Answer:
a)time t = 47.4s
b)speed v = 38.7cm/s
Explanation:
Given:
Total volume of blood = 4.93L × 1000cm^3/L = 4930cm^3
Volumetric rate of flow = 104.1cm^3/s
a) Time taken for all the blood to pass through the aorta is:
Time t = total volume/ volumetric rate
t = 4930/104.1
t = 47.4s
b) Given that the diameter of the aorta is 1.85cm.
V = Av
Where V = Volumetric rate
A = area of aorta
v = speed of blood
v = V/A ...1
Area of a circular aorta = πr^2 = (πd^2)/4
d = 1.85cm
A = (π×1.85^2)/4
A = 2.69cm^2
From equation 1.
v = V/A = 104.1/2.69
v = 38.7 cm/s
If a scuba diver fills his lungs to full capacity of 5.7 L when 8.0 m below the surface, to what volume would his lungs expand if he quickly rose to the surface? Assume he dives in the sea, thus the water is salt. Express your answer using two significant figures
To calculate the pressure in the body we will use the definition of the hydrostatic pressure for which the pressure of a body at a certain distance submerged in a liquid is defined. After calculating this relationship we will apply the equations of the relationship between the volume and the pressure to calculate the volume in state 2,
[tex]P = P_{atm} + \rho gh[/tex]
Here,
[tex]\rho[/tex]= Density of the Fluid (Water)
g = Acceleration due to gravity
h = Height
[tex]P = P_{atm} + 10^3*9.8*8[/tex]
[tex]P = 1.01*10^{5} +10^3*9.8*8[/tex]
[tex]P = 179400Pa[/tex]
Applying the equations of relationship between volume and pressure we have
[tex]P_1V_1 = P_2 V_2[/tex]
[tex]179400*5.7 = 101000*V_2[/tex]
[tex]V_2 = 10.12L[/tex]
Therefore the volume that would his lungs expand if he quickly rose to the surface is 10.12L
What region of the spectrum best corresponds to light with a wavelength equal to:_____ a. The diameter of a hydrogen atomb. The size of a virus.c. Your height?
We will make the comparison between each of the sizes against the known wavelengths.
In the case of the hydrogen atom, we know that this is equivalent to [tex]10^{-10}[/tex] m on average, which corresponds to the wavelength corresponding to X-rays.
In the case of the Virus we know that it is oscillating in a size of 30nm to 200 nm, so the size of the virus is equivalent to the range of the wavelength of an ultraviolet ray.
In the case of height, it fluctuates in a person around [tex]10 ^ 0[/tex] to [tex]10 ^ 1[/tex] m, which falls to the wavelength of a radio wave.
The region of the spectrum that best corresponds to light with a wavelength equal to the diameter of a hydrogen atom is the X-ray region. The visible light spectrum corresponds to objects around the same size as the wavelength. The size of a virus is much smaller than the wavelength of visible light
Explanation:The region of the spectrum that best corresponds to light with a wavelength equal to the diameter of a hydrogen atom is the X-ray region of the electromagnetic spectrum.
The wavelength of visible light corresponds to the size of objects that are around the same size as the wavelength. For example, if you want to use light to see a human, you would need to use a wavelength at or below 1 meter, since humans are about 1 meter in size.
The size of a virus is much smaller than the wavelength of visible light, so the wavelength is very small compared to the object's size.
An equilateral triangle with side lengths of 0.50 m has a 5.0 nC charge placed at each corner. What is the magnitude of the electric field at the midpoint of one of the three sides? (A) 240 N/C (B) 180 N/C (C) 720 N/C (D) 480 N/C (E) 120 N/C
To solve this problem we will first find the distance between each of the points 'x' and then use it as the variable of the distance in the function of the electric field. According to the graph the value of 'x' is,
[tex]x = \frac{\sqrt{3}}{2}a[/tex]
[tex]x = \frac{\sqrt{3}}{2}(0.5)[/tex]
[tex]x = 0.43301m[/tex]
The magnitude of the electric field is
E=\frac{kQ}{x^2}
Here,
k = Coulomb's Constant
Q = Charge
x = Distance
[tex]E=\frac{(9*10^9)(5*10^{-9})}{(0.43301)^2}[/tex]
[tex]E = 240.002N/C \approx 240N/C[/tex]
Therefore the correct answer is A.
A railroad freight car, mass 15 000 kg, is allowed to coast along a level track at a speed of 2.0 m/s. It collides and couples with a 50 000-kg loaded second car, initially at rest and with brakes released. What percentage of the initial kinetic energy of the 15 000-kg car is preserved in the two-coupled cars after collision
Answer:
23.0760769 %
Explanation:
[tex]m_1[/tex] = Mass of freight car = 15000 kg
[tex]m_2[/tex] = Mass of second car = 50000 kg
[tex]v_1[/tex] = Velocity of freight car = 2 m/s
[tex]v_2[/tex] = Velocity of second car = 0
v = Combined mass velocity
As the linear momentum of the system is conserved we have
[tex]m_1v_1+m_2v_2=(m_1+m_2)v\\\Rightarrow v=\dfrac{m_1v_1+m_2v_2}{m_1+m_2}\\\Rightarrow v=\dfrac{15000\times 2+50000\times 0}{15000+50000}\\\Rightarrow v=0.46153\ m/s[/tex]
The initial kinetic energy
[tex]K_i=\dfrac{1}{2}15000\times 2^2\\\Rightarrow K_i=30000\ J[/tex]
Final kinetic energy
[tex]K_f=\dfrac{1}{2}(15000+50000)\times 0.46153^2\\\Rightarrow K_f=6922.82307\ J[/tex]
The percentage is given by
[tex]\dfrac{6922.82307}{30000}\times 100=23.0760769\ \%[/tex]
The change in percentage of initial kinetic energy is 23.0760769 %
A particle has a charge of -4.25 nC.
Part A
Find the magnitude of the electric field due to this particle at a point 0.250 m directly above it.
Part B
Find the direction of the field
up, away from the particle
down, toward the particle
Part C
At what distance from this particle does its electric field have magnitude of 13.0 N/C?
Answer:
Explanation:
q = - 4.25 nC = - 4.5 x 10^-9 C
(A) d = 0.250 m
The formula for the electric field is given by
[tex]E = \frac{1}{4\pi \epsilon _{0}}\frac{q}{d^{2}}[/tex]
By substituting the values
[tex]E = \frac{9\times 10^{9}\times 4.5\times10^{-9}}{0.25\times 0.25}[/tex]
E = 648 N/C
(B) As the charge is negative in nature so the direction of electric field is towards the charge and downwards.
(a) The magnitude of the electric field is 612 N/C.
(b) The direction of the electric field will be up, away from the particle.
(c) The distance from the particle is 1.71 m.
Magnitude of the electric fieldThe magnitude of the electric field is calculated as follows;
E = (kq)/r²
where;
k is Coulomb's constant
q is the charge
r is distance
E = ( 9 x 10⁹ x 4.25 x 10⁻⁹)/(0.25 x 0.25)
E = 612 N/C
Direction of the fieldThe direction of the electric field is always opposite to the direction of the negative charge.
Thus, the direction of the electric field will be up, away from the particle.
Distance from the particleThe distance from the particle is determined using the following formula;
E = (kq)/r²
r² = kq/E
r² = (9 x 10⁹ x 4.25 x 10⁻⁹) / 13
r² = 2.94
r = √2.94
r = 1.71 m
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If the length of the air column in the test tube is 14.0 cm, what is the frequency of this standing wave?
Answer:
f = 614.28 Hz
Explanation:
Given that, the length of the air column in the test tube is 14.0 cm. It can be assumed that the speed of sound in air is 344 m/s. The test tube is a kind of tube which has a closed end. The frequency in of standing wave in a closed end tube is given by :
[tex]f=\dfrac{nv}{4l}[/tex]
[tex]f=\dfrac{1\times 344}{4\times 0.14}[/tex]
f = 614.28 Hz
So, the frequency of the this standing wave is 614.28 Hz. Hence, this is the required solution.
During a tennis match, a player serves the ball at a speed s, with the center of the ball leaving the racquet at angle θ below the horizontal at a height y0. The net is a distance d away and has a height h. When the ball is directly over the net, what is the distance between them?
Answer: Maximum distance
= {s²/g} * sine(2*theta)unit
Explanation: This is a projectile motion problem. The horizontal distance between the tennis player and where the tennis reaches over the net is given by the horizontal Range.
Range = {s² * sine2*theta}/g
(s)is the initial speed of projection
Theta is the angle of projection
g is acceleration due to gravity 10m/s².
In the Boyle’s law experiment, what was used to increase the pressure on the air (gas)?
Answer:
Volume.
Explanation:
Boyle’s law experiment :
This is also known as Mariotte's law or in the other words it is also known as Boyle–Mariotte law.
This law tell us about the variation of gas pressure when the volume of the gas changes at the constant temperature.According to this law the abslute pressure is inversely proportional to the volume .
We can say that
[tex]P\alpha \dfrac{1}{V}[/tex]
Where P=Pressure
V=Volume
PV = K
K=Constant
When volume decrease then the pressure of the gas will increase.
That is why the answer is "Volume".
The error in the measurement of the radius of a sphere is 2%. What will be the error in the calculation of its volume?
To solve this problem we will apply the geometric concepts of the Volume based on the consideration made of the radius measurement. The Volume must be written in differential terms of the radius and from the formula of the margin of error the respective response will be obtained.
The error in radius of sphere is not exceeding 2%
[tex]\frac{dr}{r} = \pm 0.02[/tex]
The objective is to find the percentage error in the volume.
The volume can be defined as
[tex]V = \frac{4}{3} \pi r^3[/tex]
Differentiate with respect the radius we have,
[tex]\frac{dV}{dr} = 4\pi r^2[/tex]
[tex]dV = 4\pi r^2 \times dr[/tex]
[tex]dV = 4\pi r^2 (\pm 0.02r)[/tex]
[tex]dV = \pm 4\times 0.02 \times \pi r^3[/tex]
The percentage change in the volume is as follows
[tex]\% change = \frac{dV}{V} \times 100[/tex]
[tex]\% change = \frac{\pm 4 \times 0.02 \times \pi r^3 \times 3}{4\pi r^3}\times 100[/tex]
[tex]\% change = \pm 6\%[/tex]
Therefore the percentage change in volume is [tex]\pm 6\%[/tex]
The work function of palladium is 5.22 eV.
(a) What is the minimum frequency of light required to observe the photoelectric effect on Pd?
(b) If light with a 200 nm wavelength is absorbed by the surface, what is the velocity of the emitted electrons?
a) The minimum frequency of the light must be [tex]1.26\cdot 10^{15} Hz[/tex]
b) The maximum velocity of the electrons is [tex]5.93\cdot 10^5 m/s[/tex]
Explanation:
a)
The photoelectric effect is a phenomenon that occurs when electromagnetic radiation hits the surface of a metal causing the release of electrons from the metal's surface.
The equation of the photoelectric effect is:
[tex]hf = \phi +K_{max}[/tex]
where :
[tex]hf[/tex] is the energy of the incoming photons, where
[tex]h[/tex] is the Planck's constant
[tex]f[/tex] is the frequency of the incoming photons
[tex]\phi[/tex] is the work function of the metal, the minimum energy that the photons must have in order to be able to free electrons from the metal
[tex]K_{max}[/tex] is the maximum kinetic energy of the emitted electrons
In order to free electrons, the minimum energy of the photons must be at least equal to the work function (so that the kinetic energy of the electrons is zero, [tex]K_{max}=0[/tex]. Therefore,
[tex]h f_0 = \phi[/tex]
In this case,
[tex]\phi = 5.22 eV \cdot (1.6\cdot 10^{-19})=8.35\cdot 10^{-19} J[/tex]
Therefore, the minimum frequency of the photons must be
[tex]f_0 = \frac{\phi}{h}=\frac{8.35\cdot 10^{-19}}{6.63\cdot 10^{-34}}=1.26\cdot 10^{15} Hz[/tex]
b)
In this case, the wavelength of the incoming light is
[tex]\lambda = 200 nm = 200 \cdot 10^{-9} m[/tex]
We can find the frequency by using the wave equation:
[tex]f=\frac{c}{\lambda}=\frac{3\cdot 10^8}{200\cdot 10^{-9}}=1.5\cdot 10^{15} Hz[/tex]
Now we can use the equation of the photoelectric effect to find the maximum kinetic energy of the electrons:
[tex]K_{max} = hf-\phi = (6.63\cdot 10^{-34})(1.5\cdot 10^{15})-8.35\cdot 10^{-19}=1.60\cdot 10^{-19} J[/tex]
And therefore, we can find their velocity by using the equation for the kinetic energy:
[tex]K_{max} = \frac{1}{2}mv^2[/tex]
where
[tex]m=9.11\cdot 10^{-31} kg[/tex] is the mass of the electrons
v is their speed
Solving for v,
[tex]v=\sqrt{\frac{2K_{max}}{m}}=\sqrt{\frac{2(1.6\cdot 10^{-19})}{9.11\cdot 10^{-31}}}=5.93\cdot 10^5 m/s[/tex]
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What is the ratio of the intensities and amplitudes of an earthquake P wave passing through the Earth and detected at two points 27 km and 13 km from the source?
(a) I 27 / 13 =__________
(b) A 27 / 13 =_____________-
For both cases we will use the proportional values of the distance referring to the amplitude and intensity. Theoretically we know that the intensity is inversely proportional to the square of the distance, while the amplitude is inversely proportional to the distance, therefore,
PART A ) Intensity is inversely proportional to the square of the distance
[tex]Intensity \propto \frac{1}{distance^2}[/tex]
Therefore the intensity of the two values would be
[tex]\frac{I_{27}}{I_{13}} = \frac{(13km)^2}{(27km)^2}[/tex]
[tex]\mathbf{\therefore \frac{I_{27}}{I_{13}} = 0.232 }[/tex]
PART B) Amplitude is inversely proportional to the distance
[tex]Amplitude \propto \frac{1}{distance}[/tex]
[tex]\frac{A_{27}}{A_{13}}= \frac{(13km)}{(27km)}[/tex]
[tex]\mathbf{\therefore\frac{A_{27}}{A_{13}}= 0.4815}[/tex]
The intensity ratio of an earthquake P wave passing through the Earth and detected at two points is equal to the square of the amplitude ratio.
Explanation:The ratio of the intensities of an earthquake P wave passing through the Earth and detected at two points is equal to the square of the ratio of their amplitudes.
Let's assume the amplitudes of the earthquake P wave at the two points are given by A27 and A13.
The intensity of a wave is given by the square of its amplitude. Therefore, the ratio of the intensities I27/I13 is equal to the square of the ratio of the amplitudes A27/A13.
So, the answer is:
(a) I27/I13 = (A27/A13)2
(b) A27/A13
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Calculate the percent weight reduction of the plane as it moves from the sea level to 11,719.342 m.
Answer: %(∆W) = 0.37%
Explanation:
According to Newton's law of gravitation which states that every particle attracts every other particle in the universe with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers
F = Gm1m2/r^2
Where
F = force between the masses
G universal gravitational constant
m1 and m2 = mass of the two particles
r = distance between the centre of the two mass
Therefore, weigh of an object on earth is inversely proportional to the square of its distance from the centre of the earth
W₁/W₂ = r₂²/r₁² .....1
W₂ = W₁r₁²/r₂²
At sea level the weight of the plane is W1 and at distance r₁ from the centre of the earth which is equal to the radius of the earth.
The radius of the earth is = 6378.1km
r₁ = radius of the earth = 6 378.1km = 6,378,100m
r₂ = r₁ + 11,719.342m = 6,378,100m + 11,719.342m
r₂ = 6,389,819.342m
W₂ = W₁r₁²/r₂²
W₂ = W₁[(6378100)²/(6,389,819.342)²]
W₂ = W₁[0.996335234422]
W₂/W₁ = 0.9963
fraction reduction of the weight is
ΔW/W₁ = 1 - W₂/W₁ = 1 - 0.9963 = 0.0037
percentage change :
%(∆W) = 0.0037 × 100% = 0.37%
Therefore, the percentage weight loss is 0.37%
How far is the center of mass of the Earth-Moon system from thecenter of the Earth? (Appendix C gives the masses of the Earth andthe Moon and the distance between the two.)
Answer:
[tex]\bar x=4679496.086\ m=4679.496086\ km[/tex] from the center of the earth.
Explanation:
We have a system of Earth & Moon:
we have the mass of earth, [tex]m_e=5.972\times 10^{24}\ kg[/tex]mass of the moon, [tex]m_m=7.348\times 10^{22}\ kg[/tex]distance between the center of the earth and the moon [tex]d=385000\ km[/tex]Now we assume the origin of the system to be at the center of the earth.
Now for the center of mass of this system:
[tex]\bar x=\frac{m_e.x_e+m_m.x_m}{m_e+m_m}[/tex]
here:
[tex]x_e\ \&\ x_m[/tex] are the distance of the centers (center of masses) of the Earth and the Moon from the origin of the system.
[tex]x_e=0[/tex] ∵ since we have taken the point as the origin of the system.
[tex]x_m=d[/tex]
now putting the values in the above equation:
[tex]\bar x=\frac{(5.972\times 10^{24}\times 0)+(7.348\times 10^{22}\times 385000\times 1000)}{5.972\times 10^{24}+7.348\times 10^{22}}[/tex]
[tex]\bar x=4679496.086\ m=4679.496086\ km[/tex] from the center of the earth.
Describe the difference between technology based effluent standards and water quality based effluent standards under the Clean Water Act. Also, indicate which of these two different standards is likely to be controlling on a small stream designated as a cold water fishery and why.
Explanation:
Technology-based:
1.As the name implies technology, no technology will be clarified about it. It only depends on the variables, which describes them.
2. That is based on a single facility's findings.
3. It takes into account the contaminants type and volume, and their equations to monitor them.
4. This is reserved for city or urban wastewater treatment plants only.
5. It takes into account the pH, need for oxygen and the suspended solids.
Water quality based on:
1.This is enforced if there is a need to apply stricter limits to pollutants that are not pleased with the limits of technology.
2. All basing on water quality was risk-based.
3. They placed some less importance on the technologies which is used in the technology based limit.
Water quality dependent restrictions are used as cool water fishery for streams.
The act on clean water will also include bodies of water belonging to wildlife, agriculture and others. The law also included that the physical chemical and biological variables of all the state water bodies must be controlled by these water quality based limits.
the density of gold is 19.3 g/cm3. suppose a certain gold wedding ring deplaced 0.55mL of liquid when dropped in a glass of sparkling cider. how much did this wedding ring cost
The question pertains to physics, specifically the concept of density. The mass of the gold in the ring is computed to be 10.615 g based on the provided volume displacement and density of gold. To determine the cost of the ring, additional information such as the current market price for gold and labor costs would be necessary.
The subject matter of this question relates to the physical property of density and its application in determining the mass of gold in a wedding ring. Given that the density of gold is 19.3 g/cm³ and the ring displaced 0.55 mL of liquid, one can calculate the mass of the ring using the rule that 1 mL equals 1 cm³. Hence, if the ring displaces 0.55 mL of the liquid, it means the volume of the ring is 0.55 cm³.
To find mass, we use Density = Mass/Volume, so Mass = Density * Volume. Therefore, the mass of the ring will be 19.3 g/cm³ * 0.55 cm³ = 10.615 g.
This calculation gives us the mass of the gold in the ring.
However, to determine the cost of the ring, we need more information such as the market price of gold per gram and the cost of workmanship and other potential elements in the ring, which are not provided in the question.
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An engineer wishes to design a curved exit ramp for a toll road in such a way that a car will not have to rely on friction to round the curve without skidding. Suppose that a typical car rounds the curve with a speed of 11.7m/s and that the radius of the curve is 50.0m. At what angle should the curve be banked?
To solve this problem we will make a graph that allows us to understand the components acting on the body. In this way we will have the centripetal Force and the Force by gravity generating a total component. If we take both forces and get the trigonometric ratio of the tangent we would have the angle is,
[tex]T_x = nsinA = \frac{mv^2}{r}[/tex]
[tex]T_y = ncosA = mg[/tex]
Dividing both.
[tex]tan A = \frac{v^2}{rg}[/tex]
[tex]tan A = \frac{11.7^2}{50*9.8}[/tex]
[tex]A = tan^{-1} (0.279367)[/tex]
[tex]A = 15.608\°[/tex]
Therefore the angle that should the curve be banked is 15.608°
An astronomer looks at the Andromeda galaxy (the other large galaxy in the Local Group) through her telescope. How long ago did that light leave Andromeda?
Answer:
[tex]2.537\times 10^{6}\ years[/tex]
Explanation:
Distance to Andromeda galaxy
[tex]2.537\times 10^{6}\ ly=2.537\times 10^{6}\times c\times y[/tex]
Speed of light is
[tex]c=3\times 10^8[/tex]
Time is given by
[tex]t=\dfrac{Distance}{Speed}\\\Rightarrow t=\dfrac{2.537\times 10^{6}\times c\times y}{c}\\\Rightarrow t=2.537\times 10^{6}\ y[/tex]
Hence, the light from Andromeda left [tex]2.537\times 10^{6}\ years[/tex] ago
The light observed from the Andromeda galaxy by the astronomer left Andromeda around 2 million years ago, showcasing the vast distances in space and providing insights into the universe's evolution.
The light from Andromeda galaxy that the astronomer observes left Andromeda approximately 2 million years ago. This is because the distance in light years is the same as the time it takes for the light to reach us.
This phenomenon is due to the vast distances in space. The Andromeda galaxy is the closest large galaxy to the Milky Way, located 2 million light years away from us.
Understanding the age of the light we observe helps us gain insights into the history of distant galaxies and the evolution of the universe.
A circular loop has radius R and carries current I2 in a clockwise direction. The center of the loop is a distance D above a long, straight wire.
What is the magnitude of the current I1 in the wire if the magnetic field at the center loop is zero? Express your answer in terms of the variables I2, R, D, and appropriate constants (μ0 and π).
Answer:
[tex]I_{1}[/tex] = (πD[tex]I_{2}[/tex])/R
Explanation:
If we define the magnitude of the field as B, then we have:
Total magnitude of the field [tex]B_{t}[/tex] = magnitude of the field B_loop + magnitude of the field B_wire. The total magnitude is equivalent to zero. Therefore, the field B_loop has an inward direction while the field B_wire has an outward direction.
B_loop = (μ0)*([tex]I_{2}[/tex])/2*R
B_wire = (μ0)*([tex]I_{1}[/tex])/2*π*D
Thus:
B_loop = B_wire at the center of the loop.
(μ0)*([tex]I_{2}[/tex])/2*R = (μ0)*([tex]I_{1}[/tex])/2*π*D
[tex]I_{1}[/tex] = (πD[tex]I_{2}[/tex])/R
The magnitude of current at the center of the loop is [tex]I_1 = \frac{\pi D I_2}{R}[/tex].
The given parameters;
radius of the loop = Rcurrent in the loop, I = I₂distance of the loop from the wire, = DThe magnetic field at the center of the loop is calculated as follows;
[tex]B_o = \frac{\mu_o I_2}{2R}[/tex]
The magnetic field at the distance of the wire is calculated as follows;
[tex]B_o = \frac{\mu_o I_1}{2\pi D}[/tex]
The magnitude of current at the center of the loop is calculated as follows;
[tex]\frac{\mu_o I_2 }{2R} = \frac{\mu_o I_1}{2\pi D} \\\\I_1 = \frac{\pi D I_2}{R}[/tex]
Thus, the magnitude of current at the center of the loop is [tex]I_1 = \frac{\pi D I_2}{R}[/tex].
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Light from the star Betelgeuse takes 640 years to reach Earth. How far away is Betelgeuse in units of light-years? Name any historical event that was occurring on Earth at about the time the light left Betelgeuse. Is the distance to Betelgeuse unusual compared with other stars?
Answer:
The distance is 641.8207 light years, and the star Betelgeuse is further away when compared to other stars
Historical event: Benedict XI succeeds Boniface VIII as pope (1302)
Explanation:
the solution is in the attached Word file
For each of the cases listed below, decide whether or not the motion described is an example of acceleration: (T/F)
1) a roller coaster as it starts to roll down the track
2) a ball tossed straight up, at the peak of its trajectory
3) a planet tracing out a circular orbit at a constant speed
4) a block sliding down a straight ramp at a constant speed
5) an airplane skidding to a stop on a runway
6) a dump truck carrying a load straight forward at a constant speed
Answer:
1. True
2. True
3. False
4. False
5. True
6. False
Explanation:
Acceleration: It refers to the change in velocity/speed of an object with respect to time. When the speed increases with time we call it acceleration and when its decreases it is called as deceleration. Let us analyze each instance individually:
1. When roller coaster starts to roll down the track its speed will increase with time. That means it is accelerating.
2. When the ball reaches at the peak of its trajectory, it comes to a stop for a fraction of a second that means it decelerates.
3. Since the velocity remains constant there is no acceleration.
4. Since the speed is no changing with time, there is no acceleration.
5. Since the moving plane comes to a stop, it is a case of deceleration.
6. Since the truck is moving at a constant speed so the acceleration is zero.
A roller coaster as it starts to roll down a track and an airplane skidding to a stop are examples of acceleration, while a ball tossed straight up, a planet tracing a circular orbit, a block sliding down a ramp, and a dump truck carrying a load at a constant speed are not examples of acceleration.
Explanation:1) True. When a roller coaster starts to roll down the track, it experiences a change in velocity, which means it is accelerating.
2) False. At the peak of its trajectory, a ball tossed straight up has zero velocity and is momentarily at rest, so it is not accelerating.
3) False. A planet tracing out a circular orbit at a constant speed is not experiencing a change in velocity, so it is not accelerating.
4) False. The block sliding down a ramp at a constant speed is not experiencing a change in velocity, so it is not accelerating.
5) True. An airplane skidding to a stop on a runway experiences a change in velocity, so it is accelerating.
6) False. A dump truck carrying a load straight forward at a constant speed is not experiencing a change in velocity, so it is not accelerating.
A river barge, whose cross section is approximately rectangular, carries a load of grain. The barge is 28 ft wide and 93 ft long. When unloaded its draft (depth of submergence) is 6 ft, and with the load of grain the draft is 9 ft. Determine: (a) the unloaded weight of the barge, and (b) the weight of the grain.
Answer:
a) [tex] W_B = F_B = 62.4 \frac{lb}{ft^3} (6ft*28ft*93ft)= 974937.6 lb[/tex]
b) [tex] W_g= 62.4 \frac{lb}{ft^3} * (9ft*28ft*93ft) -974937.6 lb =487468.8 lb[/tex]
Explanation:
Part a
For this case we have the situation illustrated on Figure 1. We will have two forces involved in equilibrium the weight [tex] W_B[/tex] and the Bouyance force[tex] F_B[/tex], and since the system is on equilibrium we have:
[tex] \sum F_{vertical}=0[/tex]
So then we have:
[tex] W_B = F_B = \gamma_{w} V_s[/tex]
Where [tex] V_s[/tex] represent the submerged volume. [tex]\gamma_w[/tex] represent the specific weight for the fluid. So we can replace and we have:
[tex] W_B = F_B = 62.4 \frac{lb}{ft^3} (6ft*28ft*93ft)= 974937.6 lb[/tex]
Part b
As we can see on figure 2 attached we have the illustration for this case. We add the weight for the grain and now the depth is 9ft.
W can do the balance of forces in the vertical and we got again:
[tex] W_B +W_g = F_B[/tex]
Where [tex] W_g[/tex] represent the weight for the grain.
And if we solve for [tex] W_g[/tex] we got:
[tex] W_g = F_B -W_B[/tex]
[tex] W_g =\gamma_w V_S -W_B[/tex]
Where [tex] \gamma_w[/tex] represent the specific weight of rthe water and [tex] V_s[/tex] the submerged volume. If we replace we got:
[tex] W_g= 62.4 \frac{lb}{ft^3} * (9ft*28ft*93ft) -974937.6 lb =487468.8 lb[/tex]
The weight of the unloaded barge is 975,769.6 lbs, and the weight of the grain is 486,572.8 lbs. These calculations were made based on Archimedes' principle, which states that the buoyant force (the weight of water displaced) is equal to the weight of the object.
Explanation:This question is about calculating the weight of a river barge and its load based on the principles of fluid mechanics. Here, we are examining the fact that the weight of the water displaced by the barge equals the weight of the barge according to Archimedes' principle.
Firstly, we calculate the unloaded weight of the barge. The volume of water displaced by the barge when it is empty is the volume of a rectangular prism with dimensions 28ft x 93ft x 6ft, which gives us 15,624 cubic feet. Given that the density of water is 62.4 lbs/ft³, the weight of this water, which is equal to the weight of the empty barge, would be (Volume x Density) = 15,624ft³ x 62.4 lbs/ft³ = 975,769.6 lbs.
Secondly, let's calculate the weight of the grain. The volume of water displaced when the barge is loaded is the volume of a rectangular prism with dimensions 28ft x 93ft x 9ft, which equals 23,436 cubic feet. The weight of this water, which is equal to the weight of the loaded barge, is (Volume x Density) = 23,436ft³ x 62.4 lbs/ft³ = 1,462,342.4 lbs. Hence, the weight of the grain is the weight of the loaded barge minus the weight of the barge itself = 1,462,342.4 lbs - 975,769.6 lbs = 486,572.8 lbs.
Learn more about Archimedes' Principle here:https://brainly.com/question/13106989
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What is the correct answer
Answer:
D and compound
Explanation:
because N2 is = to a compound
A rubber ball with a mass of 0.30 kg is dropped onto a steel plate. The ball's velocity just before impact is 4.5 m/s and just after impact is 4.2 m/s and just after impact is 4.2 m/s. What is the change in the ball's momentum?
Answer:
Change in momentum will be -2.61 kgm/sec
Explanation:
We have given mass of the rubber ball m = 0.30 kg
Velocity of the ball before the impact [tex]v_1=4.5m/sec[/tex]
Velocity of ball after impact [tex]v_2=-4.2m/sec[/tex] ( negative sign is due to opposite direction of motion )
Change in momentum is given by [tex]m(v_2-v_1)=0.3\times (-4.2-4.5)=0.3\times =0.3\times -8.7=-2.61kgm/sec[/tex] ( negative sign shows the direction of change in momentum )
Answer:
-0.09 kg m/s
Explanation:
The magnitude of the electric force between two protons is 2.30 x 10^-26 N. How far apart are they?
a) 0.0220 mb) 0.100 mc) 0.480 md) 0.000570 me) 3.10 m
To solve this problem we will use the concepts given by Coulomb's law defined for force, said law is mathematically described as
[tex]F = \frac{kq_1 q_2}{r^2}[/tex]
Here,
k = Coulomb's constant
[tex]q_{1,2}[/tex]= Charge of two protons
r = Distance between them
F = Force
Our values are given as,
[tex]F =2.3*10^{-26} N[/tex]
[tex]q_1 =q_2 = 1.6*10^{-19} C[/tex]
[tex]k =9*10^9 Nm^2/C2[/tex]
Rearrenging to find the distance and replacing we have that
[tex]r^2=\frac{(9*10^9 )(1.6*10^{-19})^2 }{2.3*10^{-26} }[/tex]
[tex]r^2=10.01*10^{-3} m^2[/tex]
[tex]r =\sqrt{10.01*10^{-3} m^2}[/tex]
[tex]r = 0.100m[/tex]
Therefore the correct option is B.
Vector vector A has a magnitude A and is directed at an angle theta measured with respect to the positive x-axis. What is the magnitude of vector A sub x, the x-component of vector A?
Answer:[tex]A_x[/tex] = Acos[tex]\theta[/tex]
Explanation:
A vector in this situation have two components
1) The X-component
2) The Y-component
So as we put [tex]cos\theta[/tex] with the x-axis while [tex]sin\theta[/tex] with the y- axis and this case our answer should be
[tex]A_x[/tex] = Acos[tex]\theta[/tex]
I hope this will answer your question,
An image is also provided please have a look at that.
Thank you.
Answer: Ax = Acos(theta)
Therefore, the x component of A: Ax = Acos(theta)
Explanation:
The attached image is a pictorial representation of the question:
From the attached image,
Cos(theta) = adjacent/hypothenus
Cos(theta) = Ax/A
Making Ax the subject of formula,
Ax = Acos(theta)
Therefore, the x component of A: Ax = Acos(theta)
A wheel accelerates from rest to 59 rad/s^2 at a uniform rate of 58 rad/s^2. Through what angle (in radians) did the wheel turn while accelerating?
A) 24 rad
B) 38 rad
C) 30 rad
D) 60 rad
To solve this problem we will apply the physical equations of the angular kinematic movement, for which it defines the square of the final angular velocity as the sum between the square of the initial angular velocity and the product between 2 times the angular acceleration and angular displacement. We will clear said angular displacement to find the correct response
Using,
[tex]\omega^2 = \omega_0^2 +2\alpha \theta[/tex]
Here,
[tex]\omega[/tex] = Final angular velocity
[tex]\omega_0[/tex] = Initial angular velocity
[tex]\alpha =[/tex] Angular acceleration
[tex]\theta =[/tex] Angular displacement
Replacing,
[tex]59^2 = 0+2*58\theta[/tex]
[tex]\theta = 30rad[/tex]
Therefore the correct answer is C.
The angle at which the wheel turns while accelerating is 30 radians and this can be determined by using the kinematics equation.
Given :
A wheel accelerates from rest to 59 rad/[tex]\rm s^2[/tex] at a uniform rate of 58 rad/[tex]\rm s^2[/tex].
The equation of kinematics is used in order to determine the angle at which the wheel turn while accelerating.
[tex]\omega^2 = \omega^2_0+2\alpha \theta[/tex]
where [tex]\omega[/tex] is the final angular velocity, [tex]\omega_0[/tex] is the initial angular velocity, [tex]\alpha[/tex] is the angular acceleration, and [tex]\theta[/tex] is the angular displacement.
Now, substitute the values of the known terms in the above formula.
[tex]59^2 =0+2\times 58 \times \theta[/tex]
Simplify the above expression.
[tex]\rm \theta = 30\; rad[/tex]
Therefore, the correct option is C).
For more information, refer to the link given below:
https://brainly.com/question/408236
Verify that for values of n less than 8, the system goes to a stable equilibrium, but as n passes 8, the equilibrium point becomes unstable, and a stable oscillation is created.
Answer:
Biological system is one of the major causes of oscillation due to sensitive negative feedback loops. For instance, imagine a father teaching his son how to drive, the teen is trying to keep the car in the centre lane and his father tell him to go right or go left as the case may be. This is a example of a negative feedback loop of a biological system. If the father's sensitivity to the car's position on the road is reasonable, the car will travel in a fairly straight line down the centre of the road. On the other hand, what happens if the father raise his voice at the son "go right" or when the car drifts a bit to the left? The startled the son will over correct, taking the car too far to the right. The father will then starts yelling "go left" then the boy will over correct again and the car will definitely oscillate back and forth. A scenario that indicates the behavior of a car driver under a very steep feedback control mechanism. Since the driver over corrects in each direction. Therefore causes oscillations.
Explanation:
In physics, a system's equilibrium point is considered stable if the system returns to that point after being slightly disturbed. For an n value less than 8, the system remains in stable equilibrium; once n exceeds 8, stable oscillations indicate an unstable equilibrium. This concept can be visualized by the marble in a bowl analogy, where the bowl's orientation determines the stability of the marble's equilibrium.
Explanation:The stability of an equilibrium point is determined by the response of a system to a disturbance. If an object at a stable equilibrium point is slightly disturbed, it will oscillate around that point. The stable equilibrium point is characterized by forces that are directed toward it on either side. On the contrary, an unstable equilibrium point will not allow the object to return to its initial position after a slight disturbance, since the forces are directed away from that point. For values of n less than 8, the system finds a stable equilibrium. However, when n surpasses 8, the system exhibits unstable equilibrium, leading to stable oscillations.
As an example, consider a marble in a bowl. When the bowl is right-side up, the marble represents a stable equilibrium; when disturbed, it returns to the center. If the bowl were inverted, the marble on top would be at an unstable equilibrium point; any disturbance would cause it to roll off, as the forces on either side would be directed outwards. Extending this concept to potential energy, n in a potential energy function acts as an adjustable parameter. For n=<8, the system remains in stable equilibrium, as with NaCl, which has an n value close to 8. Beyond this value, the equilibrium becomes unstable, giving rise to oscillatory behavior.
Stability also depends on the nature of damping in the system. An overdamped system moves slowly towards equilibrium without oscillations, an underdamped system quickly returns but oscillates, and a critically damped system reaches equilibrium as quickly as possible without any oscillations.
You would like to know whether silicon will float in mercury and you know that can determine this based on their densities. Unfortunately, you have the density of mercury in units of kg.m^3 and the density of silicon in other units of 2.33 g.cm^3. You decide to convert the density of silicon into units of kg.m^3 to perform the comparison. By which combination of conversion factors will you multiply 2.33 g.cm^3 to perform the unit conversion?
Answer:
The conversion factor used here will be 1000 (kg/m^3)/(g/cm^3).
Which is a combination of two conversion factors:
1 kg = 1000 g
1 x 10^6 cm^3 = 1 m^3
Explanation:
We will use unitary method to convert g/cm^3 into kg/m^3. This is shown below:
Since, 1 kilogram is equivalent to 1000 gram and 1 meter is equivalent to 100 centimeter. Therefore:
1 g/cm^3 = (1 g/ cm^3)(1 kg/ 1000 g)(100 cm / 1 m)^3
1 g/cm^3 = 1000 kg/m^3
Hence, the conversion factor that will be multiplied is found to be 1000.
Using this in our case, we get:
Density of silicon = (2.33)(1000) kg/m^3
Density of Silicon = 2330 kg/m^3
Answer:
1000
Explanation:
Conversion from 'g' to 'kg': divide by 1000g/kg
Conversion from 'cm^3' to 'm^3': divide by 1000000cm^3/m^3
2.33g/cm^3 = [tex]\frac{2.33*1000000}{100}[/tex]
= 2330 kg/m^3
we simply multiply by 1000 to get the units converted to kg/m^3
Suppose that 10 moles of an ideal gas have a gauge pressure of 2 atm and a temperature of 200 K. If the volume of the gas is doubled and the pressure dropped to a gauge pressure of 1 atm, what is the new temperature?
Select one:
a.
267 K
b.
300 K
c.
400 K
d.
200 K
The new temperature is: d. 200 K
The Ideal Gas Law is given by:
[tex]PV = nRT[/tex]
where:
[tex]P[/tex] = pressure
[tex]V[/tex] = volume
[tex]n[/tex] = number of moles
[tex]R[/tex] = universal gas constant
[tex]T[/tex] = temperature
Given initial conditions:
[tex]P_1 = 2[/tex] atm (gauge pressure)
[tex]T_1 = 200[/tex] K
[tex]V_1 = V[/tex]
[tex]n = 10[/tex] moles
Final conditions:
[tex]P_2 = 1[/tex] atm (gauge pressure)
[tex]V_2 = 2V[/tex]
[tex]T_2 = ?[/tex]
We can use the combined gas law equation to relate the initial and final states of the gas:
[tex]\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}[/tex]
Substituting in the given values:
[tex]\frac{2V}{200} = \frac{1 \cdot 2V}{T_2}[/tex]
Solving for [tex]T_2[/tex]:
[tex]\frac{2V}{200} = \frac{2V}{T_2}[/tex]
By simplifying, we get:
[tex]\frac{2}{200} = \frac{2}{T_2}[/tex]
Cross-multiplying gives:
[tex]2 \cdot T_2 = 400[/tex]
Dividing both sides by 2 gives:
[tex]T_2 = 200[/tex] K
Therefore, the new temperature [tex](T_2)[/tex] after the volume is doubled and the pressure is halved is 200 K.
momentum A proton interacts electrically with a neutral HCl molecule located at the origin. At a certain time t, the proton’s position is h1.6 × 10−9 , 0, 0i m and the proton’s velocity is h3200, 800, 0i m/s. The force exerted on the proton by the HCl molecule is h−1.12 × 10−11 , 0, 0i N. At a time t + (2 × 10−14 s), what is the approximate velocity of the proton? answer
Answer:
[tex]<3068.2352, 800, 0>\ m/s[/tex]
Explanation:
F = Force = [tex]<-1.12\times 10^{-11}, 0, 0>[/tex]
m = Mass of proton = [tex]1.7\times 10^{-27\ kg[/tex]
t = Time taken = [tex]2\times 10^{-14}\ s[/tex]
Acceleration is given by
[tex]a=\dfrac{F}{m}\\\Rightarrow a=\dfrac{<-1.12\times 10^{-11}, 0, 0>}{1.7\times 10^{-27}}\\\Rightarrow a=<-6.58824\times 10^{15}, 0, 0>\ m/s^2[/tex]
[tex]v=u+at\\\Rightarrow v=<3200, 800, 0>+<-6.58824\times 10^{15}, 0, 0>\times 2\times 10^{-14}\\\Rightarrow v=<3200, 800, 0>+<-6.58824\times 10^{15}, 0, 0>\times 2\times 10^{-14}\\\Rightarrow v=<3200, 800, 0>+<-131.7648, 0, 0>\\\Rightarrow v=<3068.2352, 800, 0>\ m/s[/tex]
The velocity of the proton is [tex]<3068.2352, 800, 0>\ m/s[/tex]