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BigorU [14]
3 years ago
7

A 32 kg child puts a 15 kg box into a 12 kg wagon. The child then pulls horizontally on the wagon with a force of 65 N. if the b

ox does not move relative to the wagon, what is the static friction force on the box?
A 32 kg child puts a 15 kg box into a 12 kg wagon. The child then pulls horizontally on the wagon with a force of 65 N. if the box does not move relative to the wagon, what is the static friction force on the box?
Physics
1 answer:
snow_tiger [21]3 years ago
6 0
150/gen add 55to it then it is 5689
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If earth's mass were half its actual value but its radius stayed the same, the escape velocity of earth would be:________
siniylev [52]

If the earth's mass were half its actual value but its radius stayed the same, the escape velocity of the earth would be V_e = \sqrt{\dfrac{GM}{r}}.

<h3>What is an escape velocity?</h3>

The ratio of the object's travel distance over a specific period of time is known as its velocity. As a vector quantity, the velocity requires both the magnitude and the direction. the slowest possible speed at which a body can break out of the gravitational pull of a certain planet or another object.

The formula to calculate the escape velocity of earth is given below:-

V_e=\sqrt{\dfrac{2GM}{r}}

Given that earth's mass was half its actual value but its radius stayed the same. The escape velocity will be calculated as below:-

V_e=\sqrt{\dfrac{2GM}{r\times 2}}

V_e = \sqrt{\dfrac{GM}{r}}.

Therefore, If the earth's mass were half its actual value but its radius stayed the same, the escape velocity of the earth would be V_e = \sqrt{\dfrac{GM}{r}}.

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How does unequal solar heating lead to the Gulf Stream?
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3 years ago
Why it is difficult to run fast in sand​
balu736 [363]
Running on sand requires 1.6 times more energy spent than running on hard surface, so the force applied by our foot on sand is less.
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si se deja caer un carrito desde el punto mas alto de ua psta de coches cuya altura es de 1.4m cual es la velocidad maxima que p
forsale [732]

Answer:

v = 5.24[m/s]

Explanation:

Este problema se puede resolver por medio del principio de la conservación de la energía, donde la energía potencial es igual a la energía cinética. Es decir a medida que el carrito desciende su energía potencial disminuye, pero su energía cinética aumenta.

E_{kin}=E_{pot}

Donde:

E_{kin}=\frac{1}{2} *m*v^{2} \\\\E_{pot}=m*g*h

Ahora reemplazando:

\frac{1}{2} *m*v^{2}=m*g*h\\\\0.5*v^{2}=9.81*1.4\\v=\sqrt{\frac{9.81*1.4}{0.5} }   \\\\v=5.24[m/s]

6 0
3 years ago
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