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Ann [662]
3 years ago
7

A 403.0-kg copper bar is melted in a smelter. The initial temperature of the copper is 320.0 K. How much heat must the smelter p

roduce to completely melt the copper bar? For solid copper, the specific heat is 386 J/kg • K, the heat of fusion is 205 kJ/kg, and the melting point is 1357 K.
Physics
2 answers:
Liono4ka [1.6K]3 years ago
8 0

Answer:

2.42 x 10^8 J

Explanation:

m = 403 kg

T1 = 320 K

c = 386 J/Kg C

L = 205 kJ / kg = 205000 J/Kg

T2 = 1357 K

Heat used to increase the temperature till 1357 K

H1 = m x c x (T2 - T1)

H1 = 403 x 386 x ( 1357 - 320)

H1 = 1.6 x 10^8 J

Heat used to melt

H2 = m x L

H2 = 403 x 205000 = 8.2 x 10^7 J

Total heat used to melt completely

H = H1 + H2

H = 1.6 x 10^8 + 0.82 x 10^8 = 2.42 x 10^8 J

aleksklad [387]3 years ago
7 0
The answer is 2.49 x 10^5 KJ. This was obtained (1) use the formula for specific heat to achieve Q or heat then (2) get the energy to melt the copper lastly (3) Subtract both work and the total energy required to completely melt the copper bar is achieved.
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When velocity is graphed with respect to time, what does the area under
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Answer:

the answer is D

Explanation:

Since the velocity of the object is the derivative of the position graph, the area under the line in the velocity vs. time graph is the displacement of the object.

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3 years ago
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Calculate the first and second velocities of the car with four washers attached to the pulley, using the formulas v1 = 0. 25 m /
castortr0y [4]

The first and second velocities of the car are 0.25/t1 and 0.50/t2 respectively.

Given data:

The value of the first distance is, d1 = 0.25 m.

The value of the second distance is, d2 = 0.50 m.

The ratio of distance covered by an object in a specific direction and the time taken to cover the distance is known as the velocity of the object. Mathematically, the expression for the velocity is,

v = d/t

Here,

d is the distance covered.

t is the average time taken to cover the distance.

Then the first velocity of the car at 0.25 m is,

v1 = d1/t1

v1 = 0.25 / t1

here, t1 is the average time for first distance.

And the second velocity of the car with four washers at the 0. 50 m mark is,

v2 = d2/t2

v2 = 0.50 /t2

here, t2 is the average time for the second distance.

Thus, we can conclude that the first and second velocities of the car are 0.25/t1 and 0.50/t2 respectively.

Learn more about the velocity here:

brainly.com/question/862972

6 0
2 years ago
Two electrons are initially at rest separated by a distance of 2nm. At time t=0, they start to move apart due to Coulombic repul
Gnom [1K]

Answer:

t=2.5\times 10^{-14}\ s

Explanation:

We know that charge on electron

q=1.6\times 10^{-19}\ C

r= 2 nm

We know that force between two charge given

F=K\dfrac{Q_1Q_2}{r^2}

Now by putting the value

F=9\times10^9\dfrac{1.6\times 10^{-19}\times 1.6\times 10^{-19}}{(2\times 10^{-9})^2}

F=5.67\times 10^{-11}\ N

We know that mass of electron

The mass of electron

m=9.1\times 10^{-31}\ kg

F= m a

a= Acceleration of electron

a= F/m

a=\dfrac{5.67\times 10^{-11}}{9.1\times 10^{-31}}\ m/s^2

a=6.2\times 10^{19} m/s^2

S=ut+\dfrac{1}{2}at^2

initial velocity given that zero ,u=0

20\times 10^{-9}=\dfrac{1}{2}\times 6.2\times 10^{19} t^2

t=\sqrt {\dfrac{40\times 10^{-9}}{6.2\times 10^{19}}}

t=2.5\times 10^{-14}\ s

3 0
3 years ago
A 5 kg block is sliding down a plane inclined at 30^0 with a constant velocity of 4 m/s. To determine the coefficient of frictio
Leya [2.2K]

Answer:

The necessary information is if the forces acting on the block are in equilibrium

The coefficient of friction is 0.577

Explanation:

Where the forces acting on the object are in equilibrium, we have;

At constant velocity, the net force acting on the particle = 0

However, the frictional force is then given as

F = mg sinθ

Where:

m = Mass of the block

g = Acceleration due to gravity and

θ = Angle of inclination of the slope

F = 5×9.81×sin 30 = 24.525 N

Therefore, the coefficient of friction is given as

24.525 N = μ×m×g × cos θ = μ × 5 × 9.81 × cos 30 = μ × 42.479

μ × 42.479 N= 24.525 N

∴ μ = 24.525 N ÷ 42.479 N = 0.577

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Answer:

90 kilometers an hour

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