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Ray Of Light [21]
3 years ago
12

The type of energy that depends on position is called

Physics
2 answers:
swat323 years ago
6 0
I think the answer is A
Maurinko [17]3 years ago
3 0

Answer:

a. potential energy

Explanation:

ur welcome

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A string is wrapped around a pulley with a radius of 2.0 cm. The pulley is initially at rest. A constant force of 50 N is applie
Ray Of Light [21]

Answer:

0.20kg-m^2

Explanation:

Let the linear velocity of the rope(=of pulley) is v m/s

Using kinematic equation

=> v = u + at

=>v = 0 + 4.9a

=>v = 4.9a ------------ eq1

By v^2 = u^2 + 2as

=>v^2 = 0 + 2 x v/4.9 x 1.2

=>4.9v^2 - 2.4v = 0

=>v(4.9v - 2.4) = 0

=>v = 2.4/4.9 = 0.49 m/s

Thus by v = r x omega

=>omega = v/r = 0.49/0.02 = 24.49 rad/sec

BY W = F x s = 50 x 1.2 = 60 J

=>KE(rotational) = W = 1/2 x I x omega^2

=>60 = 1/2 x I x (24.49)^2

=>I = 0.20 kg-m^2

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
You guys cant answer a dam question I ask it fricking 6th grade questions
nexus9112 [7]

Answer:

calm down please its not that serious maybe no one saw it yet

Explanation:

7 0
3 years ago
Assume that the blocks are accelerating, and the x components of their accelerations at a certain moment are a1x and a2x. find t
Serga [27]

we know that center of mass is given as

r = (m₁ r_{1x} + m₂ r_{2x})/(m₁ + m₂)

taking derivative both side relative to "t"

dr/dt = (m₁ dr_{1x}/dt + m₂ dr_{2x}/dt)/(m₁ + m₂)

v = (m₁ v_{1x} + m₂ v_{2x})/(m₁ + m₂)

taking derivative again relative to "t" both side

dv/dt = (m₁ dv_{1x}/dt + m₂ dv_{2x}/dt)/(m₁ + m₂)

a= (m₁ a_{1x} + m₂ a_{2x})/(m₁ + m₂)

3 0
3 years ago
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After scientists analyze the results of their experiments they
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They communicate their result to the scientific community- so to speak
8 0
3 years ago
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