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kvv77 [185]
3 years ago
8

Suppose that you have been chosen for a space mission to a distant planet. Due to the length of time you'll be away from Earth y

ou must carry out physical activity every day. On earth your, strength and conditioning trainer has determined you must do 90 minutes of exercise every day. If the vehicle is travelling at 0.80 c how much time, according to a timer on the space vehicle should you be active to meet your physical activity requirement?
Physics
1 answer:
Contact [7]3 years ago
8 0

Answer:

I should be active for 15 hours to meet the physical activity requirement.

Explanation:

Since time dilates in moving objects, we use the formula t = t₀/√(1 - β²) where t = time in space vehicle, t₀ = time on earth = 9 hours and β = v/c where v = speed of space vehicle = 0.8c.

So, t = t₀/√(1 - β²)

t = 9/√(1 - (v/c)²)

= 9/√(1 - (0.8c/c)²)

= 9/√(1 - (0.8)²)

= 9/√(1 - (0.64)

= 9/√0.36

= 9/0.6

= 15 hr

So, according to a timer on the space vehicle, I should be active for 15 hours to meet the physical activity requirement.

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A slab of glass has a 0.600 cm thick layer of water on top of it. A light ray strikes the water at an incident angle of 59.0°. A
Dimas [21]

Answer:

49.63 degree

Explanation:

thickness of glass slab, t = 0.6 cm

angle of incidence = 59 degree

Let r be the angle of refraction

The refractive index of glass, ng = 3/2

refractive index of water, nw = 4/3

refarctive index of glass with respect to water = ng / nw = 3 /2 ÷ 4 /3 = 9 / 8

So, by use of Snell's law

Refractive index of glass with respect to water = Sin i / Sin r

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Lets se

And

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\\ \rm\Rrightarrow \sqrt{k}T=2\pi\sqrt{m}

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If spring constant is doubled mass must be doubled

8 0
2 years ago
compare to visible light, the wavelength of x-rays is shorter longer or the same and the frequency is lower higher or the same
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Simora [160]

Answer:

Force = 35 N

Explanation:

From Newton's third law of motion, the boy must apply a force greater than the weight of the sled to lift it.

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Since the force applied on the sled by the boy is lesser than the weight of the sled, then;

Force that the sled exerts on the student = 50 - 15

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