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2 years ago

an object of 30kg is in free fall in a vacuum where there is no air resistance. Determine the acceleration of the object.

1 answer:

Ivan2 years ago

4 0

We're missing one essential piece of information that we need in order to answer this question. You have not specified <em>what planet</em> the object is falling on. The answer depends on the gravitational acceleration on that planet, and they're all different.

Without that information, we'll just go ahead and assume that the object is falling to the surface of the Earth. Wherever on Earth this tense drama is unfolding, the acceleration of gravity is going to be around <em>9.8 m/s²</em> everywhere.

So THAT's the object's acceleration if there is no air resistance. The object's MASS makes no difference. It doesn't matter whether the object is a sparrow feather or a school bus. Heavier objects DO NOT fall faster than light objects.

If there is no air resistance, then ALL objects fall with the same acceleration. It's called the "acceleration of gravity" on that planet or moon, and you can easily look it up. It's 9.8 m/s² on Earth, 1.62 m/s² on the Moon, 3.71 m/s² on Mars, 8.87 m/s² on Venus, and 24.8 m/s² on Jupiter.

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The tension in the rope is constant and equal to 40 N as the block is pulled. What is the instantaneous power (in W) supplied by

Eduardwww [97]

Complete Question:

A 10 kg block is pulled across a horizontal surface by a rope that is oriented at 60° relative to the horizontal surface.

The tension in the rope is constant and equal to 40 N as the block is pulled. What is the instantaneous power (in W) supplied by the tension in the rope if the block when the block is 5 m away from its starting point? The coefficient of kinetic friction between the block and the floor is 0.2 and you may assume that the block starting at rest.

Answer:

Power = 54.07 W

Explanation:

Mass of the block = 10 kg

Angle made with the horizontal, θ = 60°

Distance covered, d = 5 m

Tension in the rope, T = 40 N

Coefficient of kinetic friction, $\mu = 0.2$

Let the Normal reaction = N

The weight of the block acting downwards = mg

The vertical resolution of the 40 N force, $f_{y} = 40sin \theta$

$\sum f(y) = 0$

$N + 40 sin \theta - mg = 0\\N = -40sin60 + 10*9.81 = 0\\N = 63.46 N$

$\sum f(x) = 0$

$40 cos 60 - f_{r} - ma = 0\\ f_{r} = \mu N\\ f_{r} = 0.2 * 63.46\\ f_{r} = 12.69 N\\40cos 60 - 12.69-10a = 0\\7.31 = 10a\\a = 0.731 m/s^{2}$

$v^{2} = u^{2} + 2as\\u = 0 m/s\\v^{2} = 2 * 0.731 * 5\\v^{2} = 7.31\\v = \sqrt{7.31} \\v = 2.704 m/s$

Power, $P = Fvcos \theta$

$P = 40 *2.704 cos60\\P = 54.074 W$

7 0

3 years ago

A 40kg load is raised to a height of 25m. If the operation requires 1 min, find the power required

OlgaM077 [116]

Answer:

163.33 Watts

Explanation:

From the question given above, the following data were obtained:

Mass (m) = 40 Kg

Height (h) = 25 m

Time (t) = 1 min

Power (P) =..?

Next, we shall determine the energy. This can be obtained as follow:

Mass (m) = 40 Kg

Height (h) = 25 m

Acceleration due to gravity (g) = 9.8 m/s²

Energy (E) =?

E = mgh

E = 40 × 9.8 × 255

E = 9800 J

Finally, we shall determine the power. This can be obtained as illustrated below:

Time (t) = 1 min = 60 s

Energy (E) = 9800 J

Power (P) =?

P = E/t

P = 9800 / 60

P = 163.33 Watts

Thus, the power required is 163.33 Watts

8 0

2 years ago

When you look at yourself in the mirror, what is the approximate angle of incidence of the light rays?

Kitty [74]

46 degrees
I hope this helps :)

3 0

2 years ago

A block with mass 0.5 kg is forced against a horizontal spring of negligible mass, compressing the spring a distance of 0.2 m. W

lianna [129]

Answer:

So coefficient of kinetic friction will be equal to 0.4081

Explanation:

We have given mass of the block m = 0.5 kg

The spring is compressed by length x = 0.2 m

Spring constant of the sprig k = 100 N/m

Blocks moves a horizontal distance of s = 1 m

Work done in stretching the spring is equal to $W=\frac{1}{2}kx^2=\frac{1}{2}\times 100\times 0.2^2=2J$

This energy will be equal to kinetic energy of the block

And this kinetic energy must be equal to work done by the frictional force

So $\mu mg\times s=2$

$\mu\times 0.5\times 9.8\times 1=2$

$\mu =0.4081$

So coefficient of kinetic friction will be equal to 0.4081

5 0

2 years ago

What affect does friction have on motion

satela [25.4K]

Answer: friction reduces the speed during motion

Explanation:

The more the friction, the lesser the speed during motion

3 0

3 years ago