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JulijaS [17]
2 years ago
7

) Explain why the foil is attracted at first by the charged rod. Consider any charge that exists in the uncharged foil. (You wil

l consider later interaction between the rod and foil in the last question. Just think about the first interaction when answering this question.) [
Physics
1 answer:
Arisa [49]2 years ago
5 0

Answer:

The attraction is due to the induced charge.

Explanation:

When we approach a charged rod to a sheet, an induced load is produced in the sheet that is of the same magnitude as the rod of opposite sign, this is because the charges of different sign attract each other, this explains the initial attraction.

This induced load occurs if importing the plate load

The attraction is due to the induced charge.

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In a lab experiment, a student is trying to apply the conservation of momentum. Two identical balls, each with a mass of 1.0 kg,
Studentka2010 [4]

Answer:

Second Trial satisfy principle of conservation of momentum

Explanation:

Given mass of ball A and ball B =\ 1.0\ Kg.

Let mass of ball A and B\ is\ m  

Final velocity of ball A\ is\ v_1

Final velocity of ball B\ is\ v_2

initial velocity of ball A\ is\ u_1

Initial velocity of ball B\ is\ u_2

Momentum after collision =mv_1+mv_2

Momentum before collision = mu_1+mu_2

Conservation of momentum in a closed system states that, moment before collision should be equal to moment after collision.

Now, mu_1+mu_2=mv_1+mv_2

Plugging each trial in this equation we get,

First Trial

mu_1+mu_2=mv_1+mv_2\\1(1)+1(-2)=1(-2)+1(-1)\\1-2=-2-1\\-1=-3

momentum before collision \neq moment after collision

Second Trial

mu_1+mu_2=mv_1+mv_2\\1(.5)+1(-1.5)=1(-.5)+1(-.5)\\.5-1.5=-.5-.5\\-1=-1

moment before collision = moment after collision

Third Trial

mu_1+mu_2=mv_1+mv_2\\1(2)+1(1)=1(1)+1(-2)\\2+1=1-2\\3=-1

momentum before collision \neq moment after collision

Fourth Trial

mu_1+mu_2=mv_1+mv_2\\1(.5)+1(-1)=1(1.5)+1(-1.5)\\.5-1=1.5-1.5\\-.5=0

momentum before collision \neq moment after collision

We can see only Trial- 2 shows the conservation of momentum in a closed system.

4 0
3 years ago
Read 2 more answers
A sample of nitrogen gas is inside a sealed container. The volume of the container decreases while the temperature is kept const
frez [133]
I guess it’s d) isobaric mate correct me if I am wrong :D
7 0
3 years ago
I dont know how to do any of this so someone please help (the way it was solved has to be present for each thing) i will give br
Alexus [3.1K]
I cannot see all the questions, what is 18,19 and 21? (:
7 0
2 years ago
A 4000kg truck has a head-on inelastic collision with a 2500kg truck.
iogann1982 [59]

Answer:it could be B

Explanation:

im not sure

7 0
2 years ago
A train started from rest and moved with constant acceleration. At one time it was traveling 27 m/s, and 150 m farther on it was
AlekseyPX

Explanation:

(a) Given:

Δx = 150 m

v₀ = 27 m/s

v = 54 m/s

Find: a

v² = v₀² + 2aΔx

(54 m/s)² = (27 m/s)² + 2a (150 m)

a = 7.29 m/s²

(b) Given:

Δx = 150 m

v₀ = 0 m/s

a = 7.29 m/s²

Find: t

Δx = v₀ t + ½ at²

150 m = (0 m/s) t + ½ (7.29 m/s²) t²

t = 6.42 s

(c) Given:

v₀ = 0 m/s

v = 27 m/s

a = 7.29 m/s²

Find: t

v = at + v₀

27 m/s = (7.29 m/s²) t + 0 m/s

t = 3.70 s

(d) Given:

v₀ = 0 m/s

v = 27 m/s

a = 7.29 m/s²

Find: Δx

v² = v₀² + 2aΔx

(27 m/s)² = (0 m/s)² + 2 (7.29 m/s²) Δx

Δx = 50 m

7 0
3 years ago
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