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elena55 [62]
3 years ago
13

A uniformly charged thin ring has radius 15.0 cm and total charge 20.0 nC . An electron is placed on the ring's axis a distance

30.0 cm from the center of the ring and is constrained to stay on the axis of the ring. The electron is then released from rest.
Required:
a. Describe the subsequent motion of the electron.
b. Find the speed of the electron when it reaches the center of the ring.
Physics
1 answer:
Ivanshal [37]3 years ago
8 0

Answer:

b) 1.67×10^7 m/s

Explanation:

The solution is attached in the attachment section

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A plumber is going to put two pipes in a wall, one in front and one in back. The pipes will be touching once they are installed.
beks73 [17]

Answer:

They become the same exact tempature

Explanation:

Since they got connected it should be the same.

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3 years ago
As light travels from air to water to glass, it will refract. The best explanation for this would be
Lelu [443]
D) The speed of a wave slows as it travels at different speed in different media.
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A muon has a rest mass energy of 105.7 MeV, and it decays into an electron and a massless particle. If all the lost mass is conv
sergeinik [125]

Answer:

The electron’s velocity is 0.9999 c m/s.

Explanation:

Given that,

Rest mass energy of muon = 105.7 MeV

We know the rest mass of electron = 0.511 Mev

We need to calculate the value of γ

Using formula of energy

K_{rel}=(\gamma-1)mc^2

\dfrac{K_{rel}}{mc^2}=\gamma-1

Put the value into the formula

\gamma=\dfrac{105.7}{0.511}+1

\gamma=208

We need to calculate the electron’s velocity

Using formula of velocity

\gamma=\dfrac{1}{\sqrt{1-(\dfrac{v}{c})^2}}

\gamma^2=\dfrac{1}{1-\dfrac{v^2}{c^2}}

\gamma^2-\gamma^2\times\dfrac{v^2}{c^2}=1

v^2=\dfrac{1-\gamma^2}{-\gamma^2}\times c^2

Put the value into the formula

v^2=\dfrac{1-(208)^2}{-208^2}\times c^2

v=c\sqrt{\dfrac{1-(208)^2}{-208^2}}

v=0.9999 c\ m/s

Hence, The electron’s velocity is 0.9999 c m/s.

6 0
3 years ago
A pitcher exerts a force on a baseball that is 30 times the balls weight. How fast is the pitcher accelerating the ball?
iVinArrow [24]

Answer:

 Pitcher is accelerating the ball at 30 times of acceleration due to gravity = 294 m/s²

Explanation:

  Force applied on baseball = 30 times weight of the ball.

   Weight of ball = mg, where m is the mass of ball and g is acceleration due to gravity value.

  We have force applied is also equal to product of mass and acceleration.

                            F = ma = 30 x mg

                                 a = 30g

   So, pitcher is accelerating the ball at 30 times of acceleration due to gravity = 294 m/s²

8 0
3 years ago
In the parallel spring system, the springs are positioned so that the 37 N weight stretches each spring equally. The spring cons
Reptile [31]

Answer:

x = 5.29 m

Explanation:

given,

weight of stretch = 37 N

left-hand spring constant (k₁)= 2.7 N/cm

right hand spring constant(k₂)= 4.3 N/ cm

spring are connected in parallel

F = F₁ + F₂

F = k₁x + k₂x

F = (k₁+ k₂)x

37= (4.3+ 2.7)x

7 x = 37

x = 5.29 m

3 0
3 years ago
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