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2 years ago

PLS ANSWER FAST WILL GIVE BRAINLEST!!!

Physics

2 answers:

KengaRu [80]2 years ago

7 0

Answer:

$\boxed {\boxed {\sf 2500 \ Newtons }}$

Explanation:

The formula for force is the product of mass and acceleration.

$F=m*a$

The mass of the car is 25 kilograms and the acceleration is 100 meters per square second. So,

$m= 25 \ kg \\a= 100 \ m/s^2$

Substitute the values into the formula.

$F= 25 \ kg * 100 \ m/s^2$

Multiply.

$F= 2500 \ kg*m/s^2$

1 kilogram meter per square second is equal to 1 Newton
Our answer of 2500 kg*m/s² is equal to 2500 Newtons

$F= 2500 \ N$

The force of the car is <u>2500 Newtons.</u>

dexar [7]2 years ago

4 0

Answer:

2500 Newtons

Explanation:

If force equals the mass × acceleration then all should be quite simple

25kg is the mass of your car

100 m/s/s is the acceleration of you car

25kg • 100m/s/s = 2500 Newtons

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3 years ago

A factory worker pushes a crate of mass 31.0 kg a distance of 4.35 m along a level floor at constant velocity by pushing horizon

Debora [2.8K]

Answer:

a. 79.1 N

b. 344 J

c. 344 J

d. 0 J

e. 0 J

Explanation:

a. Since the crate has a constant velocity, its net force must be 0 according to Newton's 1st law. The push force $F_p$ by the worker must be equal to the friction force $F_f$ on the crate, which is the product of friction coefficient μ and normal force N:

Let g = 9.81 m/s2

$F_p = F_f = \mu N = \mu mg = 0.26 * 31 * 9.81 = 79.1 N$

b. The work is done on the crate by this force is the product of its force $F_p$ and the distance traveled s = 4.35

$W_p = F_ps = 79.1*4.35 = 344 J$

c. The work is done on the crate by friction force is also the product of friction force and the distance traveled s = 4.35

$W_f = F_fs = -79.1*4.35 = -344 J$

This work is negative because the friction vector is in the opposite direction with the distance vector

d. As both the normal force and gravity are perpendicular to the distance vector, the work done by those forces is 0. In other words, these forces do not make any work.

e. The total work done on the crate would be sum of the work done by the pushing force and the work done by friction

$W_p + W_f = 344 - 344 = 0 J$

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