All categories

3 years ago

Explain how gravity and friction work to slow down a kicked soccer ball.

Physics

1 answer:

goblinko [34]3 years ago

5 0

Use your feet to stop it since it is soccer you can't use your hands!!!! P.S. you can't use gravity.

You might be interested in

A certain nuclear power plant is capable of producing 1.2×10^9 W of electric power. During operation of the reactor, mass is con

alukav5142 [94]

Answer:

0.00016 kg

Explanation:

Given:

Power = P = 1.2 × 10⁹ Watts

Power = work done / Time

efficiency = 0.30

Input power = 1.2 × 10⁹ / 0.30 = 4 × 10⁹ W

Energy = 4 × 10⁹ x 60 x 60 = 1.44 x 10¹³ joules

E = m c² , where c is the speed of light and m is the mass.

⇒ mass = m = E / c² = (1.44 x 10¹³) / (3 × 10⁸ )²

= 0.00016 kg

6 0

3 years ago

A rock is projected upward from the surface of the moon, at time t = 0.0 s, w a velocity of 30 m/s. The acceleration due to grav

Vinvika [58]

<h2>Answer: 277.777 m</h2>

Explanation:

The situation described here is parabolic movement. However, as we are told that the rock was<u> projected upward from the surface</u>, we will only use the equations related to the Y axis.

In this sense, the movement equations in the Y axis are:

$y-y_{o}=V_{o}.t+\frac{1}{2}g.t^{2}$ (1)

$V=V_{o}-g.t$ (2)

Where:

$y$ is the rock's final position

$y_{o}=0$ is the rock's initial position

$V_{o}=30\frac{m}{s}$ is the rock's initial velocity

$V$ is the final velocity

$t$ is the time the parabolic movement lasts

$g=1.62\frac{m}{s^{2}}$ is the acceleration due to gravity at the surface of the moon

As we know $y_{o}=0$ , equation (2) is rewritten as:

$y=V_{o}.t+\frac{1}{2}g.t^{2}$ (3)

On the other hand, the maximum height is accomplished when $V=0$ :

$V=V_{o}-g.t=0$ (4)

$V_{o}-g.t=0$

$V_{o}=g.t$ (5)

Finding $t$ :

$t=\frac{V_{o}}{g}$ (6)

Substituting (6) in (3):

$y=V_{o}(\frac{V_{o}}{g})+\frac{1}{2}g(\frac{V_{o}}{g})^{2}$ (7)

$y_{max}=\frac{{V_{o}}^{2}}{2g}$ (8) Now we can calculate the maximum height of the rock

$y_{max}=\frac{{(30m/s)}^{2}}{(2)(1.62m/s^{2})}$ (9)

Finally:

$y_{max}=277.777m$

4 0

3 years ago

In a lab, a block weighing 80 N is attached to a spring scale, and both are pulled to the right on a horizontal surface. Frictio

notka56 [123]

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.

There should be an image that should accompanied your question, I was able to chcek it from other sources. the acceleration of the block when the scale reads 32N is <span>4.0 m/s*s</span>

5 0

2 years ago

Read 2 more answers

1.10) For a fixed mass of gas at constant temperature, if the volume of the gas (V) increases to twice its original amount, the

Aliun [14]

B. 1/2p
From the relation
P1V1=P2V2
P2=P1V1/V2
If everything equals one
P2=1*1/2(1)
That would equal 1/2
I hope this helps
Please mark as brainliest

7 0

3 years ago

Read 2 more answers

A machine part has the shape of a solid uniform sphere of mass 250 g and a diameter of 4.30 cm. It is spinning about a frictionl

zysi [14]

Answer: $\alpha =9.302\ rad/s^2$