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Zinaida [17]
3 years ago
9

Tsunamis are fast-moving waves often generated by underwater earthquakes. In the deep ocean their amplitude is barely noticable,

but upon reaching shore, they can rise up to the astonishing height of a six-story building. One tsunami, generated off the Aleutian islands in Alaska, had a wavelength of 612 km and traveled a distance of 3920 km in 4.09 h. (a) What was the speed (in m/s) of the wave
Physics
1 answer:
Soloha48 [4]3 years ago
7 0

Answer:

V = 266.23 m/s

Explanation:

The speed of the wave can easily be given by the following formula:

V = S/t

where,

V = Speed of the Wave = ?

S = Distance Covered by Wave = 3920 km

S = Distance Covered by Wave = (3920 km)(1000 m/1 km)

S = Distance Covered by Wave = 3.92 x 10⁶ m

t = Time taken by the wave to cover the distance = 4.09 h

t = Time taken by the wave to cover the distance = (4.09 h)(3600 s/1 h)

t = Time taken by the wave to cover the distance = 14724 s

Therefore,

V = (3.92 x 10⁶ m)/(14724 s)

<u>V = 266.23 m/s</u>

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Answer:

1.4819×10^23

Explanation:

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3 years ago
A 95-kg astronaut is stranded from his space shuttle. He throws a 2-kg hammer away from the shuttle with a velocity of 19 m/s .
geniusboy [140]

Answer:

The astronaut will be propelled towards the shuttle at the rate of 0.4 m/s.

Explanation:

Given;

mass of  the astronaut, m₁ = 95 kg

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velocity of the hammer, v₂ = 19 m/s

let the recoil velocity of the shuttle = v₁

Apply the principle of conservation of linear momentum;

m₁v₁ = m₂v₂

v₁ = m₂v₂/m₁

v₁ = (2 x 19) / 95

v₁ = 0.4 m/s

Therefore, the astronaut will be propelled towards the shuttle at the rate of 0.4 m/s.

5 0
3 years ago
A man 6 feet tall walks at a rate of 6 feet per second away from a light that is 15 feet above the ground.
Tems11 [23]

Answer:(a)10 ft/s

(b)4 ft/s

Explanation:

Given

height of light =15 feet

height of man=6 feet

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From diagram

\frac{15}{y}=\frac{6}{y-x}

5(y-x)=2y

3y=5x

differentiate both sides

3\times \frac{\mathrm{d} y}{\mathrm{d} t}=5\times \frac{\mathrm{d} x}{\mathrm{d} t}

Tip of shadow is moving at the rate of

\frac{\mathrm{d} y}{\mathrm{d} t}=\frac{5}{3}\times 6=10 ft/s

(b)rate at which length of his shadow  is changing

Length of shadow is y-x

differentiating w.r.t time

\frac{\mathrm{d} (y-x)}{\mathrm{d} t}=\frac{\mathrm{d} y}{\mathrm{d} t}-\frac{\mathrm{d} x}{\mathrm{d} t}

\frac{\mathrm{d} (y-x)}{\mathrm{d} t}=10-6=4 ft/s

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Answer:

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Explanation:

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