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Kryger [21]
3 years ago
14

Two plane mirrors are separated by 120°, as the drawing illustrates. If a ray strikes mirror M1 at a θ1 = 64° angle of incidence

, at what angle θ2 does it leave mirror M2?
Physics
1 answer:
tino4ka555 [31]3 years ago
8 0

Angle, θ2 at which the light leaves mirror 2 is 56°

<u>Explanation:</u>

Given-

θ1 = 64°

So, α will also be 64°

According to the figure:

α + β = 90°

So,

β = 90° - α

  = 90° - 64°

  = 26°

β + γ + 120° = 180°

γ = 180° - 120° - β

γ = 180° - 120° - 26°

γ = 34°

γ + δ = 90°

δ = 90° - γ

δ = 90° - 34°

δ = 56°

According to the law of reflection,

angle of incidence = angle of reflection

θ2 = δ = 56°

Therefore, angle θ2 at which the light leaves mirror 2 is 56°

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A halfback on an apparent breakaway for a touchdown is tackled from behind. If the halfback has a mass of 98 kg and was moving a
uranmaximum [27]

Answer:

The mutual speed immediately after the touchdown-saving tackle is 4.80 m/s

Explanation:

Given that,

Mass of halfback = 98 kg

Speed of halfback= 4.2 m/s

Mass of corner back = 85 kg

Speed of corner back = 5.5 m/s

We need to calculate their mutual speed immediately after the touchdown-saving tackle

Using conservation of momentum

m_{h}v_{h}+m_{c}v_{c}=m_{h+c}v_{h+c}

Where, m_{h}= mass of halfback

m_{c}=mass of corner back

v_{h}= velocity of halfback

v_{c}= velocity of corner back

Put the value into the formula

98\times4.2+85\times5.5=(98+85)\times v

v=\dfrac{98\times4.2+85\times5.5}{98+85}

v=4.80\ m/s

Hence, The mutual speed immediately after the touchdown-saving tackle is 4.80 m/s

3 0
3 years ago
A 10-kg dog is running with a speed of 5.0 m/s. what is the minimum work required to stop the dog in 2.40 s?
ankoles [38]
Given required solution

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v=5.0m/s F=mg
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S=VT
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3 years ago
A vw beetle goes from 0 to 60 mi/h with an acceleration of 2.35 m/s^2.
Vlad1618 [11]
Part a.
u = 0, the initial velocity
v = 60 mi/h, the final velocity
a = 2.35 m/s², the acceleration.

Note that
1 m = 1609.34 m.
Therefore
v = (60 mi/h)*(1609.34 m/mi)*(1/3600 h/s) = 26.822 m/s
Use the formula
v = u + at
(26.822 m/s) = (2.35 m/s²)*(t s)
t = 26.822/2.35 = 11.4 s

Answer: 11.4 s

Part b.
We already determined that v = 60 mi/h = 26.822 m/s.
t = 0.6 s
Therefore
(26.822 m/s) = (a m/s²)*(0.6 s)
a = 26.822/0.6 = 44.7 m/s²

Answer:  44.7 m/s²
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2 years ago
If a creature cats Ferrari, with an initial speed of 10 m/s, accelerate at a rate of 50 m/s/s for 3 seconds, what will it's fina
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160 m/s because you start off with 10 and go up by 50*3 which is 150 then add that with the begging speed to achieve 160m/s.

3 0
2 years ago
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