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viktelen [127]
2 years ago
14

Reposting the repost because it got deleted TvT

Physics
1 answer:
kakasveta [241]2 years ago
5 0

Answer:

Because water mixes with the dirt when it rains to make that mud when the mud is left in the sun, the sun <em>Evaporates</em><em> </em>that water, and eventualy it will turn into sand.

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Heat transfer between two substances is affected by specific heat and the
IgorLugansk [536]

The correct answer is D. Amount of time and area of physical contact between the substances.

Explanation:

Heat transfer refers to the flow of thermal energy or heat between two or more objects. This process involves multiple factors and implies heat from the hottest object goes to the coldest one until there is an equilibrium. To begin, heat transfer depends on the amount of thermal energy in the objects because objects must have a different amount of thermal energy for heat to flow.

Besides this, the amount of energy that flows depends on the time and the contact between the substances of objects. Indeed, objects need to be in contact or close to each other for heat to transfer, and the time needs to be enough for the process to occur. For example, if you place a pot over the fire just for a few seconds it is likely the heat transferred is minimal, which does not occur if you leave the pot more time. At the same time if the pot is in close contact with fire more heat will be transferred.-

6 0
3 years ago
If you were trying to build a soundproof room, which of the following materials would you choose in order to absorb the most sou
leonid [27]

Explanation :  

Absorption coefficient of a material determines how much sound is absorbed by the material.

To build a soundproof room, Heavy curtains and carpet can be used. They reduce reverberation.

Reverberation means an echoing sound which persists for some time. For example, when we bang on a huge piece of metal, we hear the reverberation even after we stop banging.  

Hence, option (A) and (D) are correct.

5 0
3 years ago
Read 2 more answers
An object is lifted from the surface of aspherical planet to an altitude equal to the radius of the planet.As a result, what hap
serg [7]

An object is lifted from the surface of a spherical planet to an altitude equal to the radius of the planet.  

As a result, the object's <em>mass remains the same</em>, and its <em>weight decreases</em> to 1/4 of whatever it is when the object is on the planet's surface.

8 0
3 years ago
The electric field strength E₀ is measured at a perpendicular distance R from an infinitely large, thin sheet that contains a un
Troyanec [42]

Answer:

Explanation:

E=(σ/ε0)

As noted by Dirac the field is the same no matter how far you are from the sheet. When your charge covers a conducting plane, as in your case, the field is, D/eo ,(D is charge density). Because the field inside the conductor (no matter how thin) is zero. The only time the field is, D/2eo, is when you have just a sheet of charge, by itself, not on a conducting plane."

5 0
3 years ago
61. A physics student has a single-occupancy dorm room. The student has a small refrigerator that runs with a current of 3.00 A
Mademuasel [1]

Answer:

Part a)

percentage = 21.3%

Part b)

percentage = 2.13 \times 10^{-5}%

Explanation:

As we know that total power used in the room is given as

P = P_1 + P_2 + P_3 + P_4

here we have

P_1 = (110)(3) = 330 W

P_2 = 100 W

P_3 = 60 W

P_4 = 3 W

P = 330 + 100 + 60 + 3

P = 493 W

Part a)

Since power supply is at 110 Volt so the current obtained from this supply is given as

110\times i = 493

i = 4.48 A

now resistance of transmission line

R = \frac{\rho L}{A}

R = \frac{(2.8 \times 10^{-8})(10\times 10^3)}{\pi(4.126\times 10^{-3})^2}

R = 5.23 \ohm

now power loss in line is given as

P = i^2 R

P = (4.48)^2(5.23)

P = 105 W

Now percentage loss is given as

percentage = \frac{loss}{supply} \times 100

percentage = \frac{105}{493} \times 100

percentage = 21.3%

Part b)

now same power must have been supplied from the supply station at 110 kV, so we have

110 \times 10^3 (i ) = 493

i = 4.48\times 10^{-3} A

now power loss in line is given as

P = i^2 R

P = (4.48 \times 10^{-3})^2(5.23)

P = 1.05 \times 10^{-4} W

Now percentage loss is given as

percentage = \frac{loss}{supply} \times 100

percentage = \frac{1.05 \times 10^{-4}}{493} \times 100

percentage = 2.13 \times 10^{-5}%

6 0
3 years ago
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