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nikdorinn [45]
4 years ago
8

Consider three planets. All have the same mass as Earth, but with different radii (from largest to smallest: Planet 1, 2, 3). Fo

r which planet is the escape velocity from the surface the largest? A) Planet 1 B) Planet 2 C) Planet 3 D) The escape velocity is the same for each.
Physics
1 answer:
LuckyWell [14K]4 years ago
3 0

Answer:

option C

Explanation:

given,

mass of the three planet is same

radius of the planets are

R₁ > R₂ > R₃

expression of escape velocity

v = \sqrt{\dfrac{2GM}{R}}

G is the gravitational constant

M is the mass of the planet

R is the radius of the planet

from the above expression we can clearly conclude that the escape velocity is inversely proportional to the radius of the Planet.

radius of planet increases escape velocity decreases.

Hence planet 3 has the smallest radius so the escape velocity of the third planet will be maximum.

The correct answer is option C

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A stone is thrown vertically upward with a speed of 18 m/s. (a) How long does it take the stone to reach a height of 11 m? (b) h
bagirrra123 [75]

Answer:

a) It takes the stone 0.7743 s to reach a height of 11 m for the first time on its way up and 2.899 s to reach again that height on its way down.

b) At t = 0.7743 s the velocity is 10.41 m/s and at t = 2.899 s the velocity is -10.41 m/s.

c) There are two answers because the stone reaches the height of 11 m one time on its way up and one more time again on its way down.

Please, see the attached figures and the explanation for a description of the figures.

Explanation:

Hi there!

The equations for the height and velocity of the stone are as follows:

y = y0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

y = height

y0 = initial height

v0 = initial velocity

t = time

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive)

v = velocity at time t

a) Let´s calculate the time it takes the stone to reach a height of 11 m. The origin of the frame of reference is at the throwing point so that y0 = 0:

y = y0 + v0 · t + 1/2 · g · t²        

11 m = 18 m/s · t - 1/2 · 9.8 m/s² · t²    

0 = -4.9 m/s² · t² + 18 m/s · t - 11 m

Solving the quadratic equation:

t = 0.7743 s and t = 2.899 s

(Notice that I have used more significant figures to avoid error by rounding)

The stone will be two times at a height of 11 m, one on its way up (at 0.7743 s) and one on its way down  (at 2.899 s). Then, it takes the stone 0.7743 s to reach a height of  11 m for the first time.

b)  Let´s use the equation of velocity:

v = v0 + g · t

at t = 0.77443 s

v = 18 m/s - 9.8 m/s² · 0.77443 s

v = 10.41 m/s

at t = 2.899 s

v = 18 m/s - 9.8 m/s² · 2.899 s

v = - 10.41 m/s

(Both velocities have to be of the same magnitude but of different sign, that´s why I haven´t rounded the time.)

c) There are two answers because the stone reaches the height of 11 m one time on its way up and one more time again on its way down. On its way up, the velocity is 10.41 m/s and on its way down it is -10.41 m/s.

Figures

The functions to plot are the following:

height in function of time (figure 1, x-axis: time. y-axis: height)

y = -4.9t² + 18t

velocity in function of time (figure 2, x-axis: time. y-axis velocity)

v = -9.8t + 18

Acceleration in function of time (figure 3, x-axis: time. y-axis: acceleration)

a = -9.8

5 0
4 years ago
If a car moves with constant velocity, does it also move with a constant speed?
GarryVolchara [31]

Answer:

Yes. The fact that an object moves at constant velocity implies that its speed is also constant. Note that the converse statement isn't necessarily true.

Explanation:

Velocity is a vector. For two vectors to be equal to each other,

  • their magnitudes (sizes) need be the same, and
  • they need to point in the same direction.

In motions, the magnitude of an object's velocity is the same as its speed.

If the car moves with a constant velocity, that means that

  • the magnitude of its velocity, the speed of the car, is constant;
  • also, the direction of the car's motion is also constant.

In other words,

\text{Velocity } \vec{v}\text{ is constant} \implies \text{Speed }v\text{ is constant}.

Note that the arrow here points only from the velocity side to the speed side. It doesn't point backward because knowing that the speed of an object is constant won't be sufficient to prove that the velocity of the object is also constant. For example, for an object in a uniform circular motion, the speed is constant but the direction keeps changing. Hence the velocity isn't constant.

3 0
3 years ago
Capacitors C1 = 5.85 µF and C2 = 2.80 µF are charged as a parallel combination across a 250 V battery. The capacitors are discon
posledela

Answer:

Q1_new = 515.68 µC

Q2_new = 246.82 µC

Explanation:

Since the capacitors are charged in parallel and not in series, then both are at 250 V when they are disconnected from the battery.

Then it is only necessary to calculate the charge on each capacitor:

Q1 = 5.85 µF * 250 V = 1462.5 µC

Q2 = 2.8 µF * 250 V = 700 µC

Now, we will look at 1462.5 µC as excess negative charges on one plate, and 1462.5 µC as excess positive charges on the other plate. Now, we will use this same logic for the smaller capacitor.

When there is a connection of positive plate of C1 to the negative plate of C2, and also a connection of the negative plate of C1 to the positive plate of C2, some of these excess opposite charges will combine and cancel each other. The result is that of a net charge:

1462.5 µC - 700 µC = 762.5 µC

Thus,762.5 µC of net charge will remain in the 'new' positive and negative plates of the resulting capacitor system.

This 762.5 µC will be divided proportionately between the two capacitors.

Q1_new = 762.5 µC * (5.85/(5.85 + 2.8)) = 515.68 µC

Q2_new = 762.5 µC * (2.8/(5.85 + 2.8) = 246.82 µC

4 0
4 years ago
Light travels _______ in a material with a higher index of refraction
exis [7]
Light will travel more slowly in a material with a higher index of refraction 
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3 years ago
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Describe and explain the potential energy changes that occur during the formation of a covalent bond
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