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3 years ago

9

A rubber band has a spring constant of 45 N/m. A classmate stretches it a length of 0.2 m before firing it at Mr. C. Ouch. Calcu

late the elastic
potential energy of the rubber band.

1 answer:

nata0808 [166]3 years ago

5 0

Answer:

<h2>0.9 Joules</h2>

Explanation:

Elastic potential energy of a spring= 1/2 × Spring constant × displacement²

following calculations you will get ur answer!!

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Physics question, please help?

Ludmilka [50]

0.4823 m/s

The initial velocity u1 of the ball=0

From the law of conservation of linear momentum.

m1u1+m2u2=m1v1+m2v2

(160×0)+(170×u1)=(160×0.3)+(170×0.2)

u1=0.4823m/s

6 0

3 years ago

An Atwood machine consists of two masses, mA = 6.8 kg and mB = 8.0 kg , connected by a cord that passes over a pulley free to ro

Lisa [10]

To solve this problem it is necessary to apply the concewptos related to Torque, kinetic movement and Newton's second Law.

By definition Newton's second law is described as

F= ma

Where,

m= mass

a = Acceleration

Part A) According to the information (and as can be seen in the attached graph) a sum of forces is carried out in mass B, it is obtained that,

$\sum F = m_b a$

$m_Bg-T_B = m_Ba$

$T_B = m_Bg-m_Ba$

In the case of mass A,

$\sum F = m_A a$

$T_A = m_Ag-m_Aa$

Making summation of Torques in the Pulley we have to

$\sum\tau = I\alpha$

$T_BR_0-T_AR_0=I\alpha$

$T_B-T_A=I\frac{a}{R^2_0}$

Replacing the values previously found,

$(m_Bg-m_Ba )-(m_Ag-m_Aa )=I\frac{a}{R^2_0}$

$(m_B-m_A)g-(m_B+m_A)a=I\frac{a}{R^2_0}$

$a = \frac{(m_B-m_A)g}{\frac{I}{R_0^2}+(m_B+m_A)}$

$a = \frac{(m_B-m_A)g}{\frac{MR^2_0^2/2}{R_0^2}+(m_B+m_A)}$

$a =\frac{(m_B-m_A)g}{\frac{M}{2}+(m_B+m_A)}$

Replacing with our values

$a =\frac{(8-6.8)(9.8)}{\frac{0.8}{2}+(8+6.8)}$

$a=0.7736m/s^2$

PART B) Ignoring the moment of inertia the acceleration would be given by

$a' =\frac{(m_B-m_A)g}{(m_B+m_A)}$

$a' =\frac{(8-6.8)(9.8)}{(8+6.8)}$

$a' = 0.7945$

Therefore the error would be,

$\%error = \frac{a'-a}{a}*100$

$\%error = \frac{0.7945-0.7736}{0.7736}*100$

$\%error = 2.7%$

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3 years ago

A grinding wheel with a moment of inertia of 2 kg-m2 has a 2.50 N-m torque applied to it. What is its final kinetic energy 10 se

grigory [225]

First you do 2.5 n-m / 2 kg-m2
This equals 1.25 rad/s^2

Then multiply rad/s^2 by 10 because of the amount of seconds
This equals 12.5 rad/s^2

Then do this following equation : (1/2)Iwf^2
Do (0.5) 12.5 * 12.5 * 2
This equals 156.25 J (meaning that the answer is C: 156 J)

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3 years ago

Phoebe's insulated foam cup is filled with 0.15 kg of coffee (mostly water) that is too hot to drink, so she adds

Nesterboy [21]

Answer:

65 Celsius.

Explanation:

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3 years ago

Plrase help a little stuck here

kkurt [141]

(I'm lucky to have a computer ... It was only through the miracle of
modern digital technology that I was able to flip your photo right-
side-up to where I could read it.)

Here's how to figure out things like this:

The circle on the left side labeled ' <em>G</em> ' is the <em><u>G</u></em>enerator or battery
that powers this whole circuit and all the devices in it. In order for
any device to work, you need to be able to set your pencil down at
the top of the Generator, and find a path through the circuit and
through that device, where current can flow all the way around to
the bottom of the Generator. If you ever come to an open switch,
then current stops there, and you have to find another way through.
If the path you found takes you back to the bottom of the generator but
it doesn't go through one of the devices, then that device doesn't work.

Look at the picture. If you open switch S-4, then Device-4 can't work,
because current can't go through it from one end of the Generator to
the other end. But all of the other devices still work.

I can see 2 ways to turn off Device-3 with a single switch ... either
open switch S-5, or else open switch S-1. Unfortunately, I think
either way will shut off all 5 devices.

7 0

3 years ago