The skater is has a mass of 75 kg. Find the total Potential energy of the skater at the top of the ramp at 6 m.

Suppose you lived in the crater Copernicus on the side of the Moon facing Earth.

Wittaler [7]

Answer and Explanation:

In case, I lived in the Crater Copernicus on Moon's side facing the Earth then:

On moon, as it orbits the Earth, it takes 29 and a half days to complete one cycle of sunrise and sunset. So, the would rise once in 29 and a half day.
The moon orbits the Earth in such a way that the its same side faces the Earth always which concludes that the Earth is overhead every time resulting in no Earth rise or Earth set on Moon.
Due to the thin atmosphere, it can't impart any specific color to the sky on the moon as a result of which the moon's sky is always black and hence, stars can be seen every time.

5 0

3 years ago

The 25-lb slender rod has a length of 6 ft. using a collar of negligible mass, its end a is confined to move along the smooth ci

sp2606 [1]

Which of the following is NOT a factor in efficiency?A.metabolismB.type of movementC.muscle efficiencyD.digestion

3 0

3 years ago

An L-R-C series circuit has L = 0.450 H, C=2.50×10^−5F, and resistance R.

Alex777 [14]

Answer:

298rad/s and 116.96 ohms

Explanation:

Given an L-R-C series circuit where

L = 0.450 H,

C=2.50×10^−5F, and resistance R= 0

In this situation we have a simple LC circuit with angular frequency

Wo = 1√LC

= 1/√(0.450)(2.50×10^-5)

= 1/√0.00001125

= 1/0.003354

= 298rad/s

B) Now we need to find the value of R such that it gives a 10% decrease in angular frequency.

Wi/W° = (100-10)/100

Wi/W° = 90/100

Wi/W° = 0.90 ............... 1

Angular frequency of oscillation

The complete aspect of the solution is attached, please check.

6 0

3 years ago

During circular motion, the force that is perpendicular to the velocity and toward the center of the circle is the ... ?

Ivanshal [37]

That's the "centripetal" force. It produces the centripetal acceleration
that pulls the object away from a straight path into a bent path.

5 0

3 years ago

An oscillator consists of a block attached to a spring (k = 427 N/m). At some time t, the position (measured from the system's e

White raven [17]

Answer:

a) 4.49Hz

b) 0.536kg

c) 2.57s

Explanation:

This problem can be solved by using the equation for he position and velocity of an object in a mass-string system:

$x=Acos(\omega t)\\\\v=-\omega Asin(\omega t)\\\\a=-\omega^2Acos(\omega t)$

for some time t you have:

x=0.134m

v=-12.1m/s

a=-107m/s^2

If you divide the first equation and the third equation, you can calculate w:

$\frac{x}{a}=\frac{Acos(\omega t)}{-\omega^2 Acos(\omega t)}\\\\\omega=\sqrt{-\frac{a}{x}}=\sqrt{-\frac{-107m/s^2}{0.134m}}=28.25\frac{rad}{s}$

with this value you can compute the frequency:

a)

$f=\frac{\omega}{2\pi}=\frac{28.25rad/s}{2\pi}=4.49Hz$

b)

the mass of the block is given by the formula:

$f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}\\\\m=\frac{k}{4\pi^2f^2}=\frac{427N/m}{(4\pi^2)(4.49Hz)^2}=0.536kg$

c) to find the amplitude of the motion you need to know the time t. This can computed by dividing the equation for v with the equation for x and taking the arctan:

$\frac{v}{x}=-\omega tan(\omega t)\\\\t=\frac{1}{\omega}arctan(-\frac{v}{x\omega })=\frac{1}{28.25rad/s}arctan(-\frac{-12.1m/s}{(0.134m)(28.25rad/s)})=2.57s$

Finally, the amplitude is:

$x=Acos(\omega t)\\\\A=\frac{0.134m}{cos(28.25rad/s*2.57s )}=0.45m$

5 0

2 years ago