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ddd [48]
3 years ago
9

A force of 1,500 N is moved through 0.10 m in order to compress a steel cylinder. The cylinder was compressed 0.010 m by a force

of 13,000 N. The efficiency of the compression machine is what % ?
115
0.87
87
8.7
Physics
1 answer:
ValentinkaMS [17]3 years ago
4 0
We first calculate the work effectively done by the force when it compressed the steel cylinder. This is solved by multiplying the distance that the cylinder was compressed by and its causing force.
0.010 m * 13,000 N = 130 N-m

Then, we solve for the work done by the 1,500 N force:
0.10 m * 1,500 N = 150 N-m

We then take the ratio of the efficient:experimental
130 N-m / 150 N-m = 0.87 = 87%

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Explain how the mass of a planet affects the motion of the planet around the sun?
vodomira [7]

Answer:

The gravitational attraction of the Sun is what holds the planets in their elliptical orbits. So to explain this the mass effects the motion of the planets because the strength of gravitational force depends of the mass.

Explanation:

6 0
3 years ago
Tracy stands on a skateboard and tosses her backpack to her friend who is standing in front of her. Which best describes the acc
madreJ [45]

Answer:

A

Explanation:

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3 years ago
How many seconds will it take for a the International Space Station to travel 450 km at a rate of 100 m/s?
SVEN [57.7K]

Time = (distance) / (speed)

<em></em>

Time = (450 km) / (100 m/s)

Time = (450,000 m) / (100 m/s)

Time = <em>4500 seconds </em>(that's 75 minutes)

Note:

This is about HALF the speed of the passenger jet you fly in when you go to visit Grandma for Christmas.

If the International Space Station flew at this speed, it would immediately go ker-PLUNK into the ocean.

The speed of the International Space Station in its orbit is more like 3,100 m/s, not 100 m/s.

8 0
3 years ago
Recall that the spring constant is inversely proportional to the number of coils in the spring, or that shorter springs equate t
ruslelena [56]

Answer:

x_1= 0.0425m

Explanation:

Using the tension in the spring and the force of the tension can by describe by

T = kx

, T = mg

Therefore:

m*g = k*x

With two springs, let, T1 be the tension in each spring,  x1 be the extension of each spring.  The spring constant of each spring is 2k so:

T_1 = 2k*x_1

2T_1 = m*g=4k x_1

Solve to x1

x_1=\frac{m*g}{4k}

x_1=\frac{k*x}{4*k}

x_1=\frac{x}{4}

x_1 = 0.170 / 4

x_1= 0.0425m

7 0
3 years ago
A charge of 25 nC is uniformly distributed along a straight rod of length 3.0 m that is bent into a circular arc with a radius o
Greeley [361]

Answer:

E = 31.329 N/C.

Explanation:

The differential electric field dE at the center of curvature of the arc is

dE = k\dfrac{dQ}{r^2}cos(\theta ) <em>(we have a cosine because vertical components cancel, leaving only horizontal cosine components of E. )</em>

where r is the radius of curvature.

Now

dQ = \lambda rd\theta,

where \lambda is the charge per unit length, and it has the value

\lambda = \dfrac{25*10^{-9}C}{3.0m} = 8.3*10^{-9}C/m.

Thus, the electric field at the center of the curvature of the arc is:

E = \int_{\theta_1}^{\theta_2} k\dfrac{\lambda rd\theta  }{r^2} cos(\theta)

E = \dfrac{\lambda k}{r} \int_{\theta_1}^{\theta_2}cos(\theta) d\theta.

Now, we find \theta_1 and \theta_2. To do this we ask ourselves what fraction is the arc length  3.0 of the circumference of the circle:

fraction = \dfrac{3.0m}{2\pi (2.3m)}  = 0.2076

and this is  

0.2076*2\pi =1.304 radians.

Therefore,

E = \dfrac{\lambda k}{r} \int_{\theta_1}^{\theta_2} cos(\theta)d\theta= \dfrac{\lambda k}{r} \int_{0}^{1.304}cos(\theta) d\theta.

evaluating the integral, and putting in the numerical values  we get:

E = \dfrac{8.3*10^{-9} *9*10^9}{2.3} *(sin(1.304)-sin(0))\\

\boxed{ E = 31.329N/C.}

4 0
3 years ago
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