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kondaur [170]
3 years ago
9

The engineer determines that the machine increased at a constant rate the disk’s angular speed from 100 rad/s to 300 rad/s over

a time of 5 seconds. What was the angular displacement of the disk during those 5 seconds?
Physics
1 answer:
aleksklad [387]3 years ago
5 0

The disk's average angular speed during those 5 seconds was

(1/2) (100 + 300) rad/s = 200 rad/sec

Spinning for 5 sec at an average angular speed of 200 rad/s, the disk turned

(200 rad/s) x (5 s)  =  <em>1,000 radians</em>

(That's 500/pi revolutions = roughly 159.2 revolutions.)

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a 2kg marble is moving at 3 m/s when it strikes another 4kg marble moving in the opposite direction at -3 m/s. What will be the
Elan Coil [88]
Momentum is conserved after the collision 

Momentum of 2 Kg before collision = 2 * 3 = 6
Momentum of 4 kg  before collision = 4 * -3 = -12

so 6 + -12 = 2 * -4 + 4 *x   where x is velocity of 4kg marble.

4x - 8 = -6
4x = 2
x = 0.5 

Velocity of 4 kg marble is 0.5 m/s after collision

The 2 kg marble will move in the opposite direction to which it was moving before the collision.
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3 years ago
(Brainliest) How are radio waves used in cell phone wireless communication technology?
poizon [28]

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Hope this helps! Can I have brainliest please?

5 0
3 years ago
Who invented abstract geometry
Brrunno [24]

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6 0
3 years ago
Read 2 more answers
An ideal spring hangs from the ceiling. A 2.15 kg mass is hung from the spring, stretching the spring a distance d = 0.0895 m fr
Igoryamba

Answer:

The kinetic energy of the mass at the instant it passes back through the equilibrium position is 0.06500 J.

Explanation:

Given that,

Mass = 2.15 kg

Distance = 0.0895 m

Amplitude = 0.0235 m

We need to calculate the spring constant

Using newton's second law

F= mg

Where, f = restoring force

kx=mg

k=\dfrac{mg}{x}

Put the value into the formula

k=\dfrac{2.15\times9.8}{0.0895}

k=235.41\ N/m

We need to calculate the kinetic energy of the mass

Using formula of kinetic energy

K.E=\dfrac{1}{2}mv^2

Here, v = A\omega

K.E=\dfrac{1}{2}m\times(A\omega)^2

Here, \omega=\sqrt{\dfrac{k}{m}}^2

K.E=\dfrac{1}{2}m\times A^2\sqrt{\dfrac{k}{m}}^2

K.E=\dfrac{1}{2}kA^2

Put the value into the formula

K.E=\dfrac{1}{2}\times235.41\times(0.0235)^2

K.E=0.06500\ J

Hence, The kinetic energy of the mass at the instant it passes back through the equilibrium position is 0.06500 J.

8 0
3 years ago
During each cycle, a refrigerator ejects 610 kJ of energy to a high-temperature reservoir, and takes in 505 kJ of energy from a
jenyasd209 [6]

Answer

A. the work done on the refrigerant in each cycle is 105kJ

B the coefficient of performance of the refrigerator is 4.8

Explanation

Given data

Work done at high temperature T2 Qh=610kJ

Work done at low temperature T1 Ql=505kJ

We know that the net work done by the refrigerator is expressed as

Wnet= Qh-Ql

=610-505

=105kJ

Also we know that the coefficient of performance is expressed as

COP= Ql/Wnet

COP= 505/105

= 4.8

8 0
3 years ago
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