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Delicious77 [7]
3 years ago
14

While jumping on a trampoline you calculate that at the highest peak of your jump you have 900 joules of gravitational potential

energy. What will be your kinetic energy just before landing back on the trampoline?
Physics
1 answer:
BabaBlast [244]3 years ago
8 0

Jumping on a trampoline is a classic example of conservation of energy, from potential into kinetic. It also shows Hooke's laws and the spring constant. Furthermore, it verifies and illustrates each of Newton's three laws of motion.

<u>Explanation</u>

When we jump on a trampoline, our body has kinetic energy that changes over time. Our kinetic energy is greatest, just before we hit the trampoline on the way down and when you leave the trampoline surface on the way up. Our kinetic energy is 0 when you reach the height of your jump and begin to descend and when are on the trampoline, about to propel upwards.

Potential energy changes along with kinetic energy. At any time, your total energy is equal to your potential energy plus your kinetic energy. As we go up, the kinetic energy converts into potential energy.

Hooke's law is another form of potential energy. Just as the trampoline is about to propel us up, your kinetic energy is 0 but your potential energy is maximized, even though we are at a minimum height. This is because our potential energy is related to the spring constant and Hooke's Law.

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8 0
3 years ago
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explain why conductors and insulators are both required to construct the electrical wiring in your home
olchik [2.2K]
Conductors (something that allows electricity to flow easily) allow for electricity to flow easily. This would be the wires. If you don't have conductors, then you cannot have electricity flow.

Insulators (something that doesn't allow electricity to flow through it) is important because it allows us to be able to touch the cables or place them next to one another and not shock ourselves

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3 years ago
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A 124-kg balloon carrying a 22-kg basket is descending with a constant downward velocity of 20.0 m/ s. A I.O-kg stone is thrown
nadya68 [22]

(a) 296.6 m

The motion of the stone is the motion of a projectile, thrown with a horizontal speed of

v_x = 15.0 m/s

and with an initial vertical velocity of

v_{y0} = -20.0 m/s

where we have put a negative sign to indicate that the direction is downward.

The vertical position of the stone at time t is given by

y(t) = h + v_{0y} t + \frac{1}{2}gt^2 (1)

where

h is the initial height

g = -9.81 m/s^2 is the acceleration due to gravity

The stone hits the ground after a time t = 6.00 s, so at this time the vertical position is zero:

y(6.00 s) = 0

Substituting into eq.(1), we can solve to find the initial height of the stone, h:

0 = h + v_{0y} y + \frac{1}{2}gt^2\\h = -v_{0y} y - \frac{1}{2}gt^2=-(-20.0 m/s)(6.00 s) - \frac{1}{2}(9.81 m/s^2)(6.00 s)^2=296.6 m

(b) 176.6 m

The balloon is moving downward with a constant vertical speed of

v_y = -20 m/s

So the vertical position of the balloon after a time t is

y(t) = h + v_y t

and substituting t = 6.0 s and h = 296.6 m, we find the height of the balloon when the rock hits the ground:

y(t) = 296.6 m + (-20.0 m)(6.00 s)=176.6 m

(c) 198.2 m

In order to find how far is the rock from the balloon when it hits the ground, we need to find the horizontal distance covered by the rock during the time of the fall.

The horizontal speed of the rock is

v_x = 15.0 m/s

So the horizontal distance travelled in t = 6.00 s is

d_x = v_x t = (15.0 m/s)(6.00 s)=90 m

Considering also that the vertical height of the balloon after t=6.00 s is

d_y = 176.6 m

The distance between the balloon and the rock can be found by using Pythagorean theorem:

d=\sqrt{(90 m)^2+(176.6 m)^2}=198.2 m

(di) 15.0 m/s, -58.8 m/s

For an observer at rest in the basket, the rock is moving horizontally with a velocity of

v_x = 15.0 m/s

Instead, the vertical velocity of the rock for an observer at rest in the basket is

v_y (t) = gt

Substituting time t=6.00 s, we find

v_y = (-9.8 m/s)(6.00 s)=-58.8 m/s

(dii) 15.0 m/s, -78.8 m/s

For an observer at rest on the ground, the rock is still moving horizontally with a velocity of

v_x = 15.0 m/s

Instead, the vertical velocity of the rock for an observer on the ground is now given by

v_y (t) = v_{0y} + gt

Substituting time t=6.00 s, we find

v_y = (-20.0 m/s)+(-9.8 m/s)(6.00 s)=-78.8 m/s

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Explanation:

Given that,

Distance 1, r = 100 m

Intensity, I_1=1.24\times 10^{-8}\ W/m^2

If distance 2, r' = 25 m

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Let I' is the intensity at r'. So,

I'=\dfrac{P}{4\pi r'^2}............(2)

From equation (1) and (2) :

I'=\dfrac{Ir}{r'^2}

I'=\dfrac{1.24\times 10^{-8}\times 100}{25^2}

I'=1.98\times 10^{-9}\ W/m^2

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dB=10\ log(\dfrac{I'}{I_o}), I_o=10^{-12}\ W/m^2

dB=10\ log(\dfrac{1.98\times 10^{-9}}{10^{-12}})

dB = 32.96 dB

Hence, this is the required solution.

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Larger mass creates a stronger pull
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