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3 years ago

What force to be required to accelerate a car of mass 120 kg from 5 m/s to 25m/s in 2s

Physics

1 answer:

Ira Lisetskai [31]3 years ago

4 0

Answer:1200

Explanation:

F=ma =m(Vf-Vi)/t

F=120(25-5)/2 =1200N

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The answer is "friction and air resistance" gravity does some of the work by keeping the object from floating away, but friction and air resistance does the biggest part. Friction is how rough the ground it meaning on tile, dirt, grass, etc... that would slow down the object and air resistance is the gravity pushing on the object also making it stop.

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3 years ago

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2 years ago

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3 years ago

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2 years ago

On earth, two parts of a space probe weigh 14500 N and 4800 N. These parts are separated by a center-to-center distance of 18 m

Nastasia [14]

Answer:

$F = 1.489*10^{-7} N$

Explanation: Weight of space probes on earth is given by: $W= m*g$

W= weight of the object( in N)

m= mass of the object (in kg)

g=acceleration due to gravity(9.81 $\frac{m}{s^{2} }$ )

Therefore,

$m_{1} = \frac{14500}{9.81}$

$m_{1} = 1478.08 kg$

Similarly,

$m_{2} = \frac{4800}{9.81}$

$m_{2} = 489.29 kg$

Now, considering these two parts as uniform spherical objects

Also, according to Superposition principle, gravitational net force experienced by an object is sum of all individual forces on the object.

Force between these two objects is given by:

$F = \frac{Gm_{1} m_{2}}{R^{2} }$

G= gravitational constant ( $6.67 * 10^{-11} m^{3} kg^{-1} s^{-2}$ )

$m_{1} , m_{2}$ = masses of the object

R= distance between their centres (in m)(18 m)

Substituiting all these values into the above formula

$F = 1.489*10^{-7} N$

This is the magnitude of force experienced by each part in the direction towards the other part, i.e the gravitational force is attractive in nature.

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3 years ago

What force to be required to accelerate a car of mass 120 kg from 5 m/s to 25m/s in 2s​

What force to be required to accelerate a car of mass 120 kg from 5 m/s to 25m/s in 2s