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Harman [31]
2 years ago
9

An object of mass M is dropped near the surface of Earth such that the gravitational field provides a constant downward force on

the object. Which of the following describes what happens to the center of mass of the object-Earth system as the object falls downward toward Earth? a. It moves toward the center of Earth. b. It moves toward the object.c. It does not move. d. The answer cannot be determined without knowing the mass of Earth and the distance between the object and Earth’s center.
Physics
1 answer:
marysya [2.9K]2 years ago
6 0

Answer:

The answer is: c. It does not move

Explanation:

Because the gravitational force is characterized by being an internal force within the Earth-particle system, in this case, the object of mass M. And since in this system there is no external force in the system, it can be concluded that the center of mass of the system will not move.

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A slender rod is 80.0 cm long and has mass 0.370 kg . A small 0.0200-kg sphere is welded to one end of the rod, and a small 0.05
nataly862011 [7]

Answer:

1.10 m/s

Explanation:

Linear speed is given by

v=r\omega

Kinetic energy is given by

KE=0.5I\omega^{2}

Potential energy

PE= mgh

From the law of conservation of energy, KE=PE hence

0.5I\omega^{2}=mgh where m is mass, I is moment of inertia, \omega is angular velocity, g is acceleration due to gravity and h is height

Substituting m2-m1 for m and 0.5l for h, \frac {2v}{L} for \omega we obtain

0.5I(\frac {2v}{L})^{2}=0.5Lg(m2-m1)

(\frac {2v}{L})^{2}=\frac {gl(m2-m1)}{I} and making v the subject

v^{2}=\frac {gl^{3}(m2-m1)}{4I}

v=\sqrt {\frac {gl^{3}(m2-m1)}{4I}}

For the rod, moment of inertia I=\frac {ML^{2}}{12} and for sphere I=MR^{2} hence substituting 0.5L for R then I=M(0.5L)^{2}

For the sphere on the left hand side, moment of inertia I

I=m1(0.5L)^{2} while for the sphere on right hand side, I=m2(0.5L)^{2}

The total moment of inertia is therefore given by adding

I=\frac {ML^{2}}{12}+ m1(0.5L)^{2}+ m2(0.5L)^{2}=\frac {L^{2}(M+3m1+3m2)}{12}

Substituting \frac {L^{2}(M+3m1+3m2)}{12} for I in the equation v=\sqrt {\frac {gL^{3}(m2-m1)}{4I}}

Then we obtain

v=\sqrt {\frac {gL^{3}(m2-m1)}{4(\frac {L^{2}(M+3m1+3m2)}{12})}}=\sqrt {\frac {3gL^{3}(m2-m1)}{L^{2}(M+3m1+3m2)}}

This is the expression of linear speed. Substituting values given we get

v=\sqrt {\frac {3*9.81*0.8^{3}(0.05-0.02)}{0.8^{2}(0.39+3(0.02)+3(0.05))}} \approx 1.08 m/s

8 0
2 years ago
A plane traveled west for 4.0 hours and covered a distance of 4,400 kilometers. What was its velocity? 18,000 km/hr 1,800 km/hr,
Airida [17]

west 1100 km / hr    ..

8 0
2 years ago
Read 2 more answers
A ball is falling at terminal velocity. Terminal velocity occurs when the ball is in equilibrium and the forces are balanced. Wh
Greeley [361]

Answer:

A free body diagram with 2 forces: the first pointing downward labeled F Subscript g Baseline 20 N and the second pointing upward labeled F Subscript air Baseline 20 N.

Explanation:

This is because at terminal velocity, the ball stops accelerating and the net force on the ball is zero. For the net force to be zero, equal and opposite forces must act on the ball, so that their resultant force is zero. That is F₁ + F₂ = 0 ⇒ F₁ = -F₂

Since F₁ = 20 N, then F₂ = -F₁ = -20 N

So, if F₁ points upwards since it is positive, then F₂ points downwards since it is negative.

So, a free body diagram with 2 forces: the first pointing downward labeled F Subscript g Baseline 20 N and the second pointing upward labeled F Subscript air Baseline 20 N best describes the ball falling at terminal velocity.

4 0
2 years ago
Read 2 more answers
23. While sliding a couch across a floor, Andrea and Jennifer exert forces F → A and F → J on the couch. Andrea’s force is due n
PSYCHO15rus [73]

Answer:

a)  (95.4 i^ + 282.6 j^) N , b) 298.27 N  71.3º and c)   F' = 298.27 N   θ = 251.4º

Explanation:

a) Let's use trigonometry to break down Jennifer's strength

      sin θ = Fjy / Fj

      cos θ = Fjx / Fj

Analyze the angle is 32º east of the north measuring from the positive side of the x-axis would be

          T = 90 -32 = 58º

         Fjy = Fj sin 58

         Fjx = FJ cos 58

         Fjx = 180 cos 58 = 95.4 N

         Fjy = 180 sin 58 = 152.6 N

Andrea's force is

         Fa = 130.0 j ^

We perform the summary of force on each axis

X axis

       Fx = Fjx

       Fx = 95.4 N

Axis y

       Fy = Fjy + Fa

       Fy = 152.6 + 130

       Fy = 282.6 N

       F = (95.4 i ^ + 282.6 j ^) N

b) Let's use the Pythagorean theorem and trigonometry

       F² = Fx² + Fy²

       F = √ (95.4² + 282.6²)

       F = √ (88963)

       F = 298.27 N

       tan θ = Fy / Fx

       θ = tan-1 (282.6 / 95.4)

       θ = tan-1 (2,962)

       θ = 71.3º

c) To avoid the movement they must apply a force of equal magnitude, but opposite direction

       F' = 298.27 N

       θ' = 180 + 71.3

       θ = 251.4º

4 0
2 years ago
In the middle of the night you are standing a horizontal distance of 14.0 m from the high fence that surrounds the estate of you
olchik [2.2K]

PART A)

horizontal distance that will be moved = 14 m

Height of the fence = 5.0 m

height from which it is thrown = 1.60 m

angle of projection = 54 degree

So here we can say that stone will travel vertically up by distance

\Delta y = 5 - 1.6 = 3.40 m

now we will have displacement in horizontal direction

\Delta x = 14 m

now we know that

v_x = vcos54

v_y = vsin54

now we will have

\Delta x = v_x t

14 = (vcos54)t

also for y direction

\Delta y = v_y t + \frac{1}{2}at^2

3.40 = (vsin54)t - \frac{1}{2}(9.8) t^2

now from the two equations we will have

3.40 = (vsin54)(\frac{14}{vcos54}) - 4.9 t^2

3.40 = 14 tan54 - 4.9 t^2

3.40 = 19.3 - 4.9 t^2

t = 1.8 s

now from above equations

14 = vcos54 (1.8)

v = 13.2 m/s

So the minimum speed will be 13.2 m/s

Part B)

Total time of the motion after which it will land on the ground will be "t"

so its vertical displacement will be

\Delta y = -1.60 m

now we will have

-1.60 = v_y t + \frac{1}{2}at^2

-1.60 = (13.2sin54)t - \frac{1}{2}(9.8)t^2

4.9 t^2 - 10.7t - 1.60 = 0

t = 2.3 s

Now the time after which it will reach the fence will be t1 = 1.8 s

so total time after which it will fall on other side of fence

t_2 = t - t_1

t_2 = 2.3 - 1.8 = 0.5 s

now the displacement on the other side is given as

\Delta x = (vcos54) t_2

\Delta x = (13.2 cos54)(0.5)

\Delta x = 3.88 m

4 0
2 years ago
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