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Solnce55 [7]
3 years ago
15

How much heat has to be added to 508 g of copper at 22.3°C to raise the temperature of the copper to 49.8°C? (The specific heat

of copper is 0.377 J/g·°C.)
Physics
1 answer:
viva [34]3 years ago
8 0

Answer:

Q = 5267J

Explanation:

Specific heat capacity of copper (S) = 0.377 J/g·°C.

Q = MSΔT

ΔT = T2 - T1

ΔT=49.8 - 22.3 = 27.5C

Q = change in energy = ?

M = mass of substance =508g

Q = (508g) * (0.377 J/g·°C) * (27.5C)

Q= 5266.69J

Approximately, Q = 5267J

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One block rests upon a horizontal surface. A second identical block rests upon the first one. The coefficient of static friction
goblinko [34]

Answer:

The magnitud of the force is 124.8N.

Explanation:

First we have to find the value of the static friction coefficient, when the external force F is applied to upper block (i will call it A Block) we have a free body diagram as the one shown in the figure i attached, so since this block has no aceleration in any direction the force F should be equal to the friction force between A and B block, one we noticed this we can use the equation for the Friction force to find the coefficient:

0=F-FrictionAB

F=FrictionAB=Nab*μs

and again, since the block has no acceleration the normal between A and B block should be equal to the weigth of the first block, so we have:

0=Nab-W

Nab=W=mg

replacing this we have:

F=μs*Nab=μs*mg=41.6N

and  μs=41.6N/(mg)

now it's time to see the free body diagram for the b block, if we now apply the F force to the B block the diagram should look like in the figure.

the color of the arrow gives you an idea of where the force comes from, the blue ones comes from the B block, the red ones from the A block and the brown ones from the ground.

now for the B block you can see two friction forces, one for the ground and one for the A block, both of these directed bacwards, and two normal forces, again one for the ground and one for the A block but the normal force for the A block is aiming downwards.

again we use the fact that the block is not accelerating in any direction so the sum of the forces in x and y direction have to be 0, so:

F-Friction1(ground)-Friction2(AB)=0

This is the new external F force that we are looking for:

F=Friction1(ground)+Friction2(AB)

we know Friction2(AB) because we found that in the previous block so:

F=Friction1(ground)+mg*μs

for the other friction we have to use the equation:

Friction(ground)=N(ground)*μs

from y axis we have:

N(ground)-w-Normal(AB)=0

N(ground)=w+Normal(AB)

we found the value of Normal(AB) with the previous block so:

N(ground)=mg+mg=2mg

and:

Friction(ground)=2mg*μs

F=Friction(ground)+mg*μs

F=2mg*μs+μs*mg=3mg*μs

and since: μs*mg=41.6N

the new F force would be:

F=3mg*μs=41.6*3=124.8N

4 0
3 years ago
Question 14 (1 point)
mash [69]
(c) as the change in the dependent variable is in direct CORRELATION to the change in the independent variable.
7 0
3 years ago
Which of the following has kinetic energy? (1 point)
mr_godi [17]

Ans:

C

Explanation:

because it is moving down the stairs

7 0
3 years ago
If a reaction starts with 30 grams how many should it end with?
GrogVix [38]
30 grams because of conservation
6 0
3 years ago
Read 2 more answers
A schoolbus accelerates to 65 mph and enters the freeway. It travels for 2.3 hours at that speed while on the freeway. What's th
PtichkaEL [24]

Answer:

The distance travelled on the freeway is 149.5 miles.

Explanation:

The school bus travels on the freeway at constant speed. According to the statement, we need to calculate the distance travelled by the vehicle by means of the following formula:

x = v\cdot t (1)

Where:

x - Traveled distance, in miles.

v - Speed, in miles per hour.

t - Time, in hours.

If we know that v = 65\,\frac{mi}{h} and t = 2.3\,h, then the distance travelled by the school bus is:

x = v\cdot t

x =  \left(65\,\frac{mi}{h} \right)\cdot (2.3\,h)

x = 149.5\,mi

The distance travelled on the freeway is 149.5 miles.

5 0
3 years ago
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