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Phantasy [73]
4 years ago
12

The Sun exerts a gravitational force of 29.9 N on a rock that's located in a river bed here on Earth. How much gravitational for

ce does the rock exert on the Sun? State your answer with the correct mks units.
Physics
1 answer:
11Alexandr11 [23.1K]4 years ago
4 0

Answer:

F = 29.9 N

Explanation:

It is given that, The Sun exerts a gravitational force of 29.9 N on a rock that's located in a river bed here on Earth. We need to find the gravitational force the rock exert on the Sun. It is a case of Newton's third law of motion which states that the force acting from one object to another is equal to the force acing from second object to the first object and the two forces must be in opposite direction. Hence, the gravitational force the rock exert on the Sun is same i.e. 29.9 N.

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A box is being pulled across a horizontal surface by a 20 N force to the right.
Oksana_A [137]

Answer:

Im pretty sure it's C

Explanation:

The answer A talks about constant speed, which does not correspond to Net Force. The answer B, talks about constant velocity, but talks about how much force apposes the box. The answer C talks about the value of the net force acting on the box, so im pretty sure the answer is C.

6 0
3 years ago
Read 2 more answers
A 10-kg object is dropped from rest. after falling a distance of 50 m, it has a speed of 26 m/s. what is the change in mechanica
likoan [24]

The change in mechanical energy caused by the dissipative resistance force is equal to, difference between the potential energy and kinetic energy of the object.

Potential energy of the object, P.E = mgh

m is mass of the object = 10 kg

g is acceleration due to gravity = 9.8 m/s²

h= height from which it is dropped =50 m

Substituting the value we get,

P.E = 10×9.8×50 = 4900 J

Kinetic energy of the object, K.E = \frac{1}{2}mv^{2}

v is the velocity of the object = 26 m/s²

K.E = (1/2)×10×(26)²

= 3380 J

Change in mechanical energy caused by dissipative force = P.E ₋ K.E

= 4900 ₋ 3380 = 1520 J

4 0
3 years ago
A truck with a mass of 1330 kg and moving with a speed of 15.0 m/s rear-ends a 805 kg car stopped at an intersection. The collis
Dovator [93]

Answer:

The Speed of the vehicles is 9.34m/s

Explanation:

For an elastic collision the two bodies move with similar velocities after collision

Given

M1=1330kg

V1=15m/s

M2=805kg

V2=0(the car is parked on neutral)

The formula is

M1V1+M2V2=(M1+M2)V

1330*15+805*0=(1330+805)V

19950+0=2135V

2135V=19950

divide both sides by 2135

V=19950/2135

V=9.34m/s

8 0
3 years ago
A player throws a football 50.0 m at 61.0° north of west. what is the westward component of the displacement of the football?
-BARSIC- [3]

Answer: 24.24 m

Explanation:

A player throws football 50.0 m at 61° North of west. we will write this in terms of horizontal and vertical components.

Horizontal component: 50 cos 61° = 24.24 m which is westwards

Vertical component: 50 sin 61° = 43.73 m which towards North.

Refer to diagram below.

Thus, the westward component of displacement of the football is the horizontal component of the displacement = 24.24 m.

7 0
3 years ago
Air at 30°C and 2MPa flows at steady state in a horizontal pipeline with a velocity of 25 m/s. It passes through a throttle valv
erastovalidia [21]

Answer:

V_2=159.9\ m/s

T_2=290.6K

Explanation:

At initial condition

P=2 MPa

T=30°C

V=25 m/s

At final condition

P=0.3 MPa

Now from first law for open system

h_1+\dfrac{V_1^2}{2000}=h_2+\dfrac{V_2^2}{2000}

We know that for air

h= 1.010 x T  KJ/kg

1.01\times 303+\dfrac{25^2}{2000}=1.01\times T+\dfrac{V_2^2}{2000}                               -----1

Now from mass balance

\rho_1A_1V_1=\rho_2A_2V_2

We also know that

\rho=\dfrac{P}{RT}

\dfrac{P_1}{RT_1}V_1=\dfrac{P_2}{RT_2}V_2

\dfrac{2}{R\times 303}\times 25=\dfrac{0.3}{RT_2}V_2

T_2=1.81V_2                  ----------2                                                                                                

Now from equation 1 and 2

V_2^2+3673.749V-612916.49=0

So we can say that

V_2=159.9\ m/s

This is the outlet velocity.

Now by putting the values in equation 2

T_2=1.81\times 159.9  

T_2=290.6K

This is the outlet temperature.

4 0
4 years ago
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