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Zigmanuir [339]
3 years ago
6

A blue box and green box are sitting on a table not moving. A student exerts the same force on both boxes and observes how long

it takes each box to travel 2 ft.
Time to Travel 2 feet

Blue Box

7 seconds

Green Box

15 seconds
Physics
2 answers:
gavmur [86]3 years ago
7 0
The green box is heavier
dem82 [27]3 years ago
5 0

Answer:

2 possible answers. 1. The green box is heavier. or 2nd. The green box has bigger friction than blue box.

Those should be the 2 main explanations. There probably are other options

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An eagle carries a fish up 50 m into the sky using 90 N of force. How much work did the eagle do on the fish? (Work: W = Fd)
Olin [163]
W=Fd so W=50(90)=4500J
4500J is your answer
4 0
3 years ago
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A stone is dropped at t = 0. A second stone, with twice the mass of the first, is dropped from the same point at t = 181 ms. How
lara [203]

Answer:

the center of mass is 316,670m bellow the release point.

Explanation:

First, we must find the distance at which the objects are in time t = 360s

We will use the formula for vertical distance in free fall

h=v_{0}t+\frac{1}{2} gt^2

v_{0} is the initial velocity, is 0 for both stones since they were just dropped.

g is acceleration of gravity and t is time (g=9.81m/s^2)

At t=360s the first stone has been falling for the entire 360 seconds, its position h1 is:

h_{1}=\frac{1}{2} (9.82m/s^2)(360s^2)=635,688m

And at 360 seconds the second stone has been fallin fort= 360s -181 s = 179s, so its position h2 is:

h_{2}=\frac{1}{2} (9.81m/s^2)(179)^2=157,161.1m

And finally using the equation for the center of mass:

CM=\frac{m_{1}h_{1}+m_{2}h_{2}}{m_{1}+m_{2}}

We know that the mass of the second stone is twice the mass of the first stone so:

m_{1}=m\\m_{2}=2m

replacing these values in the equation for the center of mass

CM=\frac{mh_{1}+2mh_{2}}{m+2m}

CM=\frac{m(h_{1}+2h_{2})}{3m}=\frac{h_{1}+2h_{2}}{3}

Finally, replacing the values we found fot h1 and h2:

CM=\frac{635,688m+2(157,161.1m)}{3}=316,670m

the center of mass is 316,670m bellow the release point.

8 0
4 years ago
A surface receiving sound is moved from its original position to a position three times farther away from the source of the soun
murzikaleks [220]
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If the distance increases to 3 x the original distance, then the intensity
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I suppose choice-'d' is the correct one, but I have to tell you that
the phrase "nine times as low" is mathematically meaningless,
and it really grinds my gears.
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