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Talja [164]
3 years ago
15

Suppose that scientists observe violent gas eruptions on a planet with an acceleration due to gravity of 3.9 m/s2. The jets thro

w sand and dust about 75 m above the surface. What is the speed i of the material just as it leaves the surface?
Physics
1 answer:
Lina20 [59]3 years ago
4 0

Answer:

The velocity with which the sand throw is 24.2 m/s.

Explanation:

acceleration due to gravity, a =  3.9 m/s2

height, h = 75 m

final velocity, v = 0

Let the initial  velocity at the time of throw is u.

Use third equation of motion

v^{2} = u^{2} - 2 g h \\\\0 = u^{2} -2\times 3.9\times 75\\\\u=24.2m/s

The velocity with which the sand throw is 24.2 m/s.

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A wave has a velocity of 24 m/s and a period of 3.0 s. What is the frequency of the wave?
Ghella [55]
The velocity is extraneous information. Frequency is the inverse of the period, thus the frequency of the wave is 1/3 Hz.
4 0
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A 600-N box is pushed up a ramp that is 2 meters high and 5 meters long. The person exerts a force of 300-N. What is the efficie
AfilCa [17]
From information given, there are two forces in action, the vertical and the horizontal. We first need to calculate separately work done by these two forces, then determine the efficiency. Work = Force × distance work(a) → 600N × 2m = 1,200 Nm work(b) → 300N × 5m = 1,500 Nm Efficiency = work(a) / work (b) × 100 Efficiency = 1200 / 1500 × 100 Efficiency = 0.8 × 100 Efficiency = 80% Therefore the efficiency of the ramp is 80%
6 0
3 years ago
An underwater diver sees the sun 57 ∘ above horizontal. how high is the sun above the horizon to a fisherman in a boat above the
nydimaria [60]

Angle at which under water diver sees the sun from horizontal = 57^{0}

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Refractive index of water = 1

We have the equation

   1.33 * sin 33^{0}= 1 *sin \alpha

 \alpha = sin ^{-1} (\frac{1.33sin33}{1} )= 46.42^0

This angle is angle from normal, so angle seen by fisherman from horizontal = 90 - 46.42 = 43.58^{0}

5 0
3 years ago
A dog is sniffing around for a bone. He walks 3.6 m east and 6.0 m north. What are the magnitude and direction of the dog from h
Ksju [112]
Vectorial form:

     3.6 m east => 3i


     6.0 north => 6 j

     Final position = 3.6i + 6j

Magnitude = √[ (3.6)^2 + 6^2] = √[12.96 + 36] = √48.96 = 7.0 m

Direction => arc tan (6/3.6) = arc tan (1.67) = 59.0 °

Answer: the dog is at 7.0 m, 59.0° to the north of the east from the starting point.
6 0
3 years ago
A wave travels a distance of 60cm in 3s. The distance blw successive crests of thd wave is 4cm. What is the frequency?
schepotkina [342]
Data:
ΔS = 60cm
Δt = 3s
Vm = ?

Vm = ΔS/Δt 
Vm =  \frac{60}{3}
Vm = 20cm/s

We have:

Vm = v
ΔS = λ
Δt = T

v =  \frac{\lambda}{T}
20 = \frac{\lambda}{4}
\lambda = 20*4
\lambda = 80cm

Soon:

v = \lambda*f
20 = 80*f
80f = 20
f =  \frac{20}{80} simplify( \frac{\div20}{\div20} )= \frac{1}{4}
\boxed{f = 0,25Hz}


6 0
3 years ago
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