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Fed [463]
3 years ago
9

60mph(mi/h) to feet per second(ft/s)

Physics
1 answer:
Ulleksa [173]3 years ago
5 0

Hi there!

There are 5,280 ft in a mile so multiply that by 60

5,280 × 60 = 316,800 ft

There are 60 min in an hour and 60 sec in a minute so multiply 60 sec by 60 sec

60 sec × 60 min = 3600 sec

Now divide 316,800 ft by 3600 sec to get:

316,800 ft ÷ 3600 sec ≈ 88 (ft/s)

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Can someone summarize the multiverse theory? I need it for a quick homework assignment :]
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A rock is dropped from a 110-m-high cliff. How long does it take to fall (a) the first 55.0 m and (b) the second 55.0 m?
Genrish500 [490]

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a) t = 3.35[s]; b) t = 1.386[s]

Explanation:

We can solve this problem by dividing it into two parts, for the first 55 [m] and then the second part with the remaining 55 [m].

We will take the initial velocity as zero, as the problem does not mention that the Rock was thrown at initial velocity.

And using kinematics equations:

v_{f}^{2}= v_{o}^{2}+2*g*y\\where:\\v_{o}=0\\g=gravity = 9.81[m/s^2]\\y=55 [m]\\v_{f}^{2}=0+2*9.81*55\\v_{f}=\sqrt{2*9.81*55} \\v_{f}=32.85[m/s]

Now we can calculate the time:

v_{f}=v_{o}+g*t\\t=\frac{v_{f}-v_{o}}{g}\\ t=\frac{32.85-0}{9.81}\\ t=3.35[s]

Now we can calculate the second time, but using as a initial velocity 32.85[m/s].

The final velocity will be:

v_{f}^{2}= v_{o}^{2}+2*g*y\\v_{f}=\sqrt{v_{o}^{2}+2*g*y} \\v_{f}=\sqrt{32.85^{2}+2*9.81*55 } \\v_{f}=46.45[m/s]

Now we can calculate the second time:

t=\frac{46.45-32.85}{9.81} \\t= 1.386[s]

Note: The reason the second time is shorter even though it is the same distance is that the acceleration of gravity increases the speed of the rock more and more as it falls.

7 0
3 years ago
A ford escort needs a force of 3500N to accelerate at the rate of 4.9 m/s2. What is the mass of the car?
Verizon [17]

Answer:

714 kg

Explanation:

ΣF = ma

3500N = 4.9m/s^2*m

m ~= 714 kg

5 0
2 years ago
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