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OleMash [197]
2 years ago
10

The specific heat of copper is 0.39 j/g*degrees celcius how much heat is needed to raise the temperature of 1000.0 g of copper f

rom 25.0 degrees celsius to 45.0 degrees celcius
Physics
1 answer:
lapo4ka [179]2 years ago
6 0
In order to calculate the amount of energy needed, we will apply:
heat energy = mass x specific heat capacity x change in temperature
Q = mcΔT

Substituting the given values,
Q = 1000 x 0.39 x (45 - 25)
Q = 7,800 Joules of heat energy are needed
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Answer:

The speed of the block is 8.2 m/s

Explanation:

Given;

mass of block, m = 2.1 kg

height above the top of the spring, h = 5.5 m

First, we determine the spring constant based on the principle of conservation of potential energy

¹/₂Kx² = mg(h +x)

¹/₂K(0.25)² = 2.1 x 9.8(5.5 +0.25)

0.03125K = 118.335

K = 118.335 / 0.03125

K = 3786.72 N/m

Total energy stored in the block at rest is only potential energy given as:

E = U = mgh

U = 2.1 x 9.8 x 5.5 = 113.19 J

Work done in compressing the spring to 15.0 cm:

W = ¹/₂Kx² = ¹/₂ (3786.72)(0.15)² = 42.6 J

This is equal to elastic potential energy stored in the spring,

Then, kinetic energy of the spring is given as:

K.E = E - W

K.E = 113.19 J - 42.6 J

K.E = 70.59 J

To determine the speed of the block due to this energy:

KE =  ¹/₂mv²

70.59 =  ¹/₂ x 2.1 x v²

70.59 = 1.05v²

v² = 70.59 / 1.05

v² = 67.229

v = √67.229

v = 8.2 m/s

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3 years ago
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Answer:

Acceleration stress, physiological changes that occur in the human body in motion as a result of rapid increase of speed. ... A force of 3 g, for example, is equivalent to an acceleration three times that of a body falling near Earth.

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A certain car engine delivers enough force to create 630 N⋅m of torque when the engine is operating at 3200 revolutions per minu
jekas [21]

The appropriate expression for the calculation of power by relating the angular energy in a given time.

In other words the instantaneous power of an angular accelerating body is the torque times the angular velocity

P=\tau\omega

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\tau = Torque

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Our values are given by

\tau = 630Nm

\omega = 3200rev/min

The angular velocity must be transformed into radians per second then

\omega = 3200rev/min (\frac{2\pi rad}{60s})

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Replacing,

P=(630)(335.103)

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If the sum of the external forces on an object is zero, then the sum of the external torques on it
Zarrin [17]

Answer:

True.

Explanation:

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