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irina1246 [14]
3 years ago
5

Anya does 20 push-ups by applying a force to elevate her body off the ground. For each push up, she does 100 J of work and takes

2 seconds. How much power per push-up did she deliver?
A. 10 W

B. 50 W

С. 120 W

D. 200 W
Physics
1 answer:
kipiarov [429]3 years ago
7 0

Answer:

its 120 all u have to do is plus 100 and 20

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A large block attached to a spring is undergoing simple harmonic motion horizontally, with angular frequency ω and amplitude A.
Naily [24]

Answer:

Explanation:

It is a problem based on simple harmonic motion and friction . maximum friction possible between large block and mass m is µ mg . During SHM , maximum acceleration is ω² A and force is  mω² A .

friction must exceed it so that mass m may not slip over it during motion . so

µ mg ≥ mω² A

µ ≥ mω² A / mg

smallest value of µ

= mω² A / mg

= ω² A /g

3 0
4 years ago
Hellppppp! will give brainliest answer !
Alenkinab [10]

The answer is C!!!XD


Hope this helped!!!XD


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5 0
3 years ago
The Olympic record for running the 200 m dash is 19.3 seconds. What is the average speed for this record?
Hoochie [10]

Answer:

A

Explanation:

v = change of X / change of T

v = 200/19.3

8 0
3 years ago
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You are working as an expert witness for an attorney who is suing a shipping company. The company operates ships that carry crud
AnnZ [28]

Answer:

56.7 m³

Explanation:

The radius of the circular area covered by the spill is 4.25 km

You can find the area covered by the spill by applying the formula for area of a circular surface

A=π×r² where π=3.14 and r=4.25km

A= 3.14×4.25² = 56.72 km²

Remember the total area is on average at 1 micron (1μm) thick.This is to say

1μm=1m³

So the minimum volume of oil covering the ocean surface in meters will be;

Change km² to m²

56.72 km²= 56.72×1000000=5.67×10⁷ m²

Finding the volume;

Volume = Area * thickness

Volume= 5.67×10⁷ m² * 1× 10⁻⁶ = 56.7 m³

Minimum amount of oil that was spilled is 56.7 m³

3 0
3 years ago
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
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