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Licemer1 [7]
3 years ago
7

How far can a car traveling at a rate of 40 kilometer per hour travelbin 2 1/2 hours

Physics
2 answers:
madam [21]3 years ago
5 0
Supposing it stays traveling at 40 kph you'll travel exactly 40 kilometers in just one hour. So in 2 hours you'll have travel 80 Kilometers. and in half an hour you travel 20 which is the Half of 40. And 20+80 = 100 Kilometers.
CaHeK987 [17]3 years ago
4 0
90 kilometers because you need to multiply 40 by 2 and then you get 80 and finally you add 10 and get 90 kilometers
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A string is attached to a ball that has a mass of 0.11 kg. A student pulls up on the string so that the ball accelerates upward
enot [183]

Answer:

T=+1.133N

Explanation:

Tension and weight are forces that have opposite directions

Weight is negative (downward)

W=m*g= 0.11kg*(-9.8m/s^2)

W= -1.078N

Tension is possitive (upward)

The total force will be the sum of both (the difference taking in consideration the direction)

Ft= T+W

Also the total force is the product of the mass due to acceleration:

Ft=m*a

Ft= +0.11kg*0.5m/s^2

Ft=+0.055N (upward)

Tension will be the difference between Ft and W:

T= Ft-W

T=+0.055N-(-1.078N)

T=+1.133N

7 0
3 years ago
Suppose that on a routine trip around the galaxy, your dog puts his head out of the spacecraft window (as dogs often do). The so
neonofarm [45]

Answer: Yes, the dog's head will burn

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If the dog holds it's breath the heat wave would burn it's head but not outrightly because the breath has ceased for some seconds and it can't absorb the much heat of the solar wind into it's nostrils. The effect of holding its breath would still be disastrous but not as the first case.

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3 years ago
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3 years ago
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In certain ranges of a piano keyboard, more than one string is tuned to the same note to provide extra loudness. For example, th
diamong [38]

Answer:

3.77 Hz

Explanation:

The beat frequency that is heard can be calculated using the following equation:

f_{b} = f_{1} - f_{2}

<u>Where:</u>

f₁ = 110 Hz

f_{2} = \frac{f_{1}}{\sqrt{\frac{T_{1}}{T_{2}}}}

<u>With:</u>

<em>T₁ = 594 N        </em>

<em>T₂ = 554 N </em>

f_{2} = \frac{f_{1}}{\sqrt{\frac{T_{1}}{T_{2}}}} = \frac{110 Hz}{\sqrt{\frac{594 N}{554 N}}} = 106.23 Hz

Hence, the beat frequency is:        

f_{b} = f_{1} - f_{2} = 110 Hz - 106.23 Hz = 3.77 Hz      

Therefore, the beat frequency that is heard when the hammer strikes the two strings simultaneously is 3.77 Hz.

I hope it helps you!                              

3 0
4 years ago
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