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11111nata11111 [884]
3 years ago
15

A 1200 kg car accelerates from rest for 45 meters in 6.0 seconds. What is the net force on the car?​

Physics
1 answer:
Alja [10]3 years ago
8 0

Answer:

Explanation:

At first, you have to get the acceleration value from the given data.

s=ut+\frac{1}{2}at^{2}

s= 45m

u= 0 m/s

t = 6s

a = ?

45 = 0+ 0.5 × a × 6²

45 = 0.5 × 36 × a

45 = 18a

a = 2.5 m/s²

Now, you can use F=ma equation

F = ?

m = 1200kg

a = 2.5 m/s²

F = 1200 × 2.5

F = 3000 N

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Answer:

1.736m/s²

Explanation:

According to Newton's second law;

\sum F_x = ma_x\\

Fm - Ff = ma_x\\ where;

Fm is the moving force = 70.0N

Ff is the frictional force acting on the body

Ff = \mu R\\Ff = \mu mg\\

\mu is the coefficient of friction

m is the mass of the object

g is the acceleration due to gravity

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The equation becomes;

Fm - Ff = ma_x\\Fm - \mu mg = ma\\

Substitute the given parameters

Fm - \mu mg = ma\\70 - 0.18(20)(9.8) = 20a\\70-35.28 = 20a\\34.72 = 20a\\a = \frac{34.72}{20}\\a =  1.736m/s^2\\

Hence the deceleration rate of the wagon as it is caught is 1.736m/s²

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3 years ago
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Answer:

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Explanation:

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Answer:

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