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AysviL [449]
3 years ago
9

Hi! Plz help? Which term identifies the effects of the buildup of stationary electric charges on everyday objects caused by the

transfer of electrons from one object to another? A. stationary electricity B. current electricity C. magnetic electricity D. static electricity
Physics
1 answer:
erma4kov [3.2K]3 years ago
8 0
<span>The question is, which term identifies the effects of build up of stationary electric charges on every day objects caused by the transfer of electrons from one object to another. The correct option is D. Static energy is defined as an imbalance of positive and negative charges within or on the surface of an object. The charge usually remain on the object until it is moved away by discharge. </span>
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How far away is Earth away from the sun?
alexira [117]
It is 92.96 millions miles away

Hope that helped :)
3 0
3 years ago
A 52.0-kg person, running horizontally with a velocity of +3.63 m/s, jumps onto a 15.2-kg sled that is initially at rest. (a) Ig
trasher [3.6K]

Answer:

The coefficient of kinetic friction between the sled and the snow is 0.0134

Explanation:

Given that:

M = mass of person = 52 kg

m = mass of sled = 15.2 kg

U = initial velocity of person = 3.63 m/s

u = initial velocity of sled = 0 m/s

After collision, the person and the sled would move with the same velocity V.

a) According to law of momentum conservation:

Total momentum before collision = Total momentum after collision

MU + mu = (M + m)V

V=\frac{MU+mu}{M+m}

Substituting values:

V=\frac{MU+mu}{M+m}=\frac{52(3.63)+15.2(0)}{52+15.2} =2.81m/s

The velocity of the sled and person as they move away is 2.81 m/s

b) acceleration due to gravity (g) = 9.8 m/s²

d = 30 m

Using the formula:

V^2=2\mu(gd)\\\mu=\frac{V^2}{2gd} \\\mu=\frac{2.81^2}{2*9.8*30} =0.0134

The coefficient of kinetic friction between the sled and the snow is 0.0134

3 0
3 years ago
A student at a window on the second floor of a dorm sees her physics professor walking on the sidewalk beside the building. she
klasskru [66]
Refer to the diagram shown below.

In order for the balloon to strike the professor's head, th balloon should drop by 18 - 1.7 = 16.3 m in the time at the professor takes to walk 1 m.
The time for the professor to walk 1 m is
t = (1 m)/(0.45 m/s) = 2.2222 s

The initial vertical velocity of the balloon is zero.
The vertical drop of the balloon in 2.2222 s is
h = (1/2)*(9.8 m/s²)*(2.2222 s)² = 24.197 m

Because 24.97 > 16.3, the balloon lands in front of the professor, and does not hit the professor.

The time for the balloon to hit the ground is
(1/2)*(9.8)*t² = 18
t = 1.9166 s

The time difference is 2.2222 - 1.9166 = 0.3056 s
Within this time interval, the professor travels 0.45*0.3056 = 0.175 m
Therefore the balloon falls 0.175 m in front of the professor.

Answer: 
The balloon misses the professor, and falls 0.175 m in front of the professor.

8 0
3 years ago
Which of the following describes an action-reaction pair?
Vlad [161]

Answer:

Explanation: LETTER A

8 0
1 year ago
Read 2 more answers
A cyclotron is to accelerate protons to an energy of 5.4 MeV. The superconduction electromagnet of the cyclotron produces a 2.9T
mart [117]
<h3><u>Answer;</u></h3>

Radius = 0.0818 m

Angular velocity = 2.775 × 10^7 rad/sec

<h3><u>Explanation;</u></h3>

The mass of proton m=1.6748 × 10^-27 kg;  

Charge of electron e= 1.602 × 10^-19 C;  

kinetic energy E= 2.7 MeV

                          = 2.7 × 10^6 × 1.602 × 10^-19 J;

                          = 4.32 × 10^-13 Joules

But; K.E =0.5m*v^2,

Hence v=√(2K.E/m)

Velocity = 2.27 × 10^7 m/s

Angular velocity, ω = v/r

Therefore; V = ωr

Hence; V = √(2K.E/m) = ωr

r= √(2E/m)/w = √E*√(2*m)/(eB)

  = √E * √(2×1.6748×10^-27)/(1.602×10^-19 ×2.9)

but E =  4.32 × 10^-13 Joules

  r = 0.0818 m

Angular speed

Angular velocity, ω = v/r , where r is the radius and v is the velocity

Therefore;

Angular velocity = 2.27 × 10^7 / 0.0818 m

                            = 2.775 × 10^7 rad /sec

3 0
2 years ago
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