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rusak2 [61]
3 years ago
11

Which is an example of gaining a static charge by conduction?

Physics
1 answer:
lianna [129]3 years ago
7 0

Answer:

shuffling shoes

Explanation:

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The kinetic theory states that the higher the temperature, the faster the
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Answer: the higher the kinetic energy

Explanation:

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A 1.2 kg basketball is thrown upwards. What is the potential energy of the basketball at the top of its path if it reaches a hei
Dahasolnce [82]

Answer:

Answer is 183.6 J

Explanation:

Using the Physics reference sheet the formula for Potential energy is

(mass) x (gravity) x (height)

Mass= 1.2

Gravity I used is 9.81 (use 10 to get the answer most schools use)

Height= 15.6

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Which is one<br> step that geologists use to find the epicenter of an earthquake?
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triangulation

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using your compass

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Read 2 more answers
URGENT!!! DUE AT 11:59, PLEASE DON’T POST A LINK FOR THIS ANSWER AND BE STRAIGHT FORWARD! Bronco the skydiver, whose mass is 100
kirill [66]

a = 7.8 m/s^2

Explanation:

Let Fnet = net force = ma

m = mass of the skydiver

a = acceleration caused by Fnet

W = weight = mg

f(air) = frictional force due to air resistance

Fnet = W - f(air)

= (100 kg)(9.8 m/s^2) - (200 N)

= 780 N

Therefore, the acceleration of the skydiver due to Fnet is

a = Fnet/m

= (780 N)/(100 kg)

= 7.8 m/s^2

4 0
3 years ago
An electron enters a region of uniform electric field with an initial velocity of 64 km/s in the same direction as the electric
8090 [49]

Answer:

1.) 11 km/s

2.) 9.03 × 10^-5 metres

Explanation:

Given that an electron enters a region of uniform electric field with an initial velocity of 64 km/s in the same direction as the electric field, which has magnitude E = 48 N/C.

Electron q = 1.6×10^-19 C

Electron mass = 9.11×10^-31 Kg

(a) What is the speed of the electron 1.3 ns after entering this region?

E = F/q

F = Eq

Ma = Eq

M × V/t = Eq

Substitute all the parameters into the formula

9.11×10^-31 × V/1.3×10^-9 = 48 × 1.6×10^-19

V = 7.68×10^-18 /7.0×10^-22

V = 10971.43 m/s

V = 11 Km/s approximately

(b) How far does the electron travel during the 1.3 ns interval?

The initial velocity U = 64 km/s

S = ut + 1/2at^2

S = 64000×1.3×10^-6 + 1/2 × 8.4×10^12 × ( 1.3×10^-9)^2

S =8.32×10^-5 + 7.13×10^-6

S = 9.03 × 10^-5 metres

3 0
3 years ago
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