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Diano4ka-milaya [45]
4 years ago
5

Please answer i need help asap!!!

Physics
1 answer:
Alexus [3.1K]4 years ago
5 0
The time to distance ratio is 2.1:1 , making the first time 5 seconds, the first distance 18.9 m, and the second time 15 seconds. I hope this helps!
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By what factor is the intensity of sound at a rock concert louder than that of a whisper when the two intensity levels are 120 d
77julia77 [94]

Answer:

The intensity of sound at rock concert is  10¹⁰ greater than that of a whisper.

Explanation:

The intensity of sound is given by;

I(dB) = 10Log(\frac{I}{I_o} )

where;

I is the intensity of the sound

I₀ is the threshold of sound intensity = 1 x 10⁻¹² W/m²

The intensity of sound at a rock concert

120 = 10Log(\frac{I}{1*10^{-12}} )\\\\12 = Log(\frac{I}{1*10^{-12}} )\\\\\frac{I}{1*10^{-12}} = 10^{12}\\\\I = 1*10^{-12} *10^{12}\\\\I = 1*10^0\\\\I =1 \ W/m^2

The intensity of sound of a whisper

20 = 10Log(\frac{I}{1*10^{-12}} )\\\\2 = Log(\frac{I}{1*10^{-12}} )\\\\\frac{I}{1*10^{-12}} = 10^{2}\\\\I = 1*10^{-12} *10^{2}\\\\I = 1*10^{-10}\\\\I =10^{-10} \ W/m^2

Thus, the intensity of sound at rock concert is  10¹⁰ greater than that of a whisper.

4 0
3 years ago
A soccer ball is kicked from the top of one building with a height of H1 = 30.2 m to another building with a height of H2 = 12.0
viktelen [127]

Hi there!

Initially, we have gravitational potential energy and kinetic energy. If we set the zero-line at H2 (12.0m), then the ball at the second building only has kinetic energy.

We also know there was work done on the ball by air resistance that decreased the ball's total energy.

Let's do a summation using the equations:
KE = \frac{1}{2}mv^2 \\\\PE = mgh

Our initial energy consists of both kinetic and potential energy (relative to the final height of the ball)

E_i = \frac{1}{2}mv_i^2 + mg(H_1 - H_2)

Our final energy, since we set the zero-line to be at H2, is just kinetic energy.

E_f = \frac{1}{2}mv_f^2

And:
W_A = E_i - E_f

The work done by air resistance is equal to the difference between the initial energy and the final energy of the soccer ball.

Therefore:
W_A = \frac{1}{2}mv_i^2 + mg(H_1 - H_2) -  \frac{1}{2}mv_f^2

Solving for the work done by air resistance:
W_A = \frac{1}{2}(.450)(15.1^2)+ (.450)(9.8)(30.2 - 12) -  \frac{1}{2}(.450)(19.89^2)

W_A = \boxed{42.552 J}

8 0
2 years ago
Did thomson conclude from his experiment ?
Law Incorporation [45]
It would be A because he wanted to know more about the physical properties of these particles like their mass
6 0
4 years ago
Suppose you have made the following hypothesis: Sharks are most common near coral reefs, because there are more fish there to ea
Brilliant_brown [7]
I would say none of the options. This evidence does not support the hypothesis, but it doesn't contradict it, however it is related to the hypothesis. I guess what I'm trying to say is that the evidence isn't sufficient enough to make any definitive comments about the hypothesis. I don't think that you can just decide on whether to accept or reject your hypothesis based on observation alone and moreover, an observation that was made once. You need to make many observations, at certain points every day, in the same area of reef and the same area of open sea for a certain amount of time to gain a good amount of data (you could split up areas of reef and open sea on a particular coast into square meters or what ever unit you want and dedicate 3 days to each area you've split up) then you can perform a statistical test that suits the model of your data. I hope this helps in some way and I'm sorry it's so long. I couldn't think of a shorter way to say this.
3 0
4 years ago
Read 2 more answers
What is meant by "elastic collisions?
Ainat [17]

When a moving car hits a parked car, causing the parked car to move, the type of collision is elastic collision. An elastic collision is when two bodies collide and separates after collision conserving the total kinetic energy before and after collision.

7 0
3 years ago
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