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AleksandrR [38]
2 years ago
11

A 10kg object rests on a frictionless surface when it is struck by a 300N force. At what rate will it accelerate?

Physics
2 answers:
maxonik [38]2 years ago
5 0

Answer:

I believe the answer is C. 0.3m/s/s

Explanation:

hope this helps :) lemme know if I'm correct

krok68 [10]2 years ago
4 0

Answer:

Answer option B: 30m/s/s.

Explanation:

Should be the answer. I'm sorry if my answer is wrong not 100% sure. I tried my best.

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According to the second law of thermodynamics, 
the answer is 
<span>4. The entropy of the universe is increasing. </span>
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3 years ago
A car that is standing still accelerates up a hill with a slope of 6.4% at an average acceleration of 2.93 ft/s^2. The hill is 1
alexgriva [62]

Answer:

213 s

Explanation:

Slope is the ratio of change in vertical distance to change in horizontal distance.

Slope = vertical height / horizontal height

Therefore:

6.4% = vertical height / 12.42

vertical height = 6.4% * 12.42

vertical height = 0.8 miles

The distance travelled by the car (s) is:

s² = 0.8² + 12.42²

s² = 154.9

s = 12.45 miles

Acceleration (a) =  2.93 ft/s^2 = 0.00055 mile/s²

initial velocity (u) = 0, final velocity = 203 mph

Using:

s = ut + 0.5at²

12.45 = 0.5(0.00055)t²

t =213 s

5 0
2 years ago
A car has a mass of 1000 kg and accelerates at 2 meters per second per second. what is the magnitude of the net force exerted on
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F=ma
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4 0
3 years ago
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If a transmission line in a cold climate collects ice, the increased diameter tends to cause vortex formation in a passing wind.
AleksAgata [21]

Answer:

a) f_1=5.587Hz

b) f_{n+1}-f_n=5.587Hz

Explanation:

The frequency of the n^{th} harmonic of a vibrating string of length <em>L, </em>linear density \mu under a tension <em>T</em> is given by the formula:

f_n=\frac{n}{2L} \sqrt{\frac{T}{\mu}

a) So for the <em>fundamental mode</em> (n=1) we have, substituting our values:

f_1=\frac{1}{2(347m)} \sqrt{\frac{65.4\times10^6N}{4.35kg/m}}=5.587Hz

b) The <em>frequency difference</em> between successive modes is the fundamental frequency, since:

f_{n+1}-f_n=\frac{n+1}{2L} \sqrt{\frac{T}{\mu}}-\frac{n}{2L} \sqrt{\frac{T}{\mu}}=(n+1-n)\frac{1}{2L} \sqrt{\frac{T}{\mu}}=\frac{n}{2L} \sqrt{\frac{T}{\mu}}=f_1=5.587Hz

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2 years ago
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This is a solar Eclipse.  
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