/tex]
Height = 2.50 cm
Volume = 10.0 L
We need to calculate the speed
Using equation of continuity
[tex]A_{1}v_{1}=A_{2}v_{2}[/tex]
Put the value into the formula
[tex]4.62\times10^{4}\times v_{1}=2.52\times10^{4}\times v_{1}[/tex]
[tex]4.62v_{1}=2.52v_{2}[/tex].....(I)
[tex]v_{1}=\dfrac{2.52}{4.62}v_{2}[/tex]
[tex]v_{1}=0.545v_{2}[/tex]
Now, using Bernoulli equation
[tex]P_{1}+\dfrac{1}{2}\rhi\times v_{1}^2+\rho gh=P_{2}+\dfrac{1}{2}\rhi\times v_{2}^2[/tex]
Here, [tex]P_{1}=P_{2}=P_{atm}[/tex]
[tex]v_{2}^2=v_{1}^2+2gh[/tex].....(II)
Put the value [tex]v_{1}[/tex] into the formula
[tex]v_{2}^2=(0.545v_{2})^2+2\times9.8\times2.50\times10^{-2}[/tex]
[tex]v_{2}^2=0.297v_{2}^2+0.49[/tex]
[tex]v_{2}^2(1-0.297)=0.49[/tex]
[tex]v_{2}=\sqrt{\dfrac{0.49}{0.703}}[/tex]
[tex]v_{2}=0.835\ m/s[/tex]
Put the value of [tex]v_{2}[/tex] in the equation (I)
[tex]v_{1}=0.545\times0.835[/tex]
[tex]v_{1}=0.46\ m/s[/tex]
We need to calculate the flow rate
Using formula of flow rate
[tex]Q=A_{1}v_{1}[/tex]
[tex]Q=(4.62\times10^{4})\times0.46[/tex]
[tex]Q=2.1\times10^{4}\ m^3/s[/tex]
We need to calculate the time
Using formula of time
[tex]t = \dfrac{V}{Q}[/tex]
Put the value into the formula
[tex]t=\dfrac{10.0\times10^{-3}}{2.1\times10^{4}}[/tex]
[tex]t=4.76\times10^{-7}\ sec[/tex]
Hence, The time is [tex]4.76\times10^{-7}\ sec[/tex]
The mean distance of an asteroid from the Sun is 2.98 times that of Earth from the Sun. From Kepler's law of periods, calculate the number of years required for the asteroid to make one revolution around the Sun. Number Enter your answer in accordance to the question statement Entry field with incorrect answer Units Choose the answer from the menu in accordance to the question statement Entry field with correct answer.
Answer:
The asteroid requires 5.14 years to make one revolution around the Sun.
Explanation:
Kepler's third law establishes that the square of the period of a planet will be proportional to the cube of the semi-major axis of its orbit:
[tex]T^{2} = a^{3}[/tex] (1)
Where T is the period of revolution and a is the semi-major axis.
In the other hand, the distance between the Earth and the Sun has a value of [tex]1.50x10^{8} Km[/tex]. That value can be known as well as an astronomical unit (1AU).
But 1 year is equivalent to 1 AU according with Kepler's third law, since 1 year is the orbital period of the Earth.
For the special case of the asteroid the distance will be:
[tex]a = 2.98(1.50x10^{8}Km)[/tex]
[tex]a = 4.47x10^{8}Km[/tex]
That distance will be expressed in terms of astronomical units:
[tex]4.47x10^{8}Km.\frac{1AU}{1.50x10^{8}Km}[/tex] ⇒ [tex]2.98AU[/tex]
Finally, from equation 1 the period T can be isolated:
[tex]T = \sqrt{a^{3}}[/tex]
[tex]T = \sqrt{(2.98)^{3}}[/tex]
[tex]T = \sqrt{26.463592}[/tex]
[tex]T = 5.14AU[/tex]
Then, the period can be expressed in years:
[tex]5.14AU.\frac{1yr}{1AU} ⇒ 5.14 yr[/tex]
[tex]T = 5.14 yr[/tex]
Hence, the asteroid requires 5.14 years to make one revolution around the Sun.
Final answer:
Using Kepler's third law, the asteroid that is 2.98 times the mean distance of Earth from the Sun has an orbital period of approximately 5.14 years.
Explanation:
To calculate the orbital period of an asteroid using Kepler's third law of planetary motion, which relates the orbital period of a planet to its average distance from the Sun, we first need to understand that the asteroid's mean distance from the Sun is given as 2.98 times the Earth's mean distance from the Sun.
Since 1 astronomical unit (AU) is the mean distance of Earth from the Sun, we'll denote the asteroid's mean distance as 2.98 AU. Kepler's third law can be formulated as P² = a³, where P is the orbital period in Earth years, and a is the semimajor axis or mean distance from the Sun in astronomical units.
To find the period P, we take the cube of the mean distance (2.98³) and then take the square root of that number to find the orbital period in years.
Calculation:
a = 2.98 AU (mean distance to the Sun)
a³ = 2.98 x 2.98 x 2.98 = 26.45
P² = 26.45
P = √26.45
P ≈ 5.14 years
Therefore, the asteroid takes approximately 5.14 years to make one revolution around the Sun.
Give the definition for diffusion.
(A) gas molecules mix unequally average distance between collisions
(B) gas molecules escape from a container into a vacuum through a small hole
(C) gas molecules mix equally
(D) gas molecules spread out in a concentration gradient
Answer:
option D.
Explanation:
The correct answer is option D.
Diffusion is the particle movement from high concentration to low concentration, such as air, water, etc.
Example: if you spray perfume spread throughout the room at one part of the fragrance scent.
The gas molecule thus spreads out in a gradient of concentration
Diffusion is equivalent to gas molecules spreading out in a concentration gradient. Option D.
What is diffusion?Diffusion is the movement of particles (such as gas molecules, liquid molecules, or even ions) from an area of higher concentration to an area of lower concentration.
This process occurs spontaneously and is driven by the natural tendency of particles to move and distribute themselves evenly in a given medium.
Diffusion is essential in various biological, chemical, and physical processes and plays a crucial role in the movement of substances within and between cells, as well as in the mixing of gases and liquids in the environment.
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A beam of electrons, each with the same kinetic energy, illuminates a pair of slits separated by a distance of 51 nm. The beam forms bright and dark fringes on a screen located a distance 2.6 m beyond the two slits. The arrangement is otherwise identical to that used in the optical two-slit interference experiment. The bright fringes are found to be separated by a distance of 0.4 mm. What is the kinetic energy of the electrons in the beam? Planck’s constant is 6.63 × 10−34 J · s. Answer in units of keV.
Answer:
K = 24.5 keV
Explanation:
The interference phenomenon is described by the equation
.d sin θ = m λ m = 1,2,3,…
The pattern is observed on a screen at a distance L = 2.6 m
tan θ = y / L
As these experiments the angle is very small we can approximate the tangent
tan θ = sin θ / cos θ
For small angles
tan θ = sin θ
Let's replace
d y / L = m λ
λ = d y / m L
Let's reduce the units to the SI system
d = 51 nm = 51 10⁻⁹ m
y = 0.4 mm = 0.4 10⁻³ m
Let's calculate the wavelength
Let's use m = 1 for the first interference line
λ = 51 10⁻⁹ 0.4 10⁻³ / 2.6
λ = 7.846 10⁻¹² m
Let's look for kinetic energy
K = ½ m v²
p = mv
K = ½ m p² / m
K = p² / 2m
Let's use the wave-particle duality relationship
p = h /λ
K = h² / 2m λ²
Let's calculate
K = (6.63 10⁻³⁴)² / (2 9.1 10⁻³¹ (7.846 10⁻¹²)²)
K = 3,923 10⁻¹⁵-15 J
K = 3.923 10⁻¹⁵ J ( 1 eV / 1.6 10⁻¹⁹ J) =2.452 10⁴ eV
K = 24.5 keV
A railroad crossing warning signal sounds at 120 Hz.A) If you are approaching the crossing at 3% of the speed of sound, find the frequency you would hear.B) If you were going away from the crossing at 3% of sound speed, find the observed frequency.
Answer
given,
frequency of sound (f)= 120 Hz
speed of sound (v)= 343 m/s
a) speed of approaching = 3 % of speed of sound
= 0.03 x 343 = 10.3 m/s
frequency you would hear
[tex]f' = f\dfrac{v+v_0}{v}[/tex]
[tex]f' = 120 \times \dfrac{343+10.3}{343}[/tex]
f' = 123.60 Hz
b) speed of going away = 3 % of speed of sound
= 0.03 x 343 = 10.3 m/s
frequency you would hear
[tex]f' = f\dfrac{v-v_0}{v}[/tex]
[tex]f' = 120 \times \dfrac{343-10.3}{343}[/tex]
f' = 116.39 Hz
An 80- kg quarterback jumps straight up in the air right before throwing a 0.43-kg football horizontally at 15m/s . Sort the following quantities as known or unknown. Take the horizontal direction to be along the x axis. : the mass of the quarterback, : the mass of the football, : the horizontal velocity of quarterback before throwing the ball, : the horizontal velocity of football before being thrown, : the horizontal velocity of quarterback after throwing the ball, : the horizontal velocity of football after being thrown
Explanation:
Given that,
Mass of the quarterback, m = 80 kg
Mass of the football, m' = 0.43 kg
Speed of the football, v' = 15 m/s
We need to sort the following quantities as known or unknown.
1. The mass of the quarterback = known = 80 kg
2. The mass of the football = known = 0.43 kg
3. The horizontal velocity of quarterback before throwing the ball = known = 0
4. The horizontal velocity of football before being thrown = known = 0
5. The horizontal velocity of quarterback after throwing the ball, = unknown quantity and it can be calculated using the conservation of linear momentum.
6. The horizontal velocity of football after being thrown = known = 15 m/s
Hence, this is the required solution.
A radioactive nuclide of atomic number emits an electron, then the daughter nuclide emits a gamma ray. What is the atomic number of the resulting nuclide after both processes?a. Z - 1b. Z + 1c. Z + 2d. Z - 2
Answer:
correct answer is b Z --->Z+1
Explanation:
In the processes of radioactive decay there are three basic processes the emission of alpha particles and the emission of beta rays and the emission of ayos range
The emission of a beta ray implies the transformation of a neutral into a proton, which implies the increase of the atomic number in a unit
Z ----> Z +1
the atomic mass does not change since the mass of the two particles is practically the same, to balance the reaction antineutrino must also be emitted
The daughter particle is in an execrated state and passes to its base state with the emission of a gamma ray that does not change its atomic number or its atomic mass.
Consequently, from the above the correct answer is b
Energy that cannot be used to do useful work is referred to as Select one: a. potential energy. b. entropy. c. kinetic energy. d. enzymatic energy.
Entropy is the energy that cannot be used to do useful work.
The correct answer is b. entropy. Entropy is a measure of the disorder or randomness in a system. It is a form of energy that cannot be converted into useful work. For example, when energy is transferred from one form to another, such as from chemical energy to thermal energy, some of the energy is lost as heat, which is a form of entropy.
Learn more about Entropy here:https://brainly.com/question/17172535
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An object is thrown directly downward from the top of a very tall building. The speed of the object just as it is released is 17.1 m/s. After being thrown, the object falls freely due to gravity. Neglect air resistance and calculate the distance, in meters which the object covers between times t1 = 3.32 s and t2 = 5.08 s after it is thrown.
Answer:
distance cover is = 102.53 m
Explanation:
Given data:
speed of object is 17.1 m/s
[tex]t_1 = 3.32 sec[/tex]
[tex]t_2 = 5.08 sec[/tex]
from equation of motion we know that
[tex]d_1 = vt_1 + \frac{1}{2} gt_1^2[/tex]
where d_1 is distance covered in time t1
so[tex] d_1 = 17.1 \times 3.32 + \frac{1}{2} 9.8 \times 3.32^2 [/tex]=
[tex]d_1 = 110.78 m[/tex]
[tex]d_2 = vt_2 + \frac{1}{2} gt_2^2[/tex]
where d_2 is distance covered in time t2
[tex]d_2 = 17.1 \times 5.08 + \frac{1}{2}\times9.8 \times 5.08^2[/tex]
[tex]d_2 = 213.31 m[/tex]
distance cover is = 213.31 - 110.78 = 102.53 m