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olga2289 [7]
3 years ago
12

A planet of mass 7.00 1025 kg is in a circular orbit of radius 6.00 1011 m around a star. The star exerts a force on the planet

of constant magnitude 6.51 1022 N. The speed of the planet is 2.36 104 m/s.
(a) In half a "year" the planet goes half way around the star. What is the distance that the planet travels along the semicircle?
distance = m
(b) During this half "year", how much work is done on the planet by the gravitational force acting on the planet?
work = J
(c) What is the change in kinetic energy of the planet?
?K = J
(d) What is the magnitude of the change of momentum of the planet?
Physics
1 answer:
BARSIC [14]3 years ago
7 0

Answer:

A) 1.88 * 10^17 m

B) 1.22 * 10^34 J

C) 1.95 * 10^34 J

Explanation:

Parameters given:

Mass of planet = 7.00 * 10^25 kg

Radius of orbit = 6.00 * 10^11 m

Force exerted on planet = 6.51 * 10^22 N

Velocity of planet = 2.36 * 10^4 m/s

A) The distance traveled by the planet is half of the circumference of the orbit (which is circular).

The circumference of the orbit is

C = 2 * pi * R

R = radius of orbit

C = 2 * 3.142 * 6.0 * 10¹¹

C = 3.77 * 10¹² m

Hence, distance traveled will be:

D = 0.5 * 3.77 * 10¹²

D = 1.88 * 10 ¹² m/s

B) Work done is given as:

W = F * D

W = 652 * 10²² * 1.88 * 10¹¹

W = 1.22 * 10³⁴ J

C) Change in Kinetic energy is given as:

K. E. = 0.5 * m * v²

K. E. = 0.5 * 7 * 10^25 * (2.36 * 10^4)²

K. E. = 1.95 * 10³⁴ J

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Answer:

The power

Explanation:

We know that the work definition is given by the following expression:

W = F * d

where:

F = force [Newtons] [N]

d = distance [meters] [m]

W = work [Joules]

And the expression that defines the work done by unit of time is called - <u>Power</u>, therefore:

P = W/t

where:

P = power [watts] [w]

W = work [Joules] [J]

t = time [seconds] [s]

6 0
3 years ago
Compare the collision between two baseballs and a catcher's mitt.
Nookie1986 [14]

The applied force is different for the two cases

The case A with a greater force involves the greatest momentum change

The case A involves the greatest force.

<h3>What is collision?</h3>
  • This is the head-on impact between two object moving in opposite or same direction.

The initial momentum of the two ball is the same.

P = mv

where;

  • m is the mass of each
  • v is the initial velocity of each ball

Since the force applied by the arm is different, the final velocity of the balls before stopping will be different.

Thus, the final momentum of each ball will be different

The impulse experienced by each ball is different since impulse is the change in momentum of the balls.

J = ΔP

The force applied by the rigid arm is greater than the force applied by the relaxed arm because the force applied by the rigid arm will cause the ball to be brought to rest faster.

Thus, we can conclude the following;

  • The applied force is different for the two cases
  • The case A with a greater force involves the greatest momentum change
  • The case A involves the greatest force.

Learn more about impulse here: brainly.com/question/25700778

3 0
2 years ago
To take off from the ground, an airplane must reach a sufficiently high speed. The velocity required for the takeoff, the takeof
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<h2> The potential and kinetic energy of airplane are affected by these factors </h2>

Explanation:

When airplane rises up , it requires potential energy . This potential energy can be taken from the kinetic energy of airplane .

Thus if the speed of wind is larger , it can either oppose the motion of velocity or can favour the velocity of airplane  . By which its kinetic energy is effected .

If the weight of airplane is changed , it will effect the potential energy required . Thus heavier plane requires higher potential energy for attaining the same height .

Thus these two factor has important role in the flight of airplane .

6 0
3 years ago
A tank is full of water. Find the work W required to pump the water out of the spout. (Use 9.8 m/s2 for g. Use 1000 kg/m3 as the
Sergio039 [100]

Answer:

W = 1.06 MJ

Explanation:

- We will use differential calculus to solve this problem.

- Make a differential volume of water in the tank with thickness dx. We see as we traverse up or down the differential volume of water the side length is always constant, hence, its always 8.

- As for the width of the part w we see that it varies as we move up and down the differential element. We will draw a rectangle whose base axis is x and vertical axis is y. we will find the equation of the slant line that comes out to be y = 0.5*x. And the width spans towards both of the sides its going to be 2*y = x.

- Now develop and expression of Force required:

                                             F = p*V*g

                                             F = 1000*(2*0.5*x*8*dx)*g

                                             F = 78480*x*dx

- Now, the work done is given by:

                                             W = F.s

- Where, s is the distance from top of hose to the differential volume:

                                             s = (5 - x)

- We have the work as follows:

                                            dW = 78400*x*(5-x)dx

- Now integrate the following express from 0 to 3 till the tank is empty:

                                           W = 78400*(2.5*x^2 - (1/3)*x^3)

                                           W = 78400*(2.5*3^2 - (1/3)*3^3)

                                           W = 78400*13.5 = 1058400 J

 

5 0
3 years ago
A spaceprobe has an 29.0 m length when measured at rest. What length
elixir [45]

Answer:

The observer sees the space-probe 9.055m long.

Explanation:

Let L_0 be the length of the space-probe when measured at rest, and L be its length as observed by an observer moving at velocity v, then

(1).\: \: L = L_0\sqrt{1-\dfrac{v^2}{c^2} }

Now, we know that L_0 = 29.0m and v = 0.95c, and putting these into (1) we get:

L = 29\sqrt{1-\dfrac{(0.95c)^2}{c^2} }

L = 29\sqrt{1-0.95^2 }

\boxed{L = 9.055m}

Thus, an observer moving at 0.95c observes the space-probe to be 9.055m long.

3 0
3 years ago
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