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beks73 [17]
2 years ago
14

Which two objects have stored energy?

Physics
2 answers:
yan [13]2 years ago
5 0
The answer is A stretched rubber band
borishaifa [10]2 years ago
5 0
C.) A<span> stretched rubber band
A.) </span><span>a ball rolling on the ground

Hope this helps!</span>
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The waves with the lowest energy and lowest frequencies of the electromagnetic spectrum are the
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<span>The waves with the lowest energy and lowest frequencies of the electromagnetic spectrum are the "Radio waves"

So, option B is your answer

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3 years ago
A student redid the experiment of mixing room temperature water and warm
Len [333]

Answer:

I think B or C it won't lower so I'll go with B bc warm water is hotter than regular temp water

7 0
2 years ago
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Velocity hdjsksskskskdldlddld​
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4 0
2 years ago
A spring of constant 20 N/m has compressed a distance 8 m by a(n) 0.3 kg mass, then released. It skids over a frictional surface
Fiesta28 [93]

Answer:

X_2=25.27m

Explanation:

Here we will call:

1. E_1: The energy when the first spring is compress

2. E_2: The energy after the mass is liberated by the spring

3. E_3: The energy before the second string catch the mass

4. E_4: The energy when the second sping compressed

so, the law of the conservations of energy says that:

1. E_1 = E_2

2. E_2 -E_3= W_f

3.E_3 = E_4

where W_f is the work of the friction.

1. equation 1 is equal to:

\frac{1}{2}Kx^2 = \frac{1}{2}MV_2^2

where K is the constant of the spring, x is the distance compressed, M is the mass and V_2 the velocity, so:

\frac{1}{2}(20)(8)^2 = \frac{1}{2}(0.3)V_2^2

Solving for velocity, we get:

V_2 = 65.319 m/s

2. Now, equation 2 is equal to:

\frac{1}{2}MV_2^2-\frac{1}{2}MV_3^2 = U_kNd

where M is the mass, V_2 the velocity in the situation 2, V_3 is the velocity in the situation 3, U_k is the coefficient of the friction, N the normal force and d the distance, so:

\frac{1}{2}(0.3)(65.319)^2-\frac{1}{2}(0.3)V_3^2 = (0.16)(0.3*9.8)(2)

Volving for V_3, we get:

V_3 = 65.27 m/s

3. Finally, equation 3 is equal to:

\frac{1}{2}MV_3^2 = \frac{1}{2}K_2X_2^2

where K_2 is the constant of the second spring and X_2 is the compress of the second spring, so:

\frac{1}{2}(0.3)(65.27)^2 = \frac{1}{2}(2)X_2^2

solving for X_2, we get:

X_2=25.27m

3 0
3 years ago
Which name is given to the type of friction that an objects falling through the air experience
andrey2020 [161]
It is called Air resistance
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