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Thepotemich [5.8K]
3 years ago
13

The effect of mechanical work on an object is _. (complete the sentence.)

Physics
1 answer:
Advocard [28]3 years ago
8 0

Answer:

Displacement (i.e. change in its final position)

Explanation:

In classical mechanics work is described as the displacement of the body in the direction of the applied force.

<u>Mathematically given as:</u>

W=F.s\ cos\ \theta

where:

F = force on the body

s = displacement of the body

θ = angle between the force and direction of displacement

While force, acceleration, velocity can also be the answer for the blank space but the displacement effect is visible during and even after the work has been done, so one can notice the change in position of the object even when the person has not seen the process but just the initial condition of the body.

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A(n) 0.47 kg softball is pitched at a speed of 10 m/s. The batter hits it back directly at the pitcher at a speed of 29 m/s. The
nadya68 [22]

Answer:

18.33 Ns

Explanation:

As the pitch back speed has the opposite direction as before, the change in velocity would be

\Delta v = v_2 - v_1 = 29 - (-10) = 39 m/s

So the change in momentum of the ball would be the product of its velocity change and its mass

\Delta p = \Delta v m = 39 * 0.47 = 18.33 kgm/s

This is equals to the impulse acted on the ball by the bat, which is 18.33 Ns

8 0
3 years ago
A 1000 kg rocket ship is traveling at 40 m/s. If the velocity changes to 70
nydimaria [60]

Answer: A

Explanation:

6 0
3 years ago
Rgfffffffffffffffffffffffffffffffffffffffffffffffffffffff
goblinko [34]
Roygbv evjnefvvnefv ekfv k kn ke nv
5 0
2 years ago
Three forces are applied to a solid cylinder of mass 12 kg (see the drawing). The magnitudes of the forces are F1 = 15 N, F2 = 2
crimeas [40]

Answer:

α = 13.7 rad / s²

Explanation:

Let's use Newton's second law for rotational motion

         ∑ τ = I α

         

we will assume that the counterclockwise turns are positive

         F₁  0 + F₂ R₂ - F₃ R₃ = I α

give us the cylinder moment of inertia

        I = ½ M R₂²

         

        α = (F₂ R₂ - F₃ R₃)  \frac{2}{M R_2^2}

let's calculate

        α = (24  0.22 - 13  0.10) \frac{2}{12 \ 0.22^2}2/12 0.22²

        α = 13.7 rad / s²

6 0
3 years ago
A person is trying to lift a crate that has a mass of 30 kg. The normal force of the floor is currently supplying 150N of force.
alexdok [17]

Here when an object is placed on the level floor then in that case there are two forces on the object

1). Weight of object downwards (mg)

2). Normal force due to floor which will counterbalance the weight (N)

so when no force is applied on the box at that time normal force is counter balanced by weight.

Now here it is given that A person tried to lift the box upwards

So now there are two forces on the box

1). Applied force of person

2). Normal force due to ground

So now these two forces will counter balance the weight of the crate

So we can write an equation for force balance like

F_g = F_n + F_a

given that

F_g = mg

here

m = 30 kg and

g = acceleration due to gravity = 10 m/s^2

F_n = 150 N

now from above equation

30*10 = 150 + F_a

F_a = 300 - 150 = 150 N

So force applied by the person must be 150 N

7 0
3 years ago
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