Answer:
The horse father from the center has a greater tangential speed. Although both horses complete one circle in the same time period, the one farther from the center covers a greater distance during that same period.
Explanation:
I'm sure you've noticed that an airplane high in the sky, far away
from you, looks like it's moving very slowly. At the same time,
somebody passing you on a skateboard whizzes past you at
high speed. The farther away something is from you, the slower
it appears to move.
The nearest star outside the solar system is almost 32 thousand times
as far away from us as the farthest visible planet (Saturn) is, and all of the
other stars are farther than that.
That's why you have to wait a few thousand years before you notice
that the shape of a constellation has changed.
To put it a slightly different way . . . Everything is in motion. The motion is
more noticeable for nearby things, and less noticeable for farther-away things.
Objects within our solar system are the only ones near enough so that a human
lifetime is a long enough period in which to notice the change in their position.
Even Pluto moves less then 1.5° against the 'background' stars in a whole year.
This all makes me feel small. How about you ?
The measure of the quantity of matter would be mass. Mass is measured in kilograms. I hope this helped!:)
Answer:
Explanation:
Given:
- Three identical charges q.
- Two charges on x - axis separated by distance a about origin
- One on y-axis
- All three charges are vertices
Find:
- Find an expression for the electric field at points on the y-axis above the uppermost charge.
- Show that the working reduces to point charge when y >> a.
Solution
- Take a variable distance y above the top most charge.
- Then compute the distance from charges on the axis to the variable distance y:
- Then compute the angle that Force makes with the y axis:
cos(Q) = sqrt(3)*a / 2*r
- The net force due to two charges on x-axis, the vertical components from these two charges are same and directed above:
F_1,2 = 2*F_x*cos(Q)
- The total net force would be:
F_net = F_1,2 + kq / y^2
- Hence,
- Now for the limit y >>a:
- Insert limit i.e a/y = 0
Hence the Electric Field is off a point charge of magnitude 3q.
0.36 J of work is done in stretching the spring from 15 cm to 18 cm.
To find the correct answer, we need to know about the work done to strech a string.
<h3>What is the work required to strech a string?</h3>
- Mathematically, the work done to strech a string is given as 1/2 ×K×x².
- K is the spring constant.
<h3>What will be the spring constant, if 40N force is required to hold a 10 cm to 15 cm streched spring?</h3>
- The force experienced by a streched spring is given as Kx. x is the length of the spring streched from its natural length.
- Then K = Force / x.
- Here x = 15 - 10 = 5 cm = 0.05 m
- K = 40/0.05 = 800N/m.
<h3>What will be the work required to strech that spring from 15 cm to 18 cm?</h3>
- Work done = 1/2×k×x²
- Here x= 18-15=3cm or 0.03 m
- So, W= 1/2×800×0.03² = 0.36 J.
Thus, we can conclude that the work done is 0.36 J.
Learn more about the spring force here:
brainly.com/question/14970750
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