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3 years ago

If an object is accelerating at 4 m/s^2 and has a mass of 8 kg, what is the net

Physics

2 answers:

Salsk061 [2.6K]3 years ago

7 0

Answer:

The force applied is 32 N

Explanation:

F = ma

F = 8 × 4

F = 32 N

cricket20 [7]3 years ago

4 0

Answer:

$\boxed {\boxed {\sf 32 \ Newtons}}$

Explanation:

We are asked to find the net force on an object. According to Newton's Second Law of Motion, force is the product of mass and acceleration. Therefore, the formula for calculating force is:

$F= m \times a$

The mass of the object is 8 kilograms and its acceleration is 4 meters per second squared.

m= 8 kg
a= 4 m/s²

Substitute these values into the formula.

$F= 8 \ kg \times 4 \ m/s^2$

Multiply.

$F= 32 \ kg *m/s^2$

Convert the units. 1 kilogram meter per second squared is equal to 1 Newton. So, our answer of 32 kilogram meters per second squared is equal to 32 Newtons.

$F= 32 \ N$

The net force on the object is 32 Newtons.

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3 years ago

How would I workout question 4?? Please help

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Answer:

550 kg

Explanation:

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3 years ago

A piece of rocky debris in space has a semi major axis of 45.0 AU. What is its orbital period?

KATRIN_1 [288]

Complete Question

Planet D has a semi-major axis = 60 AU and an orbital period of 18.164 days. A piece of rocky debris in space has a semi major axis of 45.0 AU. What is its orbital period?

Answer:

The value is $T_R = 11.8 \ days$

Explanation:

From the question we are told that

The semi - major axis of the rocky debris $a_R = 45.0\ AU$

The semi - major axis of Planet D is $a_D = 60 \ AU$

The orbital period of planet D is $T_D = 18.164 \ days$

Generally from Kepler third law

$T \ \ \alpha \ \ a^{\frac{3}{2} }$

Here T is the orbital period while a is the semi major axis

$\frac{T_D}{T_R} = \frac{a^{\frac{3}{2} }}{a_R^{\frac{3}{2} }}$

=> $T_R = T_D * [\frac{a_R}{a_D} ]^{\frac{3}{2} }$

=> $T_R = 18.164 * [\frac{ 45}{60} ]^{\frac{3}{2} }$

=> $T_R = 11.8 \ days$

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3 years ago

Three observers watch a train pull away from a station toward the right of the platform. Observer A is in one of the train’s car

juin [17]

Observer A is moving inside the train

so here observer A will not be able to see the change in position of train as he is standing in the same reference frame

So here as per observer A the train will remain at rest and its not moving at all

Observer B is standing on the platform so here it is a stationary reference frame which is outside the moving body

So here observer B will see the actual motion of train which is moving in forward direction away from the platform

Observer C is inside other train which is moving in opposite direction on parallel track. So as per observer C the train is coming nearer to him at faster speed then the actual speed because they are moving in opposite direction

So the distance between them will decrease at faster rate

Now as per Newton's II law

F = ma

Now if train apply the brakes the net force on it will be opposite to its motion

So we can say

- F = ma

$a = \frac{-F}{m}$

so here acceleration negative will show that train will get slower and its distance with respect to us is now increasing with less rate

It is not affected by the gravity because the gravity will cause the weight of train and this weight is always counterbalanced by normal force on the train

So there is no effect on train motion

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3 years ago