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3 years ago

A force of 50 newtons causes a sled to accelerate at a rate of 5 meters per second. What is the mass of the sled.

Physics

2 answers:

frosja888 [35]3 years ago

7 0

Answer:

10 kilograms

Explanation:

notka56 [123]3 years ago

5 0

F=ma
50=m(5)
m=10kg
hence,ans is B

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An airplane engine starts from rest; and 2 seconds later, it is rotating with an angular speed of 300 rev/min. if the angular ac

elena-s [515]

First of all, we need to convert the angular speed from rev/min into rev/s:

$\omega_f=300 rev/min=5 rev/s$

The angular acceleration is the variation of angular speed divided by the time:

$\alpha=\frac{\omega_f-\omega_i}{t}=\frac{5 rev/s-0}{2 s}=2.5 rev/s^2$

And this is constant, so we can use the following equation to calculate the angle through which the engine has rotated:

$\theta(t)=\frac{1}{2}\alpha t^2 =\frac{1}{2}(2.5 rev/s^2)(2 s)^2=5 rev$

so, 5 revolutions.

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3 years ago

A ball is thrown vertically upward, which is the positive direction. A little later, it returns to its point of release. The bal

Aleks [24]

Answer:

The initial velocity of the ball is <u>39.2 m/s in the upward direction.</u>

Explanation:

Given:

Upward direction is positive. So, downward direction is negative.

Tota time the ball remains in air (t) = 8.0 s

Net displacement of the ball (S) = Final position - Initial position = 0 m

Acceleration of the ball is due to gravity. So, $a=g=-9.8\ m/s^2$ (Acting down)

Now, let the initial velocity be 'u' m/s.

From Newton's equation of motion, we have:

$S=ut+\frac{1}{2}at^2$

Plug in the given values and solve for 'u'. This gives,

$0=8u-0.5\times 9.8\times 8^2\\\\8u=4.9\times 64\\\\u=\frac{4.9\times 64}{8}\\\\u=4.9\times 8=39.2\ m/s$

Therefore, the initial velocity of the ball is 39.2 m/s in the upward direction.

3 0

4 years ago

"A ball with a mass of 0.1 kg is rolling at a velocity of 5 m/s. What is its

Mumz [18]

Answer:

velocity

Explanation:

because the si unit of mass is kg, velocity is m/s, acceleration is m/S2 , moment is kgm2/s . so 5 is given as velocity.

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2 years ago

What is the maximum acceleration the belt can have without the crate slipping? express your answer using two significant figures

Montano1993 [528]

To prevent the crate from slipping, the maximum force that the belt can exert on the crate must be equal to the static friction force.

Ff = 0.5 * 16 * 9.8 = 78.4 N

a = 4.9 m/s^2

If acceleration of the belt exceeds the value determined in the previous question, what is the acceleration of the crate?

In this situation, the kinetic friction force is causing the crate to decelerate. So the net force on the crate is 78.4 N minus the kinetic friction force.

Ff = 0.28 * 16 * 9.8 = 43.904 N

Net force = 78.4 – 43.904 = 34.496 N

To determine the acceleration, divide by the mass of the crate.

a = 34.496 ÷ 16 = 2.156 m/s^2

8 0

3 years ago

using hooke's law, f spring = k triangle x, find the elastic constant of a spring that stretches 2 cm when a 4 newton force is a

Ksivusya [100]

As we know that spring force is given as

$F = kx$

here we know that

F = 4 N

x = 2 cm = 0.02 m

now from the above equation we will have

$4 = k(0.02)$

$k = 200 N/m$

so the elastic constant of the spring will be 200 N/m

8 0

3 years ago