All categories

3 years ago

Gold-leaf electroscope uses

1 answer:

7 0

-identifies an electric charge
-it can identify its polarity (positive or negative) if you compare it to a charge that you already know
-can identify the magnitude of a charge (how big of a charge it is)

You might be interested in

Can someone help me please

TEA [102]

Answer:

Explanation:

Because it is impossible for it to show the real depth of the ocean and how deep it is

8 0

3 years ago

Convert the speed of light, 3.0 x 108 m/s, to km/s.

zhuklara [117]

Answer:

I don't know

Explanation:

I don't know

Explanation:

Answer:

I don't know

Explanation:

I don't know

3 0

3 years ago

What is the condition for an object experiencing free fall?

S_A_V [24]

Answer:

The body acts under the influence of gravity.

Explanation:

An object experiencing free fall is acting under the influence of gravitational force and the acceleration due gravity is positive for any falling object. The body is able to fall freely due to the effect of gravity on it. This gravity effect causes the body to get attracted to the earth's gravitational surface due to gravitational pull exerted on the body.

3 0

3 years ago

A baseball, which has a mass of 0.685 kg., is moving with a velocity of 38.0 m/s when it contacts the baseball bat duringwhich t

Evgen [1.6K]

Answers:

a) $65.075 kgm/s$

b) $10.526 s$

c) $61.82 N$

Explanation:

<h3>a) Impulse delivered to the ball</h3>

According to the Impulse-Momentum theorem we have the following:

$I=\Delta p=p_{2}-p_{1}$ (1)

Where:

$I$ is the impulse

$\Delta p$ is the change in momentum

$p_{2}=mV_{2}$ is the final momentum of the ball with mass $m=0.685 kg$ and final velocity (to the right) $V_{2}=57 m/s$

$p_{1}=mV_{1}$ is the initial momentum of the ball with initial velocity (to the left) $V_{1}=-38 m/s$

So:

$I=\Delta p=mV_{2}-mV_{1}$ (2)

$I=\Delta p=m(V_{2}-V_{1})$ (3)

$I=\Delta p=0.685 kg (57 m/s-(-38 m/s))$ (4)

$I=\Delta p=65.075 kg m/s$ (5)

This time can be calculated by the following equations, taking into account the ball undergoes a maximum compression of approximately $1.0 cm=0.01 m$ :