Use the data provided to calculate the gravitational

A 3kg book falls from a 2m tall bookshelf what is the speed <br><br> PICTURE included

mafiozo [28]

Answer:

39.2m/s

Explanation:

The potential energy the book has right before it falls is equal to the kinetic energy in falling.

PE = KE

mgh = (1/2)mv

2gh=v

v=(2)(9.81)(2)

v=39.24m/s

8 0

3 years ago

An object with a charge of -3.2 uC and a mass of 1.0Ã—10^(-2) kg experiences an upward electric force, due to a uniform electric

USPshnik [31]

Answer:

The magnitude of the acceleration is equal to 19.6m/s² and the acceleration is directed upwards though the magnitude of the charge has doubled. This is because the electric force is directed upwards and from newton's second law of motion the charge will have acceleration in the same direction as the electric force on the charge.

Explanation:

The detailed solution can be found in the attachment below.

Thank you for reading and I hope this is helpful to you.

6 0

3 years ago

A ball is thrown nearly vertically upward from a point near the cornice of a tall building. It just misses the cornice on the wa

vovangra [49]

Answer:

a) 48.5 ft/s

b) 36.5 ft

c) -80.3 ft/s

Explanation:

a)

The equation of motion of the ball is :

y(t) = -16.1 ft/s^2 * t^2 + Vo*t

Where Vo is the initial velocity

If y(5s) = - 160 ft:

-160 ft = -16.1 ft/s^2 * (5 s)^2 + Vo*(5s)

Solving for Vo

Vo = (16.1*25- 160) ft / 5s = 48.5 ft/s

b)

To answer this question we must first know when the velocity became zero, at this time is when the ball was at its highest point.

v(t) = -32.2 ft/s^2 * t + Vo

t = Vo/32.2ft/s^2 = 1.5 s

And now, the highest point which the ball reached is given by:

y(1.5s) = -16.1 ft/s^2 * (1.5)^2 + Vo*(1.5s)

y(1.5s) = 36.52 ft

c)

We now need the time at which y(t') = -64 ft

-64 = -16.1*t'^2 + 48.5*t'

By means of the quadratic formula, we find that

t' = 4.00498 s ≈ 4 s

And the velocity at t = 4s is:

v(4s) = -32.2 ft/s^2 * 4s +48.5 ft/s = -80.3 ft/s

3 0

3 years ago

Space scientists have a large test chamber from which all the air can be evacuated and in which they can create a horizontal uni

nexus9112 [7]

Answer:

the magnitude of the electric force on the projectile is 0.0335N

Explanation:

time of flight t = 2·V·sinθ/g

= (2 * 6.0m/s * sin35º) / 9.8m/s²

= 0.702 s

The body travels for this much time and cover horizontal displacement x from the point of lunch

So, use kinematic equation for horizontal motion

horizontal displacement

x = Vcosθ*t + ½at²

2.9 m = 6.0m/s * cos35º * 0.702s + ½a * (0.702s)²

a = -2.23 m/s²

This is the horizontal acceleration of the object.

Since the object is subject to only electric force in horizontal direction, this acceleration is due to electric force only

Therefore,the magnitude of the electric force on the projectile will be

F = m*|a|

= 0.015kg * 2.23m/s²

= 0.0335 N

Thus, the magnitude of the electric force on the projectile is 0.0335N

3 0

3 years ago

Read 2 more answers

Which two options would INCREASE the electric force between two charged particles?

atroni [7]

The electric force between two charged particles can be increased by decreasing the distance between the two particles.

<h3>How to increase electric force between two charged particles.</h3>

The technique of decreasing the separation distance between objects increases the force of attraction or repulsion between the objects. while

increasing the separation distance between objects decreases the force of attraction or repulsion between the objects.