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2 years ago

What's the equation of power

Physics

1 answer:

ad-work [718]2 years ago

4 0

None of the choices is correct.

   Power = work/time
or
      energy/time
or
   (force) x (speed).

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Calculate the power required of a 60-kg person who climbs a tree 5 meters high in 10 seconds.

Alina [70]

Answer:

Explanation:

Power = Energy/time

-Don't have energy so I'm gonna solve for it

Gravitational Potential Energy = mass x gravity x height

= 60 kg x 9.8 m/s2 x 5m

= 2940 J

Power = Energy/time

=2940 J/10 s

= 294 W

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1 year ago

An object is attached to a vertical spring and bobs up and down between points a and

Anton [14]

it would be at either A or B.

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2 years ago

A surgical microscope weighing 200 lb is hung from a ceiling by four springs with stiffness 25 lb/ft. The ceiling has a vibratio

Nikitich [7]

Answer:

If there is no damping, the amount of transmitted vibration that the microscope experienced is = $5.676*10^{-3} \ mm$

Explanation:

The motion of the ceiling is y = Y sinωt

y = 0.05 sin (2 π × 2) t

y = 0.05 sin 4 π t

K = 25 lb/ft × 4 sorings

K = 100 lb/ft

Amplitude of the microscope $\frac{X}{Y}= [\frac{1+2 \epsilon (\omega/ W_n)^2}{(1-(\frac{\omega}{W_n})^2)^2+(2 \epsilon \frac{\omega}{W_n})^2}]$

where;

$\epsilon = 0$

$W_n = \sqrt { \frac{k}{m}}$

= $\sqrt { \frac{100*32.2}{200}}$

= 4.0124

replacing them into the above equation and making X the subject of the formula:

$X =$ $Y * \frac{1}{\sqrt{(1-(\frac{\omega}{W_n})^2)^2})}}$

$X =$ $0.05 * \frac{1}{\sqrt{(1-(\frac{4 \pi}{4.0124})^2)^2})}}$

$X =$ $5.676*10^{-3} \ mm$

Therefore; If there is no damping, the amount of transmitted vibration that the microscope experienced is = $5.676*10^{-3} \ mm$

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3 years ago

Can someone spell the word hi for me I’ll give you 30 points

jasenka [17]

Answer:

hi poop

Explanation:

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2 years ago

Read 2 more answers

A system undergoes a two-step process. In the first step, the internal energy of the system increases by 222 J when 150 J of wor

joja [24]

Answer:0 J

Explanation:

Given

For first step

change in internal Energy of the system is $\Delta U_1=222 J$

Work done on the system $W_1=-150 J$

For second step

change in internal Energy of the system is $\Delta U_2=123 J$

Work done on the system $W_2=-195 J$

Work done on the system is considered as Positive and vice-versa.

and from first law of thermodynamics

$Q=\Delta U+W$

for first step

$Q_1=222-150=72 J$

$Q_2=123-195=-72 J$

overall heat added $=Q_1+Q_2$

$Q_{net}=72-72 =0$

For overall Process Heat added is 0 J

8 0

2 years ago