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Paladinen [302]
3 years ago
11

Question 2 (10 points)

Physics
2 answers:
Simora [160]3 years ago
8 0

Answer:

traping

Explanation:

salantis [7]3 years ago
6 0

Answer:

B. Trapping

Explanation:

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It must exist in p<span>lasma state.</span>
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It takes a minimum distance of 57.46 m to stop a car moving at 13.0 m/s by applying the brakes (without locking the wheels). Ass
vivado [14]

Answer:

The minimum stopping distance when the car is moving at

29.0 m/sec = 285.94 m

Explanation:

We know by equation of motion that,

v^{2}=u^{2}+2\cdot a \cdot s

Where, v= final velocity m/sec

u=initial velocity m/sec

a=Acceleration m/Sec^{2}

s= Distance traveled before stop m

Case 1

u=  13 m/sec, v=0, s= 57.46 m, a=?

0^{2} = 13^{2}  + 2 \cdot a \cdot57.46

a = -1.47 m/Sec^{2} (a is negative since final velocity is less then initial velocity)

Case 2

u=29 m/sec, v=0, s= ?, a=-1.47 m/Sec^{2} (since same friction force is applied)

v^{2} = 29^{2}  - 2 \cdot 1.47 \cdot S

s = 285.94 m

Hence the minimum stopping distance when the car is moving at

29.0 m/sec = 285.94 m

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3 years ago
According to diagonal rule, the orbital with lowest energy in the given following is ________.
Kaylis [27]
The orbital with the lowest energy is 3s.
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To initiate a nuclear reaction, an experimental nuclear physicist wants to shoot a proton into a 5.50-fm-diameter 12C nucleus. T
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Explanation:

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If v be the velocity of proton

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= .5 x 1.67 x 10⁻²⁷ x v² = 2.88 x 10⁻¹³

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If V be the potential difference required

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V x 1.6 x 10⁻¹⁹ = 2.88 x 10⁻¹³

V = 1.8 x 10⁶ volt .

3 0
3 years ago
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