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Otrada [13]
3 years ago
13

Helppp plss

Physics
1 answer:
shutvik [7]3 years ago
3 0

Answer:

a) P = 149140[w]; b) 1491400[J]; c) v = 63.06[m/s]

Explanation:

As the solution to the problem indicates, we must convert the power unit from horsepower to kilowatts.

P = 200 [hp]

200[hp] * 745.7 [\frac{watt}{1 hp}]\\149140[watt]

Now the power definition is known as the amount of work done in a given time

P = w / t

where:

w = work [J]

t = time [s]

We have the time, and the power therefore we can calculate the work done.

w = P * t

w = 149140 * 10 = 1491400 [J]

And finally, we can calculate the velocity using, the expression for kinetic energy

E_{k}=w=0.5*m*v^{2}\\  where:\\v = velocity[m/s]\\m=mass=750[kg]\\w=work=1491400[J]\\

The key to solving this problem is to recognize that work equals kinetic energy

v=\sqrt{\frac{w}{0.5*m}}  \\v=\sqrt{\frac{1491400}{0.5*750}}  \\v=63.06[m/s]

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<h2>Answer: Resonance </h2>

Resonance is a phenomenon that occurs when a body capable of vibrating is subjected to the action of a periodic force, whose frequency of vibration approaches the characteristic frequency of vibration (called resonance frequence) of said body. This is due a relatively small force applied in a repeated form, causing the amplitude of the oscillating system to become very large.

In other words, for the specific case of sound waves, this phenomenon occurs when the frequency of the  wave that is external to the system or body coincides with the resonance frequency (characteristic frequency that reaches the maximum degree of oscillation) of this system or body.

In these circumstances the body vibrates, progressively increasing the amplitude of movement after each successive actions of the force. However, this effect can be destructive in some rigid materials.

4 0
3 years ago
Which person will most likely hear the loudest sound?
Nataliya [291]

Person standing on A will hear the loudest sound

Explanation:

The intensity of a sound wave (which is proportional to the loudness of the sound) follows an inverse square law, which is:

I\propto \frac{1}{r^2}

where

I is the intensity of the wave

r is the distance from the source of the sound

This equation means that the intensity of the sound wave (and therefore, its loudness) is inversely proportional to the square of the distance  from the source: therefore,

  • As we get closer to the source of sound, the loudness increases
  • As we move away from the source of sound, the loudness decreases

Therefore, the person that will hear the loudest sound is the one standing closer to the source, and therefore person A.

Learn more about sound waves:

brainly.com/question/4899681

#LearnwithBrainly

6 0
3 years ago
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A tightrope walker more easily balances on a tightwire if his pole
cestrela7 [59]
B) droops.

Why?
To maintain balance, you do not need something short so you're balanced well... You need something long and droopy to maintain balance. The pole should be held by your waist and it should be light.

Hope this helps!~
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3 years ago
Applied force is the force of support exerted by an object that holds up another object.
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6 0
3 years ago
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A child, hunting for his favorite wooden horse, is running on the ground around the edge of a stationary merry-go-round. The ang
olga55 [171]

Answer:

9.22 s

Explanation:

One-quarter of a turn away is 1/4 of 2π, or π/2 which is approximately 1.57 rad

Let t (seconds) be the time it takes for the child to catch up with the horse. We would have the following equation of motion for the child and the horse:

For the child: s_c = \omega_ct = 0.233t

For the horse: s_h = s_0 + a_ht^2/2 = 1.57 + 0.0136t^2/2 = 1.57 + 0.0068t^2

For the child to catch up with the horse, they must cover the same angular distance within the same time t:

s_c = s_h

0.233t = 1.57 + 0.0068t^2

0.0068t^2 - 0.233t + 1.57 = 0

t= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

t= \frac{0.233\pm \sqrt{(-0.233)^2 - 4*(0.0068)*(1.57)}}{2*(0.0068)}

t= \frac{0.233\pm0.11}{0.0136}

t = 25.05 or t = 9.22

Since we are looking for the shortest time we will pick t = 9.22 s

6 0
3 years ago
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