How much work is done by a 450.0 Wengine in 15.0 s?

the momentum of a spring coil when the external compressing force is removed b the difference between the final momentum and the

Fed [463]

Answer:

the correct one is b

the difference between the final moment and the initial moment

Explanation:

The momentum is related to the moment

I = ΔP

∫ F dt = p_f - p₀

where p_f and p₀ are the final and initial moments, respectively

When checking the different answers, the correct one is b

the difference between the final moment and the initial moment

7 0

3 years ago

In the physics of motion a force acting over a distance is?

lutik1710 [3]

A force over distance is work the unite is joules

8 0

4 years ago

Read 2 more answers

A 40.0 kg beam is attached to a wall with a hi.nge and its far end is supported by a cable. The angle between the beam and the c

GalinKa [24]

The magnitude of the force that the beam exerts on the hi.nge will be,261.12N.

To find the answer, we need to know about the tension.

<h3>How to find the magnitude of the force that the beam exerts on the hi.nge?</h3>

Let's draw the free body diagram of the system using the given data.
From the diagram, we have to find the magnitude of the force that the beam exerts on the hi.nge.
For that, it is given that the horizontal component of force is equal to the 86.62N, which is same as that of the horizontal component of normal reaction that exerts by the beam on the hi.nge.

$N_x=86.62N$

We have to find the vertical component of normal reaction that exerts by the beam on the hi.nge. For this, we have to equate the total force in the vertical direction.

$N_y=F_V=mg-Tsin59\\$

To find Ny, we need to find the tension T.
For this, we can equate the net horizontal force.

$F_H=N_x=Tcos59\\\\T=\frac{F_H}{cos59} =\frac{86.62}{0.51}= 169.84N$

Thus, the vertical component of normal reaction that exerts by the beam on the hi.nge become,

$N_y= (40*9.8)-(169.8*sin59)=246.4N$

Thus, the magnitude of the force that the beam exerts on the hi.nge will be,

$N=\sqrt{N_x^2+N_y^2} =\sqrt{(86.62)^2+(246.4)^2}=261.12N$

Thus, we can conclude that, the magnitude of the force that the beam exerts on the hi.nge is 261.12N.

Learn more about the tension here:

brainly.com/question/28106871

#SPJ1

4 0

2 years ago

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A woman in a sprint race accelerates from rest to 8.7 m/s in 2.7 s. What is her displacement?

Trava [24]

Answer:

23.49m

Explanation:

Distance = velocity x time

8.7 x 2.7 = 23.49m

8 0

3 years ago

A sinusoidal wave of angular frequency 1,203 rad/s and amplitude 3.1 mm is sent along a cord with linear density 3.9 g/m and ten

kobusy [5.1K]

Answer:

18.7842493212 W

Explanation:

T = Tension = 1871 N

$\mu$ = Linear density = 3.9 g/m

y = Amplitude = 3.1 mm

$\omega$ = Angular frequency = 1203 rad/s

Average rate of energy transfer is given by

$P=\dfrac{1}{2}\sqrt{T\mu}\omega^2y^2\\\Rightarrow P=\dfrac{1}{2}\sqrt{1871\times 3.9\times 10^{-3}}\times 1203^2\times (3.1\times 10^{-3})^2\\\Rightarrow P=18.7842493212\ W$

The average rate at which energy is transported by the wave to the opposite end of the cord is 18.7842493212 W

7 0

3 years ago