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2 years ago

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17. An uncharged metal rod brought close to but not touching the cap of a charged electroscope caused decrease in the divergence

of the leaf. Explain this observation. (1 mk)

1 answer:

Furkat [3]2 years ago

3 0

Explanation:

Metals get charged by induction ,the charges are transferred from the electroscope causing earthing: Charges on the electroscope induce opposite charges on the conductor.

I want brainliest plz!!!

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A baseball leaves a bat with a horizontal velocity of 20 m/s. In a time of 0.25 s, How far will it have traveled horizontally?

Maurinko [17]

Distance traveled by the ball is given by

$distance = speed \times time$

here we know that

speed = 20 m/s

times = 0.25 s

now we have

$distance = 20 \times 0.25$

$distance = 5 m$

so ball will travel 5 m distance in the given interval of time

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3 years ago

Which is harder, air tight or water tight seals?

Vladimir [108]

Should be an air tight seal

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3 years ago

An 8.5 kg crate is pulled 5.1 m up a 30 degree incline by a rope angled 17 degrees above the incline. The tension in the rope is

Ulleksa [173]

Answer:

1. a $W_t=746.63 J$

b $E_p=212.415 J$

c $W_n=183.96J$

2. $T_e=99.71J$

Explanation:

a). The work done by the tension is:

$W_t=T*dt$

$dt=\frac{5.1}{cos(17)}$

$W_t=140N*\frac{5.1}{cos(17)}$

$W_t=746.63 J$

b). The work done potential of gravity

$E_p=m*g*h$

$h=5.1*sin(17)$

$E_p=8.5kh*9.8*5.1*sin(30)$

$E_p=212.415 J$

c). The work done by the normal force

$W_n=N*d_n$

$d_n=5.1*sin(30)=2.55$

$W_n=8.5kg*9.8*cos(30)*2.55$

$W_n=183.96J$

2. The increase in thermal energy is:

$T_e=F*d$

$F_k=u_k*m*g=0.271*8.5kg*9.8*cos(30)$

$F_k=19.5N$

$T_e=19.55*5.1m$

$T_e=99.71J$

4 0

3 years ago

In a video game, a flying coconut moves at a constant velocity of 20 meters/second. The coconut hits an obstacle and moves in th

qwelly [4]

Change in velocity = Final velocity - Initial velocity

Initial velocity = 20 m/s
Final velocity = 10 m/s

Change in velocity = 10 - 20 = -10m/s

Correct option B

Hope This Helps You!

4 0

3 years ago

What must the charge (sign and magnitude) of a particle of mass 1.45 g be for it to remain stationary when placed in a downward-

sergey [27]

$q = -21 * 10^{-6} C$

<h3>What is Free-fall acceleration?</h3>

The acceleration is constant and equal to the gravitational acceleration g which is 9.8 m/s at sea level on the Earth.

As we know that charge and its sign that remains in equilibrium is under gravity must be such that it will balance the gravitational force by electric force,

mg =qE

$1.45 * 10^{-3}(9.80)=q(600)\\\\q = 21 *10^{6} C\\\\$

and its sign must be negative so that it will have upward electric force

so it is

$q = -21 * 10^{-6} C$

The charge of a particle of mass is $-21 * 10^{-6} C$

Learn more about Free-fall acceleration: brainly.com/question/2165771

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8 0

1 year ago