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Lady_Fox [76]
2 years ago
8

When the balloon hits the ground, it rebounds slightly. What is the source of the energy for this rebound? A. When the balloon h

its the ground, the water warms; this thermal energy is converted to gravitational potential energy of the rebounding balloon. B. As the balloon hits the ground, the edges rise up; this gravitational potential energy is shared with the rest of the balloon. C. When the balloon hits the ground, the rubber envelope stretches, storing elastic potential energy; this elastic potential energy is converted to the gravitational potentiaL
Physics
1 answer:
nevsk [136]2 years ago
7 0

Answer:

The correct answer is c.    When the balloon hits the ground, the rubber envelope stretches, storing elastic potential energy; this elastic potential energy is converted to the gravitational potentiaL

Explanation:

Let's analyze the situation of the globe

When it touches the ground, the part that is in contact decreases its velocity to zero, but the upper part of the ball continues to move, which creates that the molecules approach slightly, if we approximate the spring links, a repulsive force is created that after all the particles reach zero speed. The force of the springs moves the ball up until the force decreases to zero.

We can relate this force of Hooke with an elastic energy

This energy can be stored in the deformation of the system, as elastic potential energy, which is subsequently transformed into gravitational potential energy when the balloon is lifted.

The correct answer is c

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Calculate the force between charges of 5.0 x 10^-8 c and 1.0 x 10^-7 if they are 5.0 feet apart
weqwewe [10]

Answer:

F = 19.375 x 10^-6  N

Explanation:

This problem can be solved by applying Coulomb's Law, which lets us determine the force between two electrically charged particles.

It is defined as

F = (ke * q1 * q2)/ r^2

Where,

ke =  is Coulomb's constant ≈ 9×10^9 N⋅m^2⋅C^−2

q1 = 5.0 x 10^-8 C

q2 = 1.0 x 10^-7 C

r = 5 ft = 1,524 m

F  = (9×10^9 N⋅m^2⋅C^−2)*(5.0 x 10^-8 C)*(1.0 x 10^-7 C)/ ((1,524 m)^2)

F  = (9×10^9 N⋅m^2⋅C^−2)*(5.0 x 10^-8 C)*(1.0 x 10^-7 C)/ ((1,524 m)^2)

F = 19.375 x 10^-6  N

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2 years ago
A mosquito flying over a highway strikes a windshield of a moving truck. Compared to the magnitude of the force of the truck on
krok68 [10]

Answer:

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<em>The</em><em> </em><em>Same</em><em> </em>

<em>newton's</em><em> </em><em>law</em><em> </em><em>#3</em>

Explanation:

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2 years ago
Jenny and Alyssa are members of the cross-country team. On a training run, Jenny starts off and runs at a constant 3.8 m/s. Alys
guapka [62]

Answer:

285 seconds

Explanation:

Jenny speed is 3.8 m/s

Alyssa speed in 4.0 m/s

Alyssa starts after 15 seconds

Find the distance covered by Jenny, when Alyssa starts

Distance=Speed*time

Distance covered by Jenny in 15 seconds= 3.8×15=57m

Relative speed of the two members heading same direction will be;

4.0m/s-3.8m/s=0.2m/s

To find the time Alyssa catch up with Jenny you divide the distance to be covered by Alyssa by the relative speed of the two

Distance=57m, relative speed=0.2m/s  t=57/0.2 =285 seconds

=4.75 minutes

5 0
2 years ago
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A simple circuit within a laptop has a single resistor with a resistance of 0.1 Ω and requires a current of 50 mA. Select the vo
jolli1 [7]

Voltage = (current) x (resistance)

The voltage across THIS RESISTOR is

V = (0.050 A) x (0.1 ohm)

V = 0.005 v (5 millivolts)


6 0
2 years ago
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A machine gear consists of 0.10 kg of iron and 0.16 kg of copper.
Natali5045456 [20]

Answer:

option (c)

Explanation:

mass of iron = 0.10 kg

mass of copper = 0.16 kg

rise in temperature, ΔT = 35°C

specific heat of iron = 450 J/kg°C

specific heat of copper = 390 J/kg°C

Heat by iron (H1) = mass of iron x specific heat of iron x ΔT

H1 = 0.10 x 450 x 35 = 1575 J

Heat by copper (H2) = mass of copper x specific heat of copper x ΔT

H1 = 0.16 x 390 x 35 = 2184 J

Total heat H = H1 + H2

H = 1575 + 2184 = 3759 J

by rounding off

H = 4000 J

6 0
2 years ago
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