Which two factors affect the size of the gravitational field?

You want to slide a 0.39 kg book across a table. If the coefficient of kinetic friction is .21, what force is required to move t

uranmaximum [27]

Find the force that would be required in the absence of friction first, then calculate the force of friction and add them together. This is done because the friction force is going to have to be compensated for. We will need that much more force than we otherwise would to achieve the desired acceleration:

$F_{NoFric}=ma=0.39kg \times0.18 \frac{m}{s^2} =0.0702N$

The friction force will be given by the normal force times the coefficient of friction. Here the normal force is just its weight, mg

$F_{Fric}=0.39kg \times 9.8 \frac{m}{s^2} \times 0.21=0.803N$

Now the total force required is:

0.0702N+0.803N=0.873N

5 0

3 years ago

What is the power output of a pump which can raise 60kg of water height to height of 10m every minute

sattari [20]

Answer:

<h3>Power = Work Done/time</h3>

=> Power = 60×10×10/60

=> Power = 6000/60

=> Power = 100 Watt

Hence the power output of a pump is 100 Watts.

8 0

2 years ago

if the mass is 10 gram and the volume is 7 cubic centimetre,find the density in kilogram per cubic metre

wel

Answer:

1.42

Explanation:

devide 10 by 7

7 0

3 years ago

3. In 1989, Michel Menin of France walked on a tightrope suspended under a

Tamiku [17]

Answer: 80m

Explanation:

Distance of balloon to the ground is 3150m

Let the distance of Menin's pocket to the ground be x

Let the distance between Menin's pocket to the balloon be y

Hence, x=3150-y------1

Using the equation of motion,

V^2= U^s + 2gs--------2

U= initial speed is 0m/s

g is replaced with a since the acceleration is under gravity (g) and not straight line (a), hence g is taken as 10m/s

40m/s is contant since U (the coin is at rest is 0) hence V =40m/s

Slotting our values into equation 2

40^2= 0^2 + 2 * 10* (3150-y)

1600 = 0 + 63000 - 20y

1600 - 63000 = - 20y

-61400 = - 20y minus cancel out minus on both sides of the equation

61400 = 20y

Hence y = 61400/20

3070m

Hence, recall equation 1

x = 3150 - 3070

80m

I hope this solve the problem.

6 0

3 years ago

1. A runner drops her phone as she is running at a constant speed of 3 miles per hour from point A to point B in a park. Describ

joja [24]

Answer:

Let's define the point A as our zero in the x-axis.

As the phone drops, it keeps the horizontal velocity that it had before, so the horizontal velocity is:

Vx = 3 mi/h.

Now, the only force acting on the phone is the gravitational force that acts in the vertical axis, then we have:

Ay = -g

where g = 9.8 m/s^2

It is dropped, so we do not have a vertical initial velocity, then for the vertical velocity we should integrate over time:

Vy = -g*t

And for the position again, we integrate over time, but now we have an initial position H, that is the height at which the phone is dropped.

Py = -(1/2)*g*t^2 + H

And the horizontal position can be found by integrating over time the horizontal velocity.

Px = (3mi/h)*t

This will be the two equations that describe the motion of the phone, and we can not solve it further because we do not know the initial height of the phone.

But in general, we have a linear equation in the horizontal axis and a quadratic equation with a negative leading coefficient in the vertical axis.

Position(t) = ( (3mi/h)*t, -(1/2)*g*t^2 + H)

5 0

3 years ago