Answer:
The distance separating the two particles when the moving particle is momentarily stopped is 0.947 m
Explanation:
Given that,
First charge = 40 μC
Second charge = 80 μC
Distance between the two particles = 2.0 m
Kinetic energy = 16 J
We need to calculate the distance separating the two particles when the moving particle is momentarily stopped
Using conservation of energy
Put the value into the formula
Hence, The distance separating the two particles when the moving particle is momentarily stopped is 0.947 m
Answer:
(a) ω = 1.57 rad/s
(b) ac = 4.92 m/s²
(c) μs = 0.5
Explanation:
(a)
The angular speed of the merry go-round can be found as follows:
ω = 2πf
where,
ω = angular speed = ?
f = frequency = 0.25 rev/s
Therefore,
ω = (2π)(0.25 rev/s)
<u>ω = 1.57 rad/s
</u>
(b)
The centripetal acceleration can be found as:
ac = v²/R
but,
v = Rω
Therefore,
ac = (Rω)²/R
ac = Rω²
therefore,
ac = (2 m)(1.57 rad/s)²
<u>ac = 4.92 m/s²
</u>
(c)
In order to avoid slipping the centripetal force must not exceed the frictional force between shoes and floor:
Centripetal Force = Frictional Force
m*ac = μs*R = μs*W
m*ac = μs*mg
ac = μs*g
μs = ac/g
μs = (4.92 m/s²)/(9.8 m/s²)
<u>μs = 0.5</u>
Answer:do you know how to draw a electronics configuration
Explanation:
C) studying seismic waves
AKA what is studied from earthquakes
I think 3 is 1.5...i kinda hope that helps for one
