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masya89 [10]
3 years ago
14

If a box of supplies is dropped from the cargo hold of an airplane travelling at an altitude of 4,410 meters how long will it ta

ke to reach the ground!
Physics
1 answer:
Paraphin [41]3 years ago
5 0

Answer:

It will take 30 seconds to reach the ground, and it will be travelling at 294 m/s when it does so. This means that its average velocity was 147 m/s.

Explanation:

d=v_ot+\dfrac{1}{2}at^2

Since the initial velocity of a dropped object is 0, we can make this the equation:

d=\dfrac{1}{2}at^2 \\\\4410=\dfac{1}{2}(9.8)t^2 \\\\t^2=900

t=30\text{ seconds}

The final velocity can be calculated with the formula:

v_f=v_o+at

Once again, since there is no initial velocity:

v_f=at \\\\v_f=(9.8)(30)=294m/s

Since the initial velocity is 0, the average vertical velocity is 294/2=147 m/s.

Hope this helps!

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The coefficient of static friction for the tires of a race car is 0.950 and the coefficient of kinetic friction is 0.800. The ca
Arisa [49]

Answer:

speed on planet is 0.5 times of speed on a earth

Explanation:

Given data:

coeeffcient of kinetic friction= \mu+k  =0.8

friction force is given as

f_k = \mu_k Nn -normal force = mg

for max safe speed

fricional force  =  centripetal force

\mu_k N = \frac{mV^2}{R}

\mu_k m g = \frac{mV^2}{R}

v^2 = \mu_k g R

v^2 = 0.8\times 2.45 \times 50v^2 = 98

v = 9.899 m/sec

therefore maximum speed is  [v max} = 9.899 m/sec

\frac{planet\ speed}{ earth\ speed} = \sqrt { \frac{ \mu_k g_p R}{ \mu_k g_e R}}

\frac{V_p}{V_e} = \sqrt{\frac{g_p}{g_e}}

solving for planet speed

v_p = \sqrt{\frac{2.45}{9.8}} v_e

v_p = 0.5 v_e

speed on planet is 0.5 times of speed on a earth

4 0
3 years ago
Read 2 more answers
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