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kvv77 [185]
2 years ago
13

Ways that charges can be transferred

Physics
2 answers:
Rama09 [41]2 years ago
8 0
Can I have more Details?
sashaice [31]2 years ago
3 0
Friction, Induction, and Conduction
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A force of 70 N is applied to a 28 kg rock causing it to slow down from 25 m/s to 15 m/s, a change in velocity of 10 m/s. How lo
maksim [4K]

Answer: 4 s

Explanation:

Given

The applied force is 70 N

mass of the rock is 28 kg

initial velocity u=25\ m/s

final velocity v=15\ m/s

Deceleration provided by force is

a=-\dfrac{70}{28}=-2.5\ m/s^2

using the equation of motion

v=u+at\\\Rightarrow 15=25-2.5t\\\Rightarrow 2.5t=10\\\Rightarrow t=4\ s

7 0
3 years ago
What is the great red spot? what is the great red spot? a long-lived, high-pressure storm on jupiter a place where reddish parti
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<span><u><em>A</em></u><em><u></u></em><em><u></u></em><u><em> long-lived, high-pressure storm on jupiter a place where reddish particles from io impact jupiter's surface</em></u></span><u><em> </em></u>
5 0
2 years ago
I WILL GIVE BRAINLIEST
Vaselesa [24]
<h2><u>Answer:</u></h2><h2>A. Displacement</h2><h2>B. Time</h2><h2 /><h2>Your Welcome ❤</h2>
3 0
2 years ago
Read 2 more answers
A ball is thrown straight up from the ground with speed v0. At the same instant, a second ball is dropped from rest from a heigh
dlinn [17]

Answer:

a) t = H/v0

b) H = -(v0)²/g

Explanation:

Hi there!

a)The position of the balls can be calculated using the following equation:

y = y0 + v0 · t + 1/2 · g · t²

Where:

y = height of the ball at time t.

y0 = initial height.

v0 = initial velocity.

g = acceleration due to gravity.

t = time.

For the ball that is thrown upwards, the initial height is zero, then, the equation can be written as follows:

y = v0 · t + 1/2 · g · t²

The second ball is initially at a height H and the initial velocity is zero. The equation of height for the second ball will be:

y = H + 1/2 · g · t²

When the two balls collide, their height is the same. Then, equalizing both equations we can obtain the time at which they collide:

v0 · t + 1/2 · g · t² = H + 1/2 · g · t²

v0 · t = H

t = H/v0

b) When the first ball is at the highest point its velocity is zero. Using the equation of velocity we can find the time at which the ball is at that point. The equation of velocity is the following:

v = v0 + g · t

At the highest point v = 0.

0 = v0 + g · t

Solving for t:

-v0/g = t

The time at which the first ball is at the highest point is t = -v0/g

The time at which both balls collide was calculated above:

t = H/v0

Then, equalizing both times and solving for H:

H/v0 = -v0/g

H = -v0/g · v0

H = -(v0)²/g

3 0
3 years ago
If the distance d (in meters) traveled by an object in time t (in seconds) is given by the formula d=a+bt2, the si units of a an
ss7ja [257]
Distance , d = a+bt^2

The unit of d is in meter and t is in seconds.

So the unit of a a must be meter.

Now we have unit of bt^2 is meter.

So unit of b*second^2 = meter

Unit of b = meter/second^2

So unit of a = m and unit of b = m/s^2.
8 0
2 years ago
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