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Alona [7]
4 years ago
7

This is EASY true or false Cars and factories cause pollution

Physics
1 answer:
nadezda [96]4 years ago
3 0

True .................


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What would happen to a balloon placed in the freezer?
Archy [21]
The balloon will shrink because the average kinetic energy of gas molecules in a balloon decreases with fall in temperature. Butif we warm the balloon , it will rise.

MARK IT BRANLIEST PLZ..
3 0
3 years ago
Read 2 more answers
Consider the system consisting of the box and the spring, but not Earth. How does the energy of the system when the spring is fu
BabaBlast [244]

Answer:

the energy when it reaches the ground is equal to the energy when the spring is compressed.

Explanation:

For this comparison let's use the conservation of energy theorem.

Starting point. Compressed spring

         Em₀ = K_e = ½ k x²

Final point. When the box hits the ground

         Em_f = K = ½ m v²

since friction is zero, energy is conserved

          Em₀ = Em_f

          1 / 2k x² = ½ m v²

          v = \sqrt{ \frac{k}{m} }     x

Therefore, the energy when it reaches the ground is equal to the energy when the spring is compressed.

5 0
3 years ago
What type of motion is shown with this graph? (5 points)
Vladimir [108]
It should be Constant speed. The line goes straight & doesn’t change within the graph.
7 0
3 years ago
The radius of earth is 6,370,000 m. Express this measurement in km in scientific notation with the correct number of significant
AlexFokin [52]

Answer:

6.37 x 10³ Km

Explanation:

given,

Radius of earth = 6,370,000 m

we know,

1 km = 1000 m

1 m = 0.001 Km

6,370,000 m =  6,370,000 x 0.001

                       = 6,370 Km

The number 6,370 has 3 significant figure.

To transform this to an exponential number, it is necessary to move the decimal to the left so there is only one digit in front of the decimal point.

Representing the given number in scientific notation

      = 6.37 x 10³ Km

7 0
4 years ago
Positive charge Q is placed on a conducting spherical shell with inner radius R1 and outer radius R2. The electric field at a po
gregori [183]

Answer:

E = 0    r <R₁

Explanation:

If we use Gauss's law

      Ф = ∫ E. dA = q_{int} / ε₀

in this case the charge is distributed throughout the spherical shell and as we are asked for the field for a radius smaller than the radius of the spherical shell, therefore, THERE ARE NO CHARGES INSIDE this surface.

Consequently by Gauss's law the electric field is ZERO

           E = 0    r <R₁

6 0
3 years ago
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