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3 years ago

The work accomplished (produced) by a machine is called Work _____

Physics

1 answer:

Nonamiya [84]3 years ago

6 0

I think it’s output because output work is work done by a machine

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You are an industrial engineer with a shipping company. As part of the package- handling system, a small box with mass 1.60 kg i

Kisachek [45]

Answer:

v = 0.84m/s, v(max)= 0.997m/s

Explanation:

Initial work done by the spring, where c is the compression = 0.28m:

$W_s = \frac{1}{2}kc^2$

Work lost to friction:

$W_f =\mu mg(c-x)$

Energy:

$E = W_s-W_f=\frac{1}{2}kc^2 -\mu mg(c-x)=\frac{1}{2}mv^2+\frac{1}{2}kx^2$

(a) Solve for v:

$v=\sqrt{\frac{k}{m}(c^2-x^2)-2\mu g(c-x)}$

(b) Solve $\frac{dv}{dx}=0$ for x:

$\frac{dv}{dx}=\frac{\mu g-\frac{k}{m}x}{\sqrt{\frac{k}{m}(c^2-x^2)-2\mu g(c-x)}}$

$\frac{dv}{dx}=0$ if:

$\mu g-\frac{k}{m}x = 0$

$x_{max} = \frac{\mu gm}{k}$

$v_{max}=\sqrt{\frac{k}{m}c^2-2\mu gc+(\mu g)^2\frac{m}{k}}$

6 0

3 years ago

A sled is pulled at a constant velocity across a horizontal snow surface. If a force of 100 N is being applied to the sled rope

tangare [24]

Answer:

60.18 N

Explanation:

Given that:

The force applied on the sled = 100 N

Suppose, the angle between the sled rope and the ground = 53°

The horizontal force which acts in the horizontal direction can be expressed as:

$F_x = F \ cos \theta$

$F_x = 100 \ cos (53)$

$F_x = 60.18 \ N$

But if the angle between the sled rope is parallel to the ground. Then, we use an angle on a straight line which is = 180°

$F_x = F \ cos \theta$

$F_x = 100 \ cos (180)$

= 100 × -1

= -100 N

3 0

2 years ago

I can not find the answer to the one after potassium ?

Zepler [3.9K]

Answer:

it’s halogen

Explanation:

3 0

2 years ago

Chemical formulawhat is the chemical formula

evablogger [386]

Answer:

Explanation:

what's the actual question

4 0

3 years ago

Read 2 more answers

A 26.0 kg child plays on a swing having support ropes that are 2.40 m long. A friend pulls her back until the ropes are 45.0 ∘ f

Sloan [31]

A)Ep'=mgh=mgl(1-cosa).At the bottom of the swing Ep=0(reference level),so the potential energy as the child is just released is bigger than the potential energy at the bottom of the swing.;B)The speed of the child at the bottom of the swing-->v=√(2gh)=√[2gl(1-cosa)];C)I don't think that the tension does any work.

8 0

3 years ago