A teacher pushed a 98 newton desk across a floor for a distance of 5 meters. He exerted a horizontal force of 20 N for four seco
1 answer:
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Answer:
The answer to the question is
The distance d, which locates the point where the light strikes the bottom is 29.345 m from the spotlight.
Explanation:
To solve the question we note that Snell's law states that
The product of the incident index and the sine of the angle of incident is equal to the product of the refractive index and the sine of the angle of refraction
n₁sinθ₁ = n₂sinθ₂
y = 2.2 m and strikes at x = 8.5 m, therefore tanθ₁ = 2.2/8.5 = 0.259 and
θ₁ = 14.511 °
n₁ = 1.0003 = refractive index of air
n₂ = 1.33 = refractive index of water
Therefore sinθ₂ = =
= 0.1885 and θ₂ = 10.86 °
Since the water depth is 4.0 m we have tanθ₂ = or x₂ =
=
= 20.845 m
d = x₂ + 8.5 = 20.845 m + 8.5 m = 29.345 m.
Answer:
- 27.6 m
- 13.8 m/s
Explanation:
(b) The initial velocity is added to that due to acceleration by gravity. The velocity is increased linearly by gravity at the rate of 9.8 m/s². The average velocity of the pebble will be its velocity halfway through the 2-second time period.* That is, it will be ...
4 m/s + (9.8 m/s²)(2 s)/2 = 13.8 m/s . . . . average velocity
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(a) The distance covered in 2 seconds at an average velocity of 13.8 m/s is ...
d = vt
d = (13.8 m/s)(2 s) = 27.6 m
The water is about 27.6 m below ground.
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* We have chosen to make use of the fact that the velocity curve is linear, so the average velocity is half the sum of initial and final velocities:
vAvg = (vInit + vFinal)/2 = (vInit + (vInit +at))/2 = vInit +at/2
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If you work this in a straightforward way, you would find distance as the integral of velocity, then find average velocity from the distance and time.