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Serggg [28]
1 year ago
11

A 230 kg steel crate is being pushed along a cement floor. The force of friction is 480 N to the left and the applied force is 1

869 N to the right. What is the acceleration of the crate?
a=(6.0 or 6.00 or 8.10 or 8.1)=m/s^2
Physics
1 answer:
Daniel [21]1 year ago
6 0

The acceleration of the steel crate, given the data from the question is 6.0 m/s²

<h3>How to determine the net force</h3>
  • Force to the left (Fբ) = 480 N
  • Force to the right (Fᵣ) = 1869 N
  • Net force (Fₙ) = ?

Fₙ = Fᵣ - Fբ

Fₙ = 1869 - 480

Fₙ = 1389 N

<h3>How to determine the acceleration</h3>
  • Mass (m) = 230 Kg
  • Net force (Fₙ) = 1389 N
  • Acceleration (a) = ?

Fₙ = ma

Divide both sides by m

a = Fₙ / m

a = 1389 / 230

a = 6.0 m/s²

Learn more about acceleration:

brainly.com/question/491732

#SPJ1

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At an altitude of 5000 m the rocket's acceleration has increased to 6.9 m/s2 . What mass of fuel has it burned?
sergey [27]

1) Initial upward acceleration: 6.0 m/s^2

2) Mass of burned fuel: 0.10\cdot 10^4 kg

Explanation:

1)

There are two forces acting on the rocket at the beginning:

- The force of gravity, of magnitude F_g = mg, in the downward direction, where

m=1.9\cdot 10^4 kg is the rocket's mass

g=9.8 m/s^2 is the acceleration of gravity

- The thrust of the motor, T, in the upward direction, of magnitude

T=3.0\cdot 10^5 N

According to Newton's second law of motion, the net force on the rocket must be equal to the product between its mass and its acceleration, so we can write:

T-mg=ma (1)

where a is the acceleration of the rocket.

Solving for a, we find the initial acceleration:

a=\frac{T-mg}{m}=\frac{3.0\cdot 10^5-(1.9\cdot 10^4)(9.8)}{1.9\cdot 10^4}=6.0 m/s^2

2)

When the rocket reaches an altitude of 5000 m, its acceleration has increased to

a'=6.9 m/s^2

The reason for this increase is that the mass of the rocket has decreased, because the rocket has burned some fuel.

We can therefore rewrite eq.(1) as

T-m'g=m'a'

where

m' is the new mass of the rocket

Re-arranging the equation and solving for m', we find

m'=\frac{T}{g+a}=\frac{3.0\cdot 10^5}{9.8+6.9}=1.8\cdot 10^4 kg

And since the initial mass of the rocket was

m=1.9 \cdot 10^4 kg

This means that the mass of fuel burned is

\Delta m = m-m'=1.9\cdot 10^4 - 1.80\cdot 10^4 = 0.10\cdot 10^4 kg

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Answer:

a. The moment of the 4 N force is 16 N·m clockwise

b. The moment of the 6 N force is 12 N·m anticlockwise

Explanation:

In the figure, we have;

The distance from the point 'O', to the 6 N force = 2 m

The position of the 6 N force relative to the point 'O' = To the left of 'O'

The distance from the point 'O', to the 4 N force = 4 m

The position of the 4 N force relative to the point 'O' = To the right of 'O'

a. The moment of a force about a point, M = The force, F × The perpendicular distance of the force from the point

a. The moment of the 4 N force = 4 N × 4 m = 16 N·m clockwise

b. The moment of the 6 N force = 6 N × 2 m = 12 N·m anticlockwise.

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