All categories

3 years ago

A 0.1 kg beach ball hit me moving forward at 4 m/s. After hitting me, it bounced back moving at the same speed

Physics

1 answer:

xenn [34]3 years ago

7 0

0.4 N-s is the "impulse" acted on the "beach ball".

Option: C

Explanation:

Given that,

Mass of the "beach ball" is 0.1 kg.

The speed of the ball hits is 4 m/s.

We know that,

Whenever an object is collide with other object then an impulse is acted on object, this "impulse" causes "change in momentum".

Impulse acted on the beach ball is "mass" times "velocity".

Impulse = mass × velocity

Impulse = 0.1 × 4

Impulse = 0.4 kg m/s

Impulse = 0.4 N-s

Therefore, the "impulse" acted on the ball is 0.4 N-s.

You might be interested in

Hey, earn points here

navik [9.2K]

Answer:yayyyy

Explanation:

4 0

3 years ago

Read 2 more answers

Question 13 (1 point)

KonstantinChe [14]

Answer:

Reflective clothing. So vehicles can see them and stuff.

3 0

2 years ago

What is the relative size and composition of the universe, a galaxy, and a solar system? Is the universe endless?

vichka [17]

Yas, the universe is endless/infinte

7 0

3 years ago

Read 2 more answers

The majority of the sun's energy that reaches Earth is __________. View Available Hint(s) The majority of the sun's energy that

Lera25 [3.4K]

Answer:

not captured for use by living things

Explanation:

The sun is a star with a high radiation and most of(around 55-67%) the radiation that reaches the earth is usually reflected back. The earth cannot use all the energy that reaches the earth. It uses almost around 15-30%

3 0

3 years ago

A 3.0 m tall, 40 cm diameter concrete column supports a 235,000 kg load. by how much is the column compressed? assume young's mo

statuscvo [17]

The Young modulus E is given by:
$E= \frac{F L_0}{A \Delta L}$
where
F is the force applied
A is the cross-sectional area perpendicular to the force applied
$L_0$ is the initial length of the object
$\Delta L$ is the increase (or decrease) in length of the object.

In our problem, $L_0 = 3.0 m$ is the initial length of the column, $E=3.0 \cdot 10^{10}N/m^2$ is the Young modulus. We can find the cross-sectional area by using the diameter of the column. In fact, its radius is:
$r= \frac{d}{2}= \frac{40 cm}{2}=20 cm=0.2 m$
and the cross-sectional area is
$A=\pi r^2 = \pi (0.20 m)^2=0.126 m^2$
The force applied to the column is the weight of the load:
$W=mg=(235000 kg)(9.81 m/s^2)=2.305 \cdot 10^6 N$

Now we have everything to calculate the compression of the column:
$\Delta L = \frac{F L_0}{EA}= \frac{(2.305\cdot 10^6 N)(3.0 m)}{(3.0\cdot 10^{10}N/m^2)(0.126 m^2)} =1.83\cdot 10^{-3}m$
So, the column compresses by 1.83 millimeters.

3 0

3 years ago

Read 2 more answers