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3 years ago

When a car is parked in hot sunlight, the temperature of the car’s seats _ and the thermal energy of the seats _ .

2 answers:

7 0

Answer: When a car is parked in hot sunlight, the temperature of the car’s seats increases and the thermal energy of the seats increases.

Explanation:

The radiation of the Sun heats the air trapped inside the car which eventually heats everything inside the car including the seats of the car. The temperature of the seat rises.

The thermal energy is related to the kinetic energy of the constituent particles of a body. When a body is heated up, its temperature increases which in turn increases the kinetic energy of the particles thereby increasing the thermal energy of the body.

Here, rise in temperature of the seats increases the thermal energy of the seats.

defon3 years ago

4 0

More info? I think the question is incomplete. Although, I believe the first 2 blanks are "rises"

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What does KE mean? *100 pts*

Alchen [17]

Answer:

it simply means KINETIC ENERGY

Explanation:

In physics, the kinetic energy (KE) of an object is the energy that it possesses due to its motion. It is defined as the work needed to accelerate a body of a given mass from rest to its stated velocity. Having gained this energy during its acceleration, the body maintains this kinetic energy unless its speed changes

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2 years ago

Read 2 more answers

What must be attached to each carburetor on a gasoline inboard engine?

alekssr [168]

To each carburetor on a gasoline inboard engine a backfire flame arrestor must be attached.This arrestor will prevent flames from the backfire causing a fire on board. Several things are important in order the backfire arrestor to function properly:
- should be clean and undamaged.
- If there is a hole in the grid, or oil or gasoline in the grid, or if it is not properly attached, the arrestor will not work correctly.
- must be approved by the U.S. Coast Guard

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3 years ago

A small compass is placed near a very large piece of ferromagnetic material. Before the compass is placed near the material, it

NeTakaya

Answer:

it will move towards the object's magnetic south

Explanation:

The compass pints towards the earth geographic north because the magnetic south of the earth's magnetic field is located in there, if you placed such compass neaar the piece of ferromagnetic material, the magnetic field produced by the magnet will make the compass needle point towards its south magnetic pole, in the same fashion as it points to the earth's magnetic south. It will point to the object's south pole because the magnetic field will be stronger than the earth's (which is weak) that is because of the way magnetism works, opposite poles are attracted and similar poles will tend to separate from each other

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3 years ago

12. A frain moves from rest to a speed of 25 m/s in 30.0 seconds. What is its acceleration?

ziro4ka [17]

initial velocity=u=0m/s
Final velocity=v=25m/s
Time=t=30s

$\\ \tt\longmapsto Acceleration=\dfrac{v-u}{t}$

$\\ \tt\longmapsto Acceleration=\dfrac{25-0}{30}$

$\\ \tt\longmapsto Acceleration=\dfrac{25}{30}$

$\\ \tt\longmapsto Acceleration=0.8m/s^2$

6 0

2 years ago

A space probe is fired as a projectile from the Earth's surface with an initial speed of 2.05 104 m/s. What will its speed be wh

Elanso [62]

Answer:

The value is $v = 2.3359 *10^{4} \ m/s$

Explanation:

From the question we are told that

The initial speed is $u = 2.05 *10^{4} \ m/s$

Generally the total energy possessed by the space probe when on earth is mathematically represented as

$T__{E}} = KE__{i}} + KE__{e}}$

Here $KE_i$ is the kinetic energy of the space probe due to its initial speed which is mathematically represented as

$KE_i = \frac{1}{2} * m * u^2$

=> $KE_i = \frac{1}{2} * m * (2.05 *10^{4})^2$

=> $KE_i = 2.101 *10^{8} \ \ m \ \ J$

And $KE_e$ is the kinetic energy that the space probe requires to escape the Earth's gravitational pull , this is mathematically represented as

$KE_e = \frac{1}{2} * m * v_e^2$

Here $v_e$ is the escape velocity from earth which has a value $v_e = 11.2 *10^{3} \ m/s$

=> $KE_e = \frac{1}{2} * m * (11.3 *10^{3})^2$

=> $KE_e = 6.272 *10^{7} \ \ m \ \ J$

Generally given that at a position that is very far from the earth that the is Zero, the kinetic energy at that position is mathematically represented as

$KE_p = \frac{1}{2} * m * v^2$