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Colt1911 [192]
3 years ago
5

(HELP ASAP) A student’s cell phone battery is almost dead, and she plugs it into an electrical outlet to charge. As it charges,

she notices a light is glowing red to indicate that it is charging; she also notices that the phone is warm. After the battery has been fully charged and the student unplugs her cell phone, the phone vibrates when a message arrives.

Physics
1 answer:
Leokris [45]3 years ago
6 0

Explanation:

The total energy of the system remains conserved. It just converts from one form of energy to another.

The battery of the cell phone contains chemicals. As the electricity is passed through the cell phone, the electrical energy is converted to chemical energy. As she notices that the phone is warm and after the battery has been fully charged and the student unplugs her cell phone, the phone vibrates when a message arrives, this shows the conversion of chemical energy to mechanical energy, thermal energy etc.

Hence, we can say that the correct option is (b).

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A butterfly flies from the top of a tree in the center of a garden to rest on top of a red flower at the garden's edge. The tree
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The butterfly takes a vertical perpendicular path equivalent to 9m and travels a horizontal distance of 17m. The net path between the two is equivalent to that of the hypotenuse, so we will apply the Pythagorean theorem.

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3 0
3 years ago
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A boy standing throws a penny horizontally at 7.25 m/s out of the window of his apparent buliding. If the window is 10.0 m above
Ksivusya [100]

Answer:

Explanation:

This is a 2D problem (parabolic) so we have to think that way. We have to split up the problem into its 2 dimensions to solve it. Think "y-stuff" and "x-stuff".

In the y-stuff category:

v₀ = 0 (initial upwards velocity is 0 since we are told the penny is thrown horizontally)

Δx = -10.0 m (this displacement is negative because the penny lands 10.0 m below the point from which it was thrown)

a = -9.8 m/s/s

t = ? (we need to find the time in this dimension so we can use it in the x dimension to find the displacement, our unknown)

In the "x-stuff" category:

v₀ = 7.25 m/s (this is given)

Δx = ???

a = 0 (acceleration in this dimension is ALWAYS 0)

t = (we will solve for this in the y-dimension and plug it in here).

In the y dimension:

Δx = v₀t + \frac{1}{2}at^2 and plugging in from the y-dimension info:

-10.0=0t+\frac{1}{2}(-9.8)t^2 which simplifies to

-10.0=-4.9t^2 so

t=\sqrt{\frac{-10.0}{-4.9} } which, to 2 significant digits is

t = 1.4 seconds

Now we will do the same in the x-dimension, using t = 1.4:

Δx = v₀t + \frac{1}{2}at^2 and filling in the x-stuff:

Δx = 7.25(1.4)+\frac{1}{2}(0)(1.4)^2 Notice that the stuff after the + sign goes to 0 cuz of the multiplication of 0, so what we are left with is another form of the d = rt equation:

Δx = 7.25(1.4) + 0 so

Δx = 1.0 × 10¹ m (That's rounded correctly to 2 sig dig's: 10 m from the base of the building).

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