When forklift moves a crate, it performs ______

A 0.140-kg baseball is dropped from rest from a height of 1.8 m above the ground. It rebounds to a height of 1.4 m. What change

aliina [53]

Answer:

$\Delta p=-1.56\ kg-m/s$

Explanation:

It is given that,

Mass of the baseball, m = 0.14 kg

It is dropped form a height of 1.8 m above the ground. Let u is the velocity when it hits the ground. Using the conservation of energy as :

$u=\sqrt{2gh}$

h = 1.8 m

$u=\sqrt{2\times 9.8\times 1.8}$

u = 5.93 m/s

Let v is the speed of the ball when it rebounds. Again using the conservation of energy to find it :

$v=\sqrt{2gh'}$

h' = 1.4 m

$v=-\sqrt{2\times 9.8\times 1.4}$

v = -5.23 m/s

The change in the momentum of the ball is given by :

$\Delta p=m(v-u)$

$\Delta p=0.14(-5.23-5.93)$

$\Delta p=-1.56\ kg-m/s$

So, the change in the ball's momentum occurs when the ball hits the ground is 1.56 kg-m/s. Hence, this is the required solution.

4 0

3 years ago

Question 26 (1 Point)

LUCKY_DIMON [66]

The book continue to move due to its inertia

Explanation:

We can answer this question by using Newton's first law, which states that:

"When an object is moving with constant velocity (or it is at rest), it will continue moving with constant velocity (or will stay at rest) unless acted upon an external unbalanced force".

This law is also known as law of inertia.

If we apply this law to the situation in the problem, we notice that:

- Before the car stops, the book is moving with constant velocity (25 mph) together with the car

- When the car stops, there are no external forces acting on the book, which is free to continue its motion: so, the book will continue moving at 25 mph forward, due to its inertia. The book will be eventually stopped when it hits the floor (because the floor applies an unbalanced force on it).

Learn more about inertia:

brainly.com/question/2286502

brainly.com/question/691705

#LearnwithBrainly

4 0

3 years ago

Read 2 more answers

Can you answer this for me please

lina2011 [118]

I think it might be independant variable.. but not so sure

8 0

3 years ago

A block of mass 0.252 kg is placed on top of a light, vertical spring of force constant 4 825 N/m and pushed downward so that th

svp [43]

Answer:

10.77m

Explanation:

The elastic potential energy of the spring when compressed with the mass is converted to the kinetic energy of the mass when released. This ie expressed in the following equation;

$\frac{1}{2}ke^2=\frac{1}{2}mu^2..............(1)$

where k is the force constant of the spring, e is the compressed length, m is the mass of the block and u is the velocity with which the block leaves the spring after being released.

If we make u the subject of formula from equation (1) we obtain the following;

$u=\sqrt{\frac{ke^2}{m}}................(2)$

Given;

e = 0.105m,

k = 4825N/m,

m = 0.252kg,

u = ?

Substituting all values into equation (2) we obtain the following;

$u=\sqrt{\frac{4825*0.105^2}{0.252}}................(2)\\u=14.53m/s$

The maximum height attained is then obtained from the third equation of motion as follows, taking g as $9.8m/s^2$

$v^2=u^2-2gh.........(3)$

v = 0m/s

Hence

$h=\frac{u^2}{2g}\\h=\frac{14.53^2}{2*9.8}\\h= 10.77m$

7 0

3 years ago

A 50.0-kg child stands at the rim of a merry-go-round of radius 2.00 m, rotating in such a way that it makes one revolution in 2

Vaselesa [24]

Answer:

2.01

Explanation:

First, we need to find the centripetal acceleration.

We're given that the merry go round rotates 1 revolution in 2.09 seconds. Converting to rpm, we know that it rotates 30 revolution per minute

Now this speed gotten in rpm will be converted to m/s, to ease the calculation

30 rpm = πdN/60 m/s

30 rpm = (3.142 * 4 * 30)/60

30 rpm = 377.04/60

30 rpm = 6.284 m/s

a(c) = v²/r

a(c) = 6.284²/2

a(c) = 39.49 / 2

a(c) = 19.74 m/s²

F = ma

F = 50 * 19.74

F = 987 N

Also, Normal Force, F(n) =

F(n) = mg

F(n) = 50 * 9.81

F(n) = 490.5

We then use this to find the coefficient of static friction, μ

μ = F/F(n)

μ = 987 / 490.5

μ = 2.01

7 0

3 years ago

When forklift moves a crate, it performs _______.