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3 years ago

Heater supplies 80 J of energy to a block of metal. The temperature of the block rises by 20 °C.

Physics

1 answer:

Grace [21]3 years ago

5 0

Answer:

Its internal energy decreases by 40 J.

Explanation:

The energy supplied to the metal block causes the temperature of the block to rise. Therefore,

$Heat\ Supplied = mC\Delta T\\80\ J = mC(20^oC)\\mC = 4\ J/^oC\\$

Now, we will use this value of mC to find out the energy change of block:

$Energy\ Change = mC\Delta T\\Energy\ Change = (4\ J/^oC)(-10^oC)\\Energy\ Change = -40J\\$

It shows that when the temperature of the metal block falls by 10°C,

<u>Its internal energy decreases by 40 J.</u>

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While on the moon, the Apollo astronauts enjoyed the effects of a gravity much smaller than that on

Ainat [17]

Answer:

1.628 $\frac{m}{sec^{2} }$

Explanation:

Anywhere in the universe, In a closed system, <u>Conservation of energy</u> is applicable.

In this case

Neil is initially on the surface of moon and has a velocity of 1.51 $\frac{m}{sec}$ in upward direction.

⇒He has Kinetic energy= $K_{i}$ = $\frac{1}{2} m{v^{2} }$ J

But with respect to the surface of the moon,

where m=mass of moon

v=velocity of Neil

He has Potential energy= $P_{i}$ =0 J

At the highest point of his jump, his velocity =0

⇒ Kinetic energy= $K_{f}$ =0 J

His Potential energy with respect to the surface of moon= $P_{f}$ = $m \times g\times h$

where m=mass of moon

g= gravitational acceleration on moon

h=height from moon's surface

By Conservation Energy Principle

$K_{i}$ + $P_{i}$ = $K_{f}$ + $P_{f}$

$K_{i}$ +0=0+ $P_{f}$

$\frac{1}{2} m{v^{2} }$ = $m \times g\times h$

$\frac{v^{2} }{2}$ = $g\times h$

$\frac{1.5^{2} }{2}$ J= $g\times 0.7 m$

⇒ g = $\frac{1.14}{0.7}$ = 1.628 $\frac{m}{sec^{2} }$

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3 years ago

_ NaBr + __ H3PO4 → ___ Na3PO4 + __ HBr

algol [13]

Answer:

The balanced equation is 3NaBr+1H3PO4 ----> 1Na3PO4 + 3HBr

This is a double replacement because you are switching both the Na and the Hydrogen.

Explanation:

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List the types of radiation in order from least to most damaging.

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Nuclear is the most damaging, then alpha, beta, gamma.

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3 years ago

A steel plate weighing 180 lb with center of gravity at point G is supported by a roller at point A, a bar DE, and a horizontal

melamori03 [73]

Answer:

Therefore, the force supported by the hydraulic cylinder is = -70.01 lb

Thus, the force supported by the bar = -233.84 lb

The reaction force supported by the reaction of the roller = 341.31 lb

Explanation:

The attached diagrams is shown in the file below:

From there; taking moments about point A

$\sum M__A }=0$

- 40 (180) - (80) $F_E$ Cos 37° + 55

-7200 + $F_E$ (-80 Cos 37° + 55 sin 37° ) = 0

= - 7200 - 30.79 $F_E$

- 30.79 $F_E$ = 7200

$F_E$ = $-\frac{7200}{30.79}$

$F_E$ = -233.84 lb

Thus, the force supported by the bar = -233.84 lb

Taking the equilibrium of forces in the vertical direction

$\sum f_y = 0$

$F_E$ sin 37 + $F_A$ cos 20 - $F_G$ = 0

- 233.84 sin 37 + $F_A$ cos 20 - 180 = 0

$F_A$ cos 20 = 320.73

$F_A$ = $\frac{320.73}{cos 20}$

$F_A$ = 341.31 lb

The reaction force supported by the reaction of the roller = 341.31 lb

Taking the equilibrium of forces on the horizontal direction.

$\sum f_y = 0$

$F_E$ cos 37 - $F_{CB} + F_A$ sin 20° = 0

-233.84 cos 37 - $F_{CB}$ + 341.31 sin 20° = 0

-186.75 - $F_{CB}$ + 116.74 = 0

- $F_{CB}$ -70.01 = 0

- $F_{CB}$ = 70.01

$F_{CB}$ = - 70.01 lb

Therefore, the force supported by the hydraulic cylinder is = -70.01 lb

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4 years ago