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Tems11 [23]
3 years ago
10

15 g of lead (specific heat = 0.128 J/g Celsius) at 120 Degrees Celsius is placed on a glacier of ice at 0 degrees Celsius. How

much ice melts? (Lf of ice = 333 J/g)
Physics
1 answer:
Anastaziya [24]3 years ago
4 0

This can be solve using heat balance. The amount of heat the lead released is equal to the amount of heat the ice absorb.

The heat release by the leas = m1*Cp*deltaT

Where m1 is the mass of lead

Cp is the specific heat of lead

Delta is the change of temperature

Heat absorbed by the ice = m2Lf

Where m2 is the mass of ice

Lf is the latent heat of fusion of ice

 

M1*Cp*deltaT = m2Lf

(15g) ( 0.128 j/g C) ( 120 -0 C) = m2(333/g)

Solving for m2

<span>M2 = 0.69 kg of ice melts</span>

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A student forms the following hypothesis if people wear bright colors, they will be happier. What is/are the problems with this
mars1129 [50]
There is no scientific evidence to support this claim. And it is not telling what colors are needing to be worn only the stated bright colors which can mean an assortment of things.
8 0
3 years ago
The sum of kinetic energy of molecules in a body is ______​
Margarita [4]
Mechanical energy is the answer
6 0
2 years ago
A 27 kg bear slides, from rest, 14 m down a lodgepole pine tree, moving with a speed of 6.1 m/s just before hitting the ground.
Nadusha1986 [10]
<h2>Thus the force of friction is 235 N</h2>

Explanation:

When the bear was at the height of 14 m . Its potential energy = m g h

here m is the mass of bear , g is acceleration due to gravity and h is the height .

Thus P.E =  27 x 10 x 14 = 3780 J

The K.E of the bear just before hitting = \frac{1}{2} m v²

=   \frac{1}{2} x 27 x ( 6.1 )²  = 490 J

The force of friction f = P.E - K.E = 3290 J

Because the work done = Force x Distance

Thus frictional force = \frac{3290}{14} = 235 N

3 0
3 years ago
You get hit in the head with a baseball as it is falling. What are the action and reaction forces?
german

Answer:

Action: Gravity pulls on the ball.

Reaction: The ball falls to the ground.

Explanation:

5 0
2 years ago
Three wires meet at a junction. Wire 1 has a current of 0.40 A into the junction. The current of wire 2 is 0.57 A out of the jun
AlekseyPX

Answer:

a. 1.56 × 10¹⁸ electrons per second

b. The electrons in wire 3 flow into the junction.

Explanation:

Here is the complete question

Three wires meet at a junction. Wire 1 has a current of 0.40 A into the junction. The current of wire 2 is 0.65 A out of the junction. (a) How many electrons per second move past a point in wire 3? (b) In which direction do the electrons move in wire 3 -- into or out of the junction?

Solution

(a) How many electrons per second move past a point in wire 3?

Using Kirchhoff's current law, at the junction, i₁ + i₂ + i₃ = 0 where i₁ = current in wire 1 = 0.40 A, i₂ = current in wire 2 = 0.65 A and  i₃ = = current in wire 3,

So, i₃ = -(i₁ + i₂)

taking current flowing into the junction as positive and those leaving as negative, i₁ = + 0.40 A and i₂ = -0.65 A

So, i₃ = -(i₁ + i₂)

i₃ = -(0.40 A + (-0.65 A))

i₃ = -(0.40 A - 0.65 A)

i₃ = -(-0.25 A)

i₃ = 0.25 A

Since i₃ = 0.25 C/s and we have e = 1.602 × 10⁻¹⁹ C per electron, then the number of electrons flowing in wire 3 per second is i₃/e = 0.25 C/s ÷ 1.602 × 10⁻¹⁹ C per electron = 0.1561  × 10¹⁹ electrons per second = 1.561  × 10¹⁸ electrons per second ≅ 1.56 × 10¹⁸ electrons per second

(b) In which direction do the electrons move -- into or out of the junction?

Given that i₃ = + 0.25 A and that positive flows into the junction, thus, the electrons in wire 3 flow into the junction.

8 0
3 years ago
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