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3 years ago

14

Copper has a specific heat of 0.385 j. A piece of copper absorbs 5000 J of energy and undergoes a temperature change from 100 Ce

lsius to 200 degrees Celsius. What is the mass of the piece of copper?

1 answer:

Svetach [21]3 years ago

3 0

Energy= mcT
5000 = m• 0.385 • (200-100)
m = 130kg

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An object moving with a constant

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Answer:

Acceleration:

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From third equation:

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If the coefficient of static friction between your friend and the car seat is 0.500 and you keep driving at a constant speed of

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Let g = 9.8 m/s2. The friction is the product of normal force and its coefficient, and normal force is equal to gravity

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For the friend to slide over, then the centripetal acceleration must be equal to friction acceleration.

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What is mechanical energy

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Explanation:

In physical sciences, mechanical energy is the sum of potential energy and kinetic energy. It is the macroscopic energy associated with a system. The principle of conservation of mechanical energy states that if an isolated system is subject only to conservative forces, then the mechanical energy is constant. If an object moves in the opposite direction of a conservative net force, the potential energy will increase; and if the speed (not the velocity) of the object changes, the kinetic energy of the object also changes. In all real systems, however, nonconservative forces, such as frictional forces, will be present, but if they are of negligible magnitude, the mechanical energy changes little and its conservation is a useful approximation. In elastic collisions, the kinetic energy is conserved, but in inelastic collisions some mechanical energy may be converted into thermal energy. The equivalence between lost mechanical energy (dissipation) and an increase in temperature was discovered by James Prescott Joule.

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The 20-g bullet is travelling at 400 m/s when it becomes embedded in the 2-kg stationary block. The coefficient of kinetic frict

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Answer:

The distance the block will slide before it stops is 3.3343 m

Explanation:

Given;

mass of bullet, m₁ = 20-g = 0.02 kg

speed of the bullet, u₁ = 400 m/s

mass of block, m₂ = 2-kg

coefficient of kinetic friction, μk = 0.24

Step 1:

Determine the speed of the bullet-block system:

From the principle of conservation of linear momentum;

m₁u₁ + m₂u₂ = v(m₁ + m₂)

where;

v is the speed of the bullet-block system after collision

(0.02 x 400) + (2 x 0) = v (0.02 + 2)

8 = v (2.02)

v = 8/2.02

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Step 2:

Determine the time required for the bullet-block system to stop

Apply the principle of conservation momentum of the system

$v(m_1+m_2) -F_kt = v_f(m_1 +m_2)\\\\v(m_1+m_2) -N \mu_kt = v_f(m_1 +m_2)\\\\v(m_1+m_2) -g(m_1 +m_2) \mu_kt = v_f(m_1 +m_2)\\\\3.9604(2.02)-9.8(2.02)0.24t = v_f(2.02)\\\\8 - 4.751t = 2.02v_f\\\\3.9604 - 2.352t = v_f$

when the system stops, vf = 0

3.9604 -2.352t = 0

2.352t = 3.9604

t = 3.9604/2.352

t = 1.684 s

Thus, time required for the system to stop is 1.684 s

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Now, recall that t = 1.684 s

S = 3.9604(1.684) - 1.176(1.684)²

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