Ask question

Ask question

All categories

sergij07 [2.7K]

3 years ago

3

3. A 200-g ball falls vertically downward, hitting the floor with a speed of 2.5 m/s and rebounding upward with a speed of 2.0 m

/s. a. Determine the change in momentum of the ball.. b. If the ball is in contact with the floor for 0.02 ms (milliseconds), what is the average force applied to the ball

1 answer:

vovikov84 [41]3 years ago

7 0

Answer:

Explanation:

A 200-g ball falls vertically downward, hitting the floor with a speed of 2.5 m/s and rebounding upward with a speed of 2.0 m/s. a. Determine the change in momentum of the ball.. b. If the ball is in contact with the floor for 0.02 ms (milliseconds), what is the average force applied to the ball

Given data

mass= 200g= 0.2kg

initial velocity= 2.5m/s

final velocity= 2m/s

time= 0.02ms

time= 0.00002 seconds

ΔP= mΔv

ΔP= 0.2*2.5-2

ΔP= 0.2*0.5

ΔP=0.1kgm/s

F= mv/t

F=0.1/0.00002

F=5000N

You might be interested in

How much work must be done in the car to slow it from 100km/h to 50km/h

Brilliant_brown [7]

Answer:

$-96.465m$ joule

Explanation:

Let m = mass of the car and v1 = initial velocity and v2 = final velocity

Given.

Initial velocity = 100 km/h

final velocity = 50 km/h

What is work done in the car to slow it from 100km/h to 50km/h?

$v1=100km/h=27.78m/s$

$v2=50km/h=13.89m/s$

The work done in the car to slow it from v1 to v2.

w=Δk

$w=k2-k1$

$w= \frac{1}{2}m(v2)^{2}- \frac{1}{2}m(v1)^{2}$

$w=\frac{1}{2} m(v2-v1)^{2}$

$w=\frac{1}{2}\times m(13.89 -27.78)^{2}$

$w=\frac{1}{2}\times m(-13.89)^{2}$

$w=\frac{1}{2}\times m\times (-192.93)$

$w=-96.465m$ joule.

Therefore, the work done is $-96.465m$ joule

7 0

3 years ago

A force of attraction would exist between

siniylev [52]

Answer:

A. positively charged object and a negatively charged object.

3 0

4 years ago

A man is flying in a hot-air balloon in a straight line at a constant rate of 5 feet per second, while keeping it at a constant

icang [17]

Answer:

$x = 220.85 ft$

Explanation:

Let at any moment of time the friend's car is at some horizontal distance "x" from the position of balloon.

Now if the altitude of the balloon is fixed and it is at height "h"

so here we will have

$tan \theta = \frac{h}{x}$

now we know that

initially the angle of the friend's car is 35 degree

so the horizontal distance will be

$x_1 = h cot35$

similarly if the angle after passing the car position is 36 degree

then we have

$x_2 = h cot36$

now the speed of the balloon is constant

so we have

$v = \frac{x_1 + x_2}{\Delta t}$

$5 ft/s = \frac{h cot35 + h cot36}{90 s}$

$5 ft/s = \frac{2.8h}{90}$

$h = 160.45 ft$

so the final position of friend when the angle is 36 degree

$x = \frac{h}{tan36}$

$x = \frac{160.45}{tan36}$

$x = 220.85 ft$

4 0

3 years ago

In order to reduce the amount of energy lost due to heat flow, electricity is delivered to our homes using _____.

erica [24]

In order to reduce the amount of energy lost due to heat flow, electricity is delivered to our homes using <u>d. low current</u>.

<u>Explanation</u>:

The presence of flow of electrons is said to be electricity. The electricity flows in one direction. The electricity enables function of the electrical devices and machines.

Heat is generated when the electricity flows from one place to another. So the energy is lost in the form of heat.

So to avoid the loss of energy, the electricity is delivered to home using low current. The strength of an electric current can be measured and represented in the unit of ampere.

4 0

4 years ago

Read 2 more answers

A flywheel with a very low friction bearing takes 1.6 h to stopafter the motor power

Nina [5.8K]

Answer:

(I). The initial rotation rate is 4.29 rad/s.

(II). The revolutions is 3932.

Explanation:

Given that,

Time = 1.6 h

Angular velocity = 41 rpm

(I). We need to calculate the initial rotation rate in rad/s

$\omega=41\ rev/m$

$\omega=\dfrac{41\times 2\pi}{60}\ rad/s$

$\omega=4.29\ rad/s$

(II). We need to calculate the revolutions

Using formula of revolutions

$\theta=\omega t$

$\theta=4.29\times1.6\times3600$

$\theta=24710\ rad$

$\theta=\dfrac{24710}{2\pi}$

$\theta=3932\ revolution$

Hence, (I). The initial rotation rate is 4.293 rad/s.

(II). The revolutions is 3932.

7 0

4 years ago