Explanation:
As the distance between a planet and the sun increase, the following quantities will decrease.
1. The gravitational force is given by :
[tex]F=G\dfrac{m_1m_2}{r^2}[/tex]
As the distance between a planet and the sun increase, the gravitational force decrease.
2. As the distance between a planet and the sun increase, it result in the decrease in its speed. Due to decrease in gravitational force, centripetal force acting on the planet should decrease. Since,
[tex]F_c=\dfrac{mv^2}{r}[/tex]
v is the tangential speed
So, the tangential speed must decrease.
3. The distance traveled in a day should decrease.
Hence, this is the required solution.
True or False?
A small sphere of mass 1 kg is attached to the end of a cord of length 1 m and rotated in a vertical circle about a fixed point O. If the speed at the uppermost point of the trajectory is 1 m/s, the sphere will have a circular trajectory (will move in a circle).
Use g = 9.8 m/s².
Answer:
False. the system does not complete the circle movement
Explanation:
For this exercise, we must find the rope tension at the highest point of the path,
-T - W = m a
The acceleration is centripetal
a = v² / R
T = ma - mg
T = m (v² / R - g)
The minimum tension that the rope can have is zero (T = 0)
v² / R - g = 0
v = √ g R
Let's find out what this minimum speed is
v = √ 9.8 1
v = 3.13 m / s
We see that the speed of the body is less than this, so the system does not complete the movement.
Capacitor plates have area 5.00 cm^2 and separation 2.00 mm. The plates are in a vacuum. The charging current I_c has a constant value of 1.80 mA. At t = 0 the charge on the plates is zero. (a) Calculate the charge on the plates, the electric field between the plates, and the potential difference between the plates when t = 0.500 ms. (b) Calculate dE/dt, the time rate of change of the electric field between the plates. Does dE/dt vary in time? (c) Calculate the displacement current density jD between the plates, and from this the total displacement current I_c. How do I_c and I_d compare?
Answer:
a) Q = [tex]9*10^{-10}[/tex] C; E = [tex]2.03*10^5[/tex] V/m; V = 406.8 V
b) dE/dt = [tex]4.07*10^{11}[/tex] V/(m.s); No
c) [tex]J_d[/tex] = 3.6 [tex]A/m^2[/tex]; Equal
Explanation:
Given parameters are:
Area, A = 5 cm^2
Separation, d = 2 mm
Changing current, [tex]i_c[/tex] = 1.8 mA
At time t = 0 the charge [tex]Q_0[/tex] = 0
a) Here, we are asked to find charge, Q, electric field, E, and potential difference, V at time t = 0.5 [tex]\mu s[/tex]
[tex]Q = i_ct = 1.8*10^{-3}*5*10^{-7} = 9*10^{-10}[/tex] C
[tex]E = \sigma/\epsilon_0 = (Q/A)/\epsilon_0 = (9*10^{-10}/5*10^{-4})/(8.85*10^{-12}) = 2.03*10^5[/tex] [tex]\frac{V}{m}[/tex]
[tex]V = Ed = 2.03*10^{-5}*2*10^{-3} = 406.8[/tex] V
b) [tex]E = (Q/A)/\epsilon_0[/tex]
⇒ [tex]\frac{dE}{dt} = \frac{dQ}{dt} \frac{1}{\epsilon_0 A} = \frac{i_c}{\epsilon_0 A} = \frac{1.8*10^{-3}}{5*10^{-4}*8.85*10^{-12}} = 4.07*10^{11}[/tex] V/(m.s)
No, it is constant that does not vary in time because [tex]i_c[/tex] is constant.
c) the displacement current density, [tex]J_d = \epsilon_0\frac{dE}{dt} = \epsilon_0\frac{i_c}{\epsilon_0 A} = i_c/A[/tex]
⇒ [tex]J_d = 1.8*10^{-3}/(5*10^{-4}) = 3.6[/tex] [tex]A/m^2[/tex]
[tex]i_d =J_dA = 3.6*5*10^{-4} = 1.8*10^{-3}[/tex] A
So, [tex]i_c[/tex] and [tex]i_d[/tex] are equal.
Answer:
[tex]a)[/tex]The charge on the plates[tex]$Q=9 * 10^{-10} C ; E=2.03 * 10^{5} \mathrm{~V} / \mathrm{m} ; \mathrm{V}=406.8 \mathrm{~V}$[/tex]
[tex]b)[/tex]The time rate of change of the electric field between the plates[tex]$\mathrm{dE} / \mathrm{dt}=4.07 * 10^{11} \mathrm{~V} /(\mathrm{m} . \mathrm{s}) ; \mathrm{No}$[/tex]
[tex]c)[/tex]The displacement current density jD between the plates[tex]$J_{d}=3.6 \mathrm{~A} / \mathrm{m}^{2}$[/tex]Equals
Explanation:
Given parameters are:
Area, [tex]$A=5cm^{2}[/tex]
Separation, [tex]$\mathrm{d}=2 \mathrm{~mm}$[/tex]
Changing current, [tex]$i_{c}=1.8 \mathrm{~mA}$[/tex]
At time [tex]$\mathrm{t}=0$[/tex] the charge [tex]$Q_{0}=0$[/tex]
a) Here, we are asked to find charge, [tex]$Q$[/tex], electric field, [tex]$E$[/tex] and potential difference, [tex]$\mathrm{V}$[/tex] at time [tex]$\mathrm{t}=0.5 \mu \mathrm{s}$[/tex]
[tex]$Q=i_{c} t=1.8 \times10^{-3} \times 5 \times10^{-7}=9 \times10^{-10} \mathrm{C}$[/tex]
[tex]$E=\sigma / \epsilon_{0}=(Q / A) / \epsilon_{0}=\left(9 \times10^{-10} / 5 \times10^{-4}\right) /\left(8.85\times 10^{-12}\right)=2.03 \times10^{5} \frac{\mathrm{V}}{m}$[/tex]
[tex]$V=E d=2.03 \times10^{-5} \times2\times 10^{-3}=406.8 \mathrm{~V}$[/tex]
b) [tex]$E=(Q / A) / \epsilon_{0}$[/tex]
[tex]$\Rightarrow \frac{d E}{d t}=\frac{d Q}{d t} \frac{1}{\epsilon_{0} A}=\frac{i_{c}}{\epsilon_{0} A}=\frac{1.8 \times10^{-3}}{5\times 10^{-4} \times 8.85 \times 10^{-12}}=4.07 \times 10^{11} \mathrm{~V} /(\mathrm{m} . \mathrm{S})$[/tex]
No, it is constant that does not vary in time because is constant.
c) the displacement current density, [tex]$J_{d}=\epsilon_{0} \frac{d E}{d t}=\epsilon_{0} \frac{i_{c}}{\epsilon_{0} A}=i_{c} / A$[/tex]
[tex]$\Rightarrow J_{d}=1.8 \times 10^{-3} /\left(5 \times10^{-4}\right)=3.6 \mathrm{~A} / \mathrm{m}^{2}$[/tex]
[tex]$i_{d}=J_{d} A=3.6 \times5 \times10^{-4}=1.8\times 10^{-3} \mathrm{~A}$[/tex]
So, [tex]$i_{c}$[/tex] and [tex]$i_{d}$[/tex] are equal.
To know more about
https://brainly.com/question/20343142
If an object is thrown downward with an initial velocity of v 0, then the distance it travels is given by sequals4.9tsquaredplusv 0t. An object is thrown downward from a cliff 400 m high and it travels 138.3 m in 3 sec. What was the initial velocity of the object?
Explanation:
If an object is thrown downward with an initial velocity of v₀, then the distance it travels is given by s = 4.9 t²+v₀t
Now an object is thrown downward from a cliff 400 m high and it travels 138.3 m in 3 sec. We need to find initial velocity of the object.
s = 4.9 t²+v₀t
138.3 = 4.9 x 3²+ v₀ x 3
3v₀ = 94.2
v₀ = 31.4 m/s
Initial velocity of the object = 31.4 m/s
What is the general aerobic activity recommendation for an exercise frequency that is at least five days per week?
Explanation:
Minimum of aerobic activity 150 minutes, or a mix of moderates and intensive exercise 75 minutes of vigorous aerobic activity a week. We can extend this practice throughout the week with the instructions. Mild aerobic workouts might include practices like quick strolling or swimming, while activity like running can include strong aerobic activity.
Based on Planet Z's size, orbital distance, and rotation rate, which of the following properties is it likely to have?
erosion due to liquid water
polar ice caps
a surface crowded with impact craters
seasons
active volcanoes
strong winds and violent storms
an atmosphere produced by outgassing
active tectonics
Answer:
Planet Z will have the following properties;
Active Volcanoes
Active Tectonics
An Atmosphere produced by outgassing
Explanation:
The little terrestrial worlds have heat shorter than the much bigger terrestrial worlds, so the bigger worlds tend to have active volcanism and tectonics. These active volcanism and tectonics are likely to erase ancient craters. The active volcanism and tectonics would create an atmosphere by producing gases.
It is know that the Terrestrial worlds that are not far from the star have higher surface temperature.
Fast rate of rotation can cause winds and strong storms but here it is slower compared to earth. Also, a tilt of axis causes seasons.
The properties the star have are active volcanoes, active tectonics and an atmosphere produced by outgassing.
Final answer:
Planet Z's characteristics such as geological activity, seasons, and atmosphere can be inferred from its size, orbital distance, and rotation rate. A planetary mass similar to Earth's suggests active geology and an atmosphere from outgassing, while a proper distance from the sun allows for liquid water and polar ice caps. The planet's rotation influences the presence of seasons and potential for strong winds and violent storms.
Explanation:
Based on Planet Z's size, orbital distance, and rotation rate, it is possible to infer several characteristics that this planet might have. The level of geological activity on a planet is often proportional to its mass, suggesting that planets similar in size to Earth and Venus are more likely to exhibit geological activity such as active volcanoes or tectonics. Similarly, a planet's distance from its sun can influence the presence of liquid water, with those at optimum distances having the potential for erosion due to liquid water and possibly polar ice caps. A slower rotation might lead to more extreme temperature differences between day and night, which could impact atmospheric conditions and lead to strong winds and violent storms due to the larger temperature gradient.
Planetary rotation also contributes significantly to the development of seasons; hence, how Planet Z rotates will affect whether it experiences seasons. A planet that has active geology and volcanism will likely have an atmosphere that is at least partially produced by outgassing, as seen on Earth, and could also support active tectonics. Lastly, if the planet is not geologically active, it may have a surface crowded with impact craters, similar to the Moon and Mercury, which have less geological activity to renew their surfaces.
Your car has stalled and you need to push it. You notice as the car gets going that you need less and less force to keep it going. Suppose that for the first 15 m your force decreased at a constant rate from 210 N to 45 N. How much work did you do on the car?
Answer:
675 Joules
Explanation:
Considering that work can be calculated with the following formula:
W = Fx D
Where:
W = work
F = force
D = distance
We can directly use this formula in case the applied force remains constant
In case the force does not remain constant, we can calculate the work as a change, this is :
ΔW = ΔFxΔD
For this scenario, we have:
W₁ = F₁xD₁
F₁ = 210 N, D = 0 m
W₁ = 210Nx0m = 0 Joules
W₂ = F₂xD₂
F₂ = 45 N , D₂ = 15 m
W₂ = 45Nx15m = 675 Joules
Finally: ΔW = Total work performed when moving the car 15 m
ΔW = W₂ - W₁ = 675 Joules - 0 Joules = 675 Joules
Final answer:
The amount of work done on the car while pushing it is 1575 Joules.
Explanation:
Work is defined as the product of force and displacement. In this case, the force required to keep the car moving decreases as the car gets going. Work can be calculated using the formula:
Work = Force × Distance
Given that the force decreased from 210 N to 45 N over a distance of 15 m, we can calculate the work done as follows:
Work = (210 N + 45 N) / 2 × 15 m = 1575 J
Therefore, the amount of work done on the car while pushing it is 1575 Joules.
A rat receives food when it lever presses in the presence of a 1100 Hz tone, but not in the presence of a 1300 Hz tone. The 1100 Hz tone is an ________, while the 1300 Hz tone is an ________.
Answer:
1100 Hz is a discriminitive stimulus signal (SD), while 1300 Hz is a discriminitive stimuli for extinction (S∆ )
Explanation:
This is used to describe Discrimination Training in learning and behavior. The 1100 Hz is a discriminitive stimulus signal (SD), while 1300 Hz is a discriminitive stimuli for extinction (S∆ )
The 1100 Hz tone acts as a conditioned stimulus (CS), predicting food, while the 1300 Hz tone serves as an unconditioned stimulus (UCS), with no associated response. Understanding pitch and loudness is key in differentiating stimuli in conditioning.
Explanation:The question relates to classical conditioning concepts in behavioral psychology, where the 1100 Hz tone is conditionally associated with the provision of food. In this context, the 1100 Hz tone functions as a conditioned stimulus (CS), since it predicts the arrival of food, prompting the rat to press the lever. On the other hand, the 1300 Hz tone does not forecast the coming of food and, thus, operates as an unconditioned stimulus (UCS), having no effect on the rat's lever-pressing behavior. It's also important to note the perception of frequency, which we refer to as pitch, and the perception of intensity, which is understood as loudness. The ability to discern differences in pitch allows for different responses to the 1100 Hz and 1300 Hz tones.
A mountain 10.0 km from a person exerts a gravitational force on him equal to 2.00% of his weight. (a) Calculate the mass of the mountain. (b) Compare the mountain’s mass with that of Earth. (c) What is unreasonable about these results? (d) Which premises are unreasonable or inconsistent?
The mass of the mountain is 3.002 × 10^16 kg and its fraction of Earth's mass is approximately 4.92 × 10^-8. These results are unreasonable due to the large mass of the mountain compared to Earth and the assumptions made in the question.
To calculate the mass of the mountain, we can use Newton's Law of Universal Gravitation. The gravitational force exerted by the mountain is equal to 2.00% of the person's weight. Since weight is equal to mass multiplied by acceleration due to gravity, we can set up the equation:
0.02mg = GMm / d^2
where G is the gravitational constant, M is the mass of the mountain, m is the mass of the person, and d is the distance between them. Since the person's weight is equal to mg, we can rewrite the equation as:
0.02mg = (GMm / d^2)
Dividing both sides by mg gives us:
0.02 = (GM / d^2)
Now we can solve for the mass of the mountain (M):
M = (0.02d^2 / G)
Substituting the given values (d = 10.0 km, G = 6.673 × 10^-11 Nm²/kg²),
M = (0.02 × 10000^2) / (6.673 × 10^-11)
M = 3.002 × 10^16 kg
The mass of the mountain is 3.002 × 10^16 kg.
Comparing the mass of the mountain with that of Earth, we can use the equation:
Mountain's Mass / Earth's Mass = 3.002 × 10^16 / (6 × 10^24)
Mountain's Mass / Earth's Mass = 4.92 × 10^-8
The mass of the mountain is approximately 4.92 × 10^-8 of Earth's mass.
These results are unreasonable because the mass of the mountain and its fraction of Earth's mass are too large. It is unlikely for a mountain to have such a massive mass compared to the entire Earth.
The premises that are unreasonable or inconsistent in this scenario are the assumption that the gravitational force exerted by the mountain is 2.00% of the person's weight and the assumption that the distance between the mountain and the person is 10.0 km.
Learn more about Mass and Gravitation here:
https://brainly.com/question/30429836
#SPJ6
The calculated mass of the mountain is 2.937 × 10¹⁷ kg, which is 4.91 × 10⁻⁸ of Earth's mass. This result is unreasonable because the mass and fraction are too large. The assumption about the gravitational force exerted by the mountain is inconsistent.
Let's start by calculating the mass of the mountain, given the gravitational force it exerts.
(a)
Given data:
Distance between person and mountain, r = 10.0 km = 10,000 mGravitational force exerted by mountain on person, F = 2.00% of the person's weightThe weight of the person is W = mg, where m is the mass of the person and g is the gravitational acceleration (9.8 m/s²).
The gravitational force F is given by Newton's law of gravitation: F = G * (m_p * M) / r², where:
G = gravitational constant (6.674 × 10-11 N·m²/kg²)m_p = mass of the personM = mass of the mountainGiven that F = 0.02 * mg, we can equate:
0.02 * mg = G * (m * M) / (10,000 m)²
Cancel out m and solve for M:
M = (0.02 * g * (10,000 m)²) / G
M = 2.937 × 10¹⁷ kg
(b)
Mass of Earth ME = 5.972 × 10²⁴ kg
Fraction of Earth's mass = M / ME
Fraction = 2.937 × 10¹⁷ kg / 5.972 × 10²⁴ kg = 4.91 × 10⁻⁸
(c)
The mass of the mountain and its fraction of the Earth's mass are excessively large. Such a massive mountain would be geologically and physically improbable.
(d)
The gravitational force assumed to be exerted by the mountain is too large. In reality, a mountain would exert a much smaller percentage of gravitational force on a person.
Please !!! I really need help !!! How do I understand these ?!!!!
Answer
The answer for the first one I think is false.
The second one would be true i think. I hope i got it right and have a wonderful day
Answer:
True
False
Explanation:
From 0 to E, the train moves a distance of 55 m.
From F to J, the train moves a distance of 59 m.
The total distance is 55 + 59 = 114 m.
The displacement is the difference between the final position and initial position. Here, the distance between J and 0 is -4 m.
If a force of 163 newtons is applied to the end of the jack handle the end moves down 15 cm. If lifting part of the jac raises the car bumper 1.75 cm what force is the jack exerting on the bumper?
The force exerted on the bumper is 1397 N
Explanation:
We can solve this problem by using the equilibrium of the torques: in fact, the torque exerted on one side of the jack must be equal to the torque exerted on the other side of the jack.
Therefore, we can write:
[tex]F_h d_h = F_b d_b[/tex]
where
[tex]F_h = 163 N[/tex] is the force applied to the end of the jack handle
[tex]d_h = 15 cm[/tex] is the distance between the force applied on the handle and the pivot
[tex]F_b[/tex] is the force exerted by the jack on the car bumper
[tex]d_b = 1.75 cm[/tex] is the distance between this force and the car bumper
And solving for [tex]F_b[/tex], we find:
[tex]F_b = \frac{F_h d_h}{d_b}=\frac{(163)(15)}{1.75}=1397 N[/tex]
Learn more about levers and torques:
brainly.com/question/5352966
#LearnwithBrainly
A ball of mass 24 g is attached to a cord of length 0.463 m and rotates in a vertical circle. What is the minimum speed the ball must have at the top of the circle so the cord does not become slack? The acceleration of gravity is 9.8 m/s 2 . Answer in units of m/s.
Answer:2.13 m/s
Explanation:
Given
mass of ball [tex]m=24 gm[/tex]
Length of cord [tex]L=0.463 m[/tex]
acceleration due to gravity [tex]g=9.8 m/s^2[/tex]
minimum velocity after which cord will slack
[tex]\frac{mv^2}{r}=mg[/tex]
[tex]v=\sqrt{rg}[/tex]
[tex]v=\sqrt{0.463\times 9.8}[/tex]
[tex]v=2.13 m/s[/tex]
The gold foil experiment led to the conclusion that each atom in the foil was composed mostly of empty space because most alpha particles directed at the foil
(1) passed through the foil
(2) remained trapped in the foil
(3) were deflected by the nuclei in gold atoms
(4) were deflected by the electrons in gold atoms
Answer:
(1) passed through the foil
Explanation:
Ernest Rutherford conducted an experiment using an alpha particle emitter projected towards a gold foil and the gold foil was surrounded by a fluorescent screen which glows upon being struck by an alpha particle.
When the experiment was conducted he found that most of the alpha particles went away without any deflection (due to the empty space) glowing the fluorescent screen right at the point of from where they were emitted. While a few were deflected at reflex angle because they were directed towards the center of the nucleus having the net effective charge as positive. And some were acutely deflected due to the field effect of the positive charge of the proton inside the nucleus. All these conclusions were made based upon the spot of glow on the fluorescent screen.Answer:
Option 1
Explanation:
The correct answer is option 1
The gold foil experiment was conducted by Rutherford. This experiment was conducted to study the Atom.
In the experiment Alpha rays from the emitter are passed through gold foil and there was a receiver that was present there to intercept the alpha rays.
The outcome of the result was that most of the alpha particle pass through foil undeflected and very few rays revert back on the original path from the heavy mass present at the center.
Later this heavy mass was known Nucleus.
Hence, most alpha particles passed through the foil.
The triceps muscle in the back of the upper arm is primarily used to extend the forearm. Suppose this muscle in a professional boxer exerts a force of 1.9 × 103 N with an effective perpendicular lever arm of 2.85 cm, producing an angular acceleration of the forearm of 150 rad/s2?
Answer:
I=0.0361 kg.m^2
Explanation:
Torque is the rotational equivalent of a force
Torque= perpendicular distance r X Force F
Torque T = I(moment of inertia) X α (angular acceleration)
T= Iα
r= 0.0285m
F= 1.9 x 10^3
T=0.0285 x 1.9 x 10^3
T= 54.15Nm
I=T/α
I=54.15/150
I=0.361 kg.m^2
How many vibrations per second are represented in a radio wave of 101.7 MHz. (1 MHz = 106Hz)?
Answer:
[tex]n=101.7\times 10^6[/tex]
Explanation:
It is given that,
Frequency of the radio wave, [tex]f=101.7\ MHz=101.7\times 10^6\ Hz[/tex]
We know that the number of vibrations per second is called frequency of an object. We need to find the number of vibrations per second. Clearly, the number of vibrations per second represented in a radio wave is [tex]101.7\times 10^6[/tex]. Hence, this is the required solution.
A man goes for a walk, starting from the origin of an xyz coordinate system, with the xy plane horizontal and the x axis eastward. Carrying a bad penny, he walks 1300 m east, 2400 m north, and then drops the penny from a cliff 640 m high
(a) In unit-vector notation, what is the displacement of the penny from start to its landing point?
(b) When the man returns to the origin, what is the magnitude of his displacement for the return trip?
Explanation:
a) He walks 1300 m east, 2400 m north, and then drops the penny from a cliff 640 m high.
1300 m east = 1300 i
2400 m north = 2400 j
Drops the penny from a cliff 640 m high = -640 k
Displacement of penny = 1300 i + 2400 j - 640 k
b) Displacement of man for return trip = -1300 i - 2400 j
[tex]\texttt{Magnitude = }\sqrt{(-1300)^2+(-2400)^2}=2729.47m[/tex]
Magnitude of his displacement = 2729.47 m
Answer:
Explanation:
d1 = 1300 m east
d2 = 2400 m north
d3 = 640 m downward
(a)
The displacement of penny is given by
[tex]\overrightarrow{d}=\overrightarrow{d_{1}}+\overrightarrow{d_{2}}+\overrightarrow{d_{3}}[/tex]
[tex]\overrightarrow{d}=1300\widehat{i}+2400 \widehat{j}-640\widehat{k}[/tex]
(b) For the return journey of man, the displacement is given by
[tex]\overrightarrow{d}=-\overrightarrow{d_{1}}-\overrightarrow{d_{2}}[/tex]
[tex]\overrightarrow{d}=-1300\widehat{i}-2400 \widehat{j}[/tex]
The magnitude of the displacement is given by
[tex]d=\sqrt{1300^{2}+2400^{2}}=2729.47 m[/tex]
A car is parked on a cliff overlooking the ocean on an incline that makes an angle of 22◦ below the horizontal. The negligent driver leaves the car in neutral, and the emergency brakes are defective. The car rolls from rest down the incline with a constant acceleration of 4.4 m/s 2 and travels 54 m to the edge of the cliff. The cliff is 28 m above the ocean. How long is the car in the air? The acceleration of gravity is 9.81 m/s 2 . Answer in units of s.
Answer:
The car is 3.4 s in the air.
Explanation:
Hi there!
Please, see the attached figure for a graphical description of the problem.
The vertical position of the car can be obtained by the following equation:
y = y0 + v0 · t · sin α + 1/2 · g · t²
Where:
y = vertical position of car at time t.
y0 = initial vertical position.
v0 = initial velocity.
t = time.
α = launching angle.
g = acceleration of gravity.
The vertical component of the position vector when the car reaches the ground is -28 m (considering the edge of the cliff as the origin of the system of reference) and the initial vertical position is therefore 0 m. The launching angle is 22° below the horizontal (see figure). Then, we only have to find the initial velocity to solve the equation of vertical position for the time of flight.
To find the initial velocity, we have to use two equations: the equation of velocity of the car at the time it reaches the edge of the cliff and the equation of position of the car to find that time:
x = x0 + v0 · t + 1/2 · a · t²
v = v0 + a · t
Where:
x = position of the car at time t.
x0 = initial position.
v0 = initial velocity.
t = time.
a = acceleration.
v = velocity of the car at time t.
If we place the origin of the frame of reference at the point where the car starts rolling, then the initial position is zero. Since the car starts from rest, the initial velocity, v0, is zero. Then, we can find the time it takes the car to travel the 54 m down the incline:
x = x0 + v0 · t + 1/2 · a · t² (x0 = 0 and v0 = 0)
x = 1/2 · a · t²
54 m = 1/2 · 4.4 m/s² · t²
2 · 54 m / 4.4 m/s² = t²
t = 5.0 s
With this time, we can find the velocity of the car when it reaches the edge of the cliff:
v = v0 + a · t (v0 = 0)
v = a · t
v = 4.4 m/s² · 5.0 s
v = 22 m/s
Then, the initial velocity of the falling car is 22 m/s. Using the equation of vertical position:
y = y0 + v0 · t · sin α + 1/2 · g · t² (y0 = 0)
y = v0 · t · sin α + 1/2 · g · t²
-28 m = 22 m/s · t · sin 22° - 1/2 · 9.81 m/s² · t²
0 = 28 m + 22 m/s · t · sin 22° - 4.91 m/s² · t²
Solving the quadratic equation for t using the quadratic formula:
t =3.4 s (the other values is negative and, thus, discarded).
The car is 3.4 s in the air.
Final answer:
By solving the equation h = 1/2 g t^2 for the time it takes an object to fall 28 m, we find that the car is in the air for approximately 2.39 seconds before it hits the ocean.
Explanation:
To calculate how long the car is in the air, we need to analyze the car's vertical motion separately from its horizontal motion. The car falls 28 m, which we can use with the acceleration of gravity to find the time it takes to hit the water.
We use the equation h = ½ g t^2 where h is the height the car falls (28 m), and g is the acceleration due to gravity (9.81 m/s2). We're looking for t, the time in seconds.
28 m = ½ (9.81 m/s2) t^2
Rearranging for t: t = √(2 * 28 m / 9.81 m/s2)
t = √(5.70 s2)
t = 2.39 s
The car is in the air for approximately 2.39 seconds before hitting the water.
Identify the method of thermal energy transfer.
Conduction, convection, radiation
A:
B:
C:
radiation
convection
conduction
trust me dawg
The method of thermal energy transfer in this three cases:
A: Radiation
B: Convection
C: Conduction
What are the methods of thermal energy transfer?Heat transfer is defined by thermodynamic systems as "The transfer of heat over the system boundary caused by a temperature difference between the system and its surroundings."
There are several ways for heat to go from one place to another. Conduction, convection, and radiation are some of the several ways that heat is transferred.
Conduction is the process of energy being transferred from one medium particle to another while the particles are in close proximity to one another.
The flow of fluid molecules from higher temperature regions to lower temperature regions is referred to as convection.
Radiant heat is the name for thermal radiations. Emission of electromagnetic waves results in the production of thermal radiation. These waves remove the energy from the body that is releasing them.
Learn more about thermal energy here:
https://brainly.com/question/11278589
#SPJ2
A baseball pitcher throws a baseball horizontally at a linear speed of 42.5 m/s (about 95 mi/h). Before being caught, the baseball travels a horizontal distance of 16.5 m and rotates through an angle of 49.0 rad. The baseball has a radius of 3.67 cm and is rotating about an axis as it travels, much like the earth does. What is the tangential speed of a point on the "equator" of the baseball?
Answer:
4.5m/s
Explanation:
Linear speed (v) = 42.5m/s
Distance(x) = 16.5m
θ= 49.0 rad
radius (r) = 3.67 cm
= 0.0367m
The time taken to travel = t
Recall that speed = distance / time
Time = distance / speed
t = x/v
t = 16.5/42.5
t = 0.4 secs
tangential velocity is proportional to the radius and angular velocity ω
Vt = rω
Angular velocity (ω) = θ/t
ω = 49/0.4
ω = 122.5 rad/s
Vt = rω
Vt = 0.0367 * 122.5
Vt =4.5 m/s
The tangential speed of a point on the 'equator' of the baseball is approximately 1.7983 m/s, based on the provided radius of the ball and its angular speed.
Explanation:The question is essentially asking for the tangential speed of the baseball, which describes the speed of a point on its outer edge (or 'equator') as the baseball spins or rotates. The tangential speed can be calculated using the formula v = rω, where v is the tangential speed, r is the radius, and ω is the angular speed. The angular speed is essentially the rate at which an object is rotating or spinning, measured in radians per second.
Given from the prompt, we know that ω = 49.0 rad and r = 3.67 cm = 0.0367 m (because we need the radius in meters). Inserting these values into the formula we get: v = (0.0367 m)(49.0 rad/s) = 1.7983 m/s. So, a point on the 'equator' of the baseball is moving at a tangential speed of approximately 1.7983 m/s.
Learn more about Tangential Speed here:https://brainly.com/question/17025846
#SPJ3
Did the displacement at this point reach its maximum of 2 mm before or after the interval of time when the displacement was a constant 1 mm?
Answer:
a. before
Explanation:
Did the displacement at this point reach its maximum of 2 mm before or after the interval of time when the displacement was a constant 1 mm?
from the graph given from a source. the vertical axis represents the displacement of the graph motion, whilst the horizontal side is representing the time variable of the motion .
displacement is distance in a specific direction.
before the displacement was maximum at 2mm was instant at time=0.04s.
But later was constant at 0.06s at a displacement point of 1mm
If a small number of tree frogs migrate to a mat of vegetation that is already home to an established population of tree frogs and introduce new alleles into the native population's gene pool, this change is referred to as A the bottleneck effect. B directional selection. C genetic drift. D gene flow.
Answer:
D
Explanation:
Gene flow is the transfer of genetic variation from one population to another
Compared to a 1-kg block of solid iron, a 2-kg block of solid iron has twice as much
A)volume.
B)inertia.
C)mass.
D) all of the above
E) none of the above
The 2-kg block of solid iron has twice the mass of the 1-kg block.
Explanation:The correct option is C) mass.
Mass is a measure of the amount of matter in an object, and it is not affected by its shape or size. Since both blocks are made of solid iron, they have the same density. Therefore, the 2-kg block of solid iron has twice the mass of the 1-kg block.
Learn more about solid iron blocks here:https://brainly.com/question/32813527
#SPJ3
If a small sports car collides head-on with a massive truck, which vehicle experiences the greater impact force? Which vehicle experiences the greater acceleration? Explain your answers.
Answer:
Small sports car.
Explanation:
Lets take
mass of the small car = m
mass of the truck = M
As we know that when car collide with the massive truck then due to change in the moment of the car both car as well as truck will feel force.We also know that from Third law of Newton's ,it states that every action have it reaction with same magnitude but in the opposite direction.
Therefore
F = m a
a=Acceleration of the car
[tex]a=\dfrac{F}{m}[/tex]
F= M a'
a'=Acceleration of the massive truck
[tex]a'=\dfrac{F}{M}[/tex]
Here given that M > m that is why a > a'
Therefore car will experiences more acceleration.
A flat uniform circular disk has a mass of 3.97 kg and a radius of 85.7 cm. It is suspended in a horizontal plane by a vertical wire attached to its center. If the disk is rotated 2.42 rad about the wire, a torque of 0.0688 N·m is required to maintain that orientation. Calculate
(a) the rotational inertia of the disk about the wire,
(b) the torsion constant, and
(c) the angular frequency of this torsion pendulum when it is set oscillating.
Answer:
1.457881265 kgm²
0.02842 Nm/rad
0.13962 rad/s
Explanation:
M = Mass = 3.97 kg
R = Radius = 85.7 cm
[tex]\tau[/tex] = Torque = 0.0688 Nm
[tex]\theta[/tex] = Angle of rotation = 2.42 rad
Moment of inertia about the center of the disk is given by
[tex]I=\dfrac{1}{2}MR^2\\\Rightarrow I=\dfrac{1}{2}\times 3.97\times 0.857^2\\\Rightarrow I=1.457881265\ kgm^2[/tex]
The rotational inertia of the disk about the wire is 1.457881265 kgm²
Torque is given by
[tex]\tau=\kappa \theta\\\Rightarrow \kappa=\dfrac{\tau}{\theta}\\\Rightarrow \kappa=\dfrac{0.0688}{2.42}\\\Rightarrow \kappa=0.02842\ Nm/rad[/tex]
The torsion constant is 0.02842 Nm/rad
Time period is given by
[tex]T=2\pi\sqrt{\dfrac{I}{\kappa}}[/tex]
Angular frequency is given by
[tex]\omega=\dfrac{2\pi}{T}\\\Rightarrow \omega=\sqrt{\dfrac{\kappa}{I}}\\\Rightarrow \omega=\sqrt{\dfrac{0.02842}{1.457881265}}\\\Rightarrow \omega=0.13962\ rad/s[/tex]
The angular frequency of this torsion pendulum when it is set oscillating is 0.13962 rad/s
he light energy that falls on a square meter of ground over the course of a typical sunny day is about 20 MJ. The average rate of electric energy consumption in one house is 1.0 kW. Part A On average, how much energy does one house use during each 24 h day? Express your answer in joules.
Answer:
[tex]E = 86.4 \times 10^6 J[/tex]
Explanation:
Given data:
light energy = 20 MJ
Electric consumption is 1.0 kW
Duration of energy consumption is 24 hr
Energy consumption is given as
[tex]E = Power \times time[/tex]
[tex]E = 1 \times 10^3 \times 24 \times 3600[/tex]
[tex]E = 8.64 \times 10^6 J = 86.4 MJ [/tex]
[tex]E = 86.4 \times 10^6 J[/tex]
Which technique is best for manual in-line stabilization of a person floating faceup on the surface?
Answer:
vise grip
Explanation:
Manual in-line stabilization (MILS) of the cervical spine is a type of airway management when dealing with patients in traumatic condition ..it is a means that is performed by grasping the mastoid process of the patient, so as to prevent the movement of the cervical column during intubation of the trachea
MLS provides a means of stability to the cervical column for a patient in trauma. During this technique, a patient is restricted from moving his or her cervical collar. The vise grip can be used for a patient with neck injury. The technique is used to roll a patient to face up to prevent further injuries.
Answer: the technique which is best for manual in-line stabilization of a person floating faceup on the surface is vice grip.
Explanation:
Vice grip is a rescue technique used to prevent further injury to a guest who is suspected of having suffered a spinal injury as the typical posture is floating faceup on the surface.
A 900-kg car traveling east at 15.0 m/s collides with a 750-kg car traveling north at 20.0 m/s. The cars stick together. Assume that any other unbalanced forces are negligible. (a) What is the speed of the wreckage just after the collision? (b) In what direction does the wreckage move just after the collision?
Explanation:
It is given that,
Mass of the car 1, [tex]m_1=900\ kg[/tex]
Initial speed of car 1, [tex]u_1=15i\ m/s[/tex] (east)
Mass of the car 2, [tex]m_2=750\ kg[/tex]
Initial speed of car 2, [tex]u_1=20j\ m/s[/tex] (north)
(b) As the cars stick together. It is a case of inelastic collision. Let V is the common speed after the collision. Using the conservation of momentum as :
[tex]m_1u_1+m_2u_2=(m_1+m_2)V[/tex]
[tex]900\times 15i +750\times 20j=(900+750)V[/tex]
[tex]13500i+15000j=1650V[/tex]
[tex]V=(8.18i+9.09j)\ m/s[/tex]
The magnitude of speed,
[tex]|V|=\sqrt{8.18^2+9.09^2}[/tex]
V = 12.22 m/s
(b) Let [tex]\theta[/tex] is the direction the wreckage move just after the collision. It is given by :
[tex]tan\theta=\dfrac{v_y}{v_x}[/tex]
[tex]tan\theta=\dfrac{9.09}{8.18}[/tex]
[tex]\theta=48.01^{\circ}[/tex]
Hence, this is the required solution.
To solve this problem, we can use the principle of conservation of momentum. We calculate the momentum of each car before the collision, then find the total momentum before the collision. Since the cars stick together after the collision, we can calculate the velocity of the wreckage and determine its direction using trigonometry.
Explanation:To solve this problem, we can use the principle of conservation of momentum. The momentum before the collision is equal to the momentum after the collision.
First, we need to calculate the momentum of each car before the collision. The momentum is calculated by multiplying the mass of the car by its velocity. Next, we can calculate the total momentum before the collision by adding up the individual momenta of the two cars. Since the cars stick together after the collision, we can find the velocity of the wreckage by dividing the total momentum by the total mass of the cars. Finally, we can use trigonometry to find the direction of the wreckage's motion. We have the x-component and y-component of the velocity, so we can use these values to calculate the direction. Learn more about Conservation of momentum here:https://brainly.com/question/33316833
#SPJ12
A sound is recorded at 19 decibels. What is the intensity of the sound?
1 x 10^-8.7 W/m^2
1 x 10^-10.1 W/m^2
1 x 10^-11.9 W/m^2
1 x 10^-9.4 W/m^2
[tex]1 \times 10^{-10.1} \mathrm{Wm}^{-2}[/tex] is the intensity of the sound.
Answer: Option B
Explanation:
The range of sound intensity that people can recognize is so large (including 13 magnitude levels). The intensity of the weakest audible noise is called the hearing threshold. (intensity about [tex]1 \times 10^{-12} \mathrm{Wm}^{-2}[/tex]). Because it is difficult to imagine numbers in such a large range, it is advisable to use a scale from 0 to 100.
This is the goal of the decibel scale (dB). Because logarithm has the property of recording a large number and returning a small number, the dB scale is based on a logarithmic scale. The scale is defined so that the hearing threshold has intensity level of sound as 0.
[tex]\text { Intensity }(d B)=(10 d B) \times \log _{10}\left(\frac{I}{I_{0}}\right)[/tex]
Where,
I = Intensity of the sound produced
[tex]I_{0}[/tex] = Standard Intensity of sound of 60 decibels = [tex]1 \times 10^{-12} \mathrm{Wm}^{-2}[/tex]
So for 19 decibels, determine I as follows,
[tex]19 d B=(10 d B) \times \log _{10}\left(\frac{I}{1 \times 10^{-12} W m^{-2}}\right)[/tex]
[tex]\log _{10}\left(\frac{1}{1 \times 10^{-12} \mathrm{Wm}^{-2}}\right)=\frac{19}{10}[/tex]
[tex]\log _{10}\left(\frac{1}{1 \times 10^{-12} \mathrm{Wm}^{-2}}\right)=1.9[/tex]
When log goes to other side, express in 10 to the power of that side value,
[tex]\left(\frac{I}{1 \times 10^{-12} W m^{-2}}\right)=10^{1.9}[/tex]
[tex]I=1 \times 10^{-12} \mathrm{Wm}^{-2} \times 10^{1.9}=1 \times 10^{-12-1.9}=1 \times 10^{-10.1} \mathrm{Wm}^{-2}[/tex]
A car is traveling north at 17.7m/s After 12 s its velocity is 14.1m/s in the same direction. Find the magnitude and direction of the car's average acceleration.A2.7 m/s2, north b0.30 m/s2, south c2.7 m/s2, south d0.30 m/s2, north
Answer:
Option B
0.3 m/s2 South
Explanation:
Acceleration, [tex]a=\frac {v-u}{t}[/tex] where v and u are final and initial velocities respectively, t is the time taken
Substituting 14.1 m/s for v, 17.7 m/s for u and 12 s for t then
[tex]a=\frac {14.1 m/s- 17.7 m/s}{12}=-0.3 m/s^{2}[/tex]
Since this is negative acceleration, it's direction is opposite hence 0.3 m/s2 South
Things you can do with an environmental engineering major and a broadcasting minor?
Answer:
With an Environmental Engineering and a broadcasting minor
You can work as an On Air personality that host programs that provide your audience with documentaries about the environments and project carried out by Environmental Engineer
and also you can work as a journalist that explore the world making research that will preserve the environment and leveraging the media as a broadcaster to provide this research findings as a video for you audience
Explanation:
In order to get a better understanding let define some terms
Environmental Engineer :
Environmental engineers resolve and help prevent environmental problems. They work in many areas, including air pollution control, industrial hygiene, toxic materials control, and land management. The duties of an environmental engineer range from planning and designing an effective waste treatment plant to studying the effects of acid rain on a particular area. An environmental engineer is sometimes required to work outdoors, though most of her work is done in a laboratory or office setting. Career opportunities for environmental engineers exist in consulting, research, corporate, and government positions.
Broadcasting:
Broadcasting is the distribution of audio or video content to a dispersed audience via any electronic mass communications medium, but typically one using the electromagnetic spectrum (radio waves), in a one-to-many model.
) Which one of the following statements best describes the operation of a heat engine?
a) A heat engine performs work and generates an equal amount of thermal energy in a cyclic process.
b) A heat engine decreases the entropy of the universe by generating an equal amount of heat and work.
c) A heat engine uses input heat to perform work and rejects excess thermal energy to a lower temperature reservoir.
d) A heat engine transfers thermal energy from a lower temperature reservoir to a higher temperature reservoir through work performed on the system.
e) A heat engine transfers thermal energy from a higher temperature reservoir to a lower temperature reservoir through work performed on the system
Operation of a heat engine is best described as using input heat to perform work and rejects excess thermal energy to a lower temperature reservoir.
What is Heat?This is defined as the form of energy that is transferred between two substances at different temperatures.
Heat engine uses heat to perform work which results in the rejection of excess thermal energy to a lower temperature reservoir which was why option C was chosen as the most appropriate choice.
Read more about Heat energy here https://brainly.com/question/19666326
The correct description of a heat engine's operation is that it uses heat to perform work, while also rejecting a portion of the heat to a cooler reservoir; it is a cyclical process where energy is partially converted from thermal to mechanical form according to the Second Law of Thermodynamics.
Explanation:The statement that best describes the operation of a heat engine is (c) A heat engine uses input heat to perform work and rejects excess thermal energy to a lower temperature reservoir. In thermodynamics, a heat engine is a system that converts heat or thermal energy to mechanical work. This process involves transferring energy from a high-temperature source, the hot reservoir (Qh), and partly converting this energy into work (W), while the rest is rejected as waste heat into a lower temperature sink, the cold reservoir (Qc).
Based on the Second Law of Thermodynamics, no heat engine can convert 100% of the input thermal energy into work because there is always some waste heat that must be expelled to a colder environment. The efficiency of a heat engine is measured as the ratio of work done to the heat input from the hot reservoir. The theoretical upper limit of this efficiency depends on the temperatures of the hot and cold reservoirs and is given by the Carnot efficiency, which is always less than 100%.