# 5 Physic Question I don't understand! Please help!1. The gravitational force between two objects 2.5 x 10^--1 m apart is 5.6 x10^-6 N. If the mass of one object is 2.97 x 10^-1 kg, what is the mass...

5 Physic Question I don't understand! Please help!

1. The gravitational force between two objects 2.5 x 10^--1 m apart is 5.6 x

10^-6 N. If the mass of one object is 2.97 x 10^-1 kg, what is the mass of

the other object?

2. A 7.3 kg object is accelerated at a rate of 2.5 m/s2 by a net force.

What is the magnitude of the net force acting on the object?

3. A 1.38 x 10^3 kg car accelerates uniformly from 2.0 m/s to 17 m/s in

the same direction. During this acceleration the car travels 112 m.

What is the net force acting on the car during this acceleration?

4. A net force of 3.7 x 10^3 N acts on an object for 3.95 s. During this

time an object accelerates from rest to a velocity of 40 km/h. What is

the mass of this object?

5. A 4.18 kg shopping cart is accelerated by a net force of 17.70 N.

What is the acceleration of the object?

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### 1 Answer

1) You need to remember the formula of Newton's gravitational force between two objects such that:

F =G*(m_1*m_2)/(r^2)

G expresses the gravitational constant, G = 6.673

m_1,m_2, express the masses of the objects

r expresses the distance between objects

5.6*10^(-6) = 6.673*(0.297*m_2)/((2.5*10^(-1))^2)

m_2 = (5.6*6.25*10^(-8))/(6.673*0.297)

m_2 = 17.659*10^(-8) Kg

**Hence, evaluating the mass of the second object yields m_2 = 17.659*10^(-8) Kg.**

2)You need to use the second Newton's law of motion to find the net force acting on the object such that:

F = m*a

F = 7.3*2.5 = 18.25 N

**Hence, evaluating the the net force acting on the object yields F = 18.25 N.**

3) You need to remember that derivative of distance expresses the instantaneous velocity.

d = v*t =>t = d/v average = 112/((v_initial+v_final)/2)

t = 112*2/(2+17)

t = 11.789 s

You also need to remember that v_ final = v_initial + a*t such that:

17 = 2 + a*11.789 => 11.789a = 15 => a = 15/11.789

a = 1.272m/s^2

You need to use the second Newton's law such that:

F = m*a

F = 1380*1.272

F = 1755.874N

**Hence, evaluating the the net force acting on the object yields F = 1755.874N.**

4) You need to use the relation v_ final = v_initial + a*t to find acceleration such that:

v_ final = v_initial + a*t

40*100/6 = 0 + a*3.95 s

a = 666/3.95

a = 168.776 m/s^2

You need to use the second Newton's law such that:

F = m*a =>3700 = m*168.776 => m = 3700/168.776

m =21.92 Kg

**Hence, evaluating the mass of object yields m =21.92 Kg.**

5) You need to use the second Newton's law such that:

F = m*a => a = F/m => a = 17.70/4.18 =4.23 m/s^2

**Hence, evaluating the mass of object yields a =4.23 m/s^2**