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emmainna [20.7K]
3 years ago
7

Which of the following is a high wave that crashes down onto the

Physics
2 answers:
AnnyKZ [126]3 years ago
6 0
A breaker should be the wave that you’re looking for.
andrew-mc [135]3 years ago
4 0
A breaker is a high wave that crashes onto the ocean floor.
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VashaNatasha [74]

direction ................................................................

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Which shows the correct order of processes leading up to the formation of a main sequence star?
Sphinxa [80]
C. Should be the answer it worked for me
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What is the ACCELERATION of a 50 kg object pushed with a force of 500 newtons *
Anon25 [30]

Answer:

The answer is

<h2>10 m/s²</h2>

Explanation:

To find the acceleration of an object given the force and mass we use the formula

<h3>acceleration =  \frac{force}{mass}</h3>

From the question

mass of object = 50 kg

force = 500 N

So the acceleration is

<h3>acceleration =  \frac{500}{50}  \\  =  \frac{50}{5}</h3>

We have the final answer as

<h3>10 m/s²</h3>

Hope this helps you

5 0
3 years ago
Explain why frog will not look green under the red light?
IrinaVladis [17]
A frog can be many different colours. It appears green under normal 'white' light because it absorbs all the other colours in the light's spectrum apart from green. It reflects the green light back and that is picked up by your eye.

If the light is red, there is no green in the spectrum of the light, only red. So, the red light will be absorbed and there is no green to be reflected back for you to see. Therefore, the frog will not look green.
8 0
3 years ago
In a 100 mm diameter horizontal pipe, a venturimeter of 0.5 contraction ratio has been fitted. The head of water on the meter wh
horrorfan [7]

Answer:

the rate of flow = 29.28 ×10⁻³ m³/s or 0.029 m³/s

Explanation:

Given:

Diameter of the pipe = 100mm = 0.1m

Contraction ratio = 0.5

thus, diameter at the throat of venturimeter = 0.5×0.1m = 0.05m

The formula for discharge through a venturimeter is given as:

Q=C_d\frac{A_1A_2}{\sqrt{A_1^2-A_2^2}}\sqrt{2gh}

Where,

C_d is the coefficient of discharge = 0.97 (given)

A₁ = Area of the pipe

A₁ = \frac{\pi}{4}0.1^2 = 7.85\times 10^{-3}m^2

A₂ = Area at the throat

A₂ = \frac{\pi}{4}0.05^2 = 1.96\times 10^{-3}m^2

g = acceleration due to gravity = 9.8m/s²

Now,

The gauge pressure at throat = Absolute pressure - The atmospheric pressure

⇒The gauge pressure at throat = 2 - 10.3 = -8.3 m (Atmosphric pressure = 10.3 m of water)

Thus, the pressure difference at the throat and the pipe = 3- (-8.3) = 11.3m

Substituting the values in the discharge formula we get

Q=0.97\frac{7.85\times 10^{-3}\times 1.96\times 10^{-3}}{\sqrt{7.85\times 10^{-3}^2-1.96\times 10^{-3}^2}}\sqrt{2\times 9.8\times 11.3}

or

Q=\frac{0.97\times15.42\times 10^{-6}\times 14.88}{7.605\times 10^{-3}}

or

Q = 29.28 ×10⁻³ m³/s

Hence, the rate of flow = 29.28 ×10⁻³ m³/s or 0.029 m³/s

5 0
3 years ago
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