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Crank
2 years ago
6

5. A ball weighing 10 kg rolls 200 m down a frictionless incline with a 50 degree angle to the horizontal. If the ball’s initial

velocity was 0 m/s, how much does the mechanical energy of the system change by the time the ball reaches its destination? A) It increased by 12%. B) It increases by 58%. C) It decreases by 12%. D) It does not change.
Physics
1 answer:
sdas [7]2 years ago
3 0

Answer:

D) It does not change

Explanation:

Since there is no friction in the inclined plane. Therefore, there is no loss in the total mechanical energy of the system. So according to the law of conservation of energy we can write:

Total Mechanical Energy at Start = Total Mechanical Energy at End + Frictional Loss

Total Mechanical Energy at Start = Total Mechanical Energy at End + 0

Total Mechanical Energy at Start = Total Mechanical Energy at End

It means there is no change in the total mechanical energy of the system.

Therefore, the correct option is:

<u>D) It does not change</u>

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The weight of an object is taken to be the force on the object due to gravity. The weight ( W ) is the product of the mass ( m ) of the object and the magnitude of the gravitational acceleration ( g ).
On Earth: g = 9.81 m/s²
m = 20 kg
W = m · g = 20 kg · 9.81 m/s² = 196.2 N
6 0
3 years ago
An LC circuit consists of a 3.4-µF capacitor and a coil with a self-inductance 0.080 H and no appreciable resistance. At t = 0 t
alexira [117]

Answer

given,

capacitance = C = 3.4-µF

inductance = L = 0.08 H

frequency is expressed as

f = \dfrac{1}{2\pi\sqrt{LC}}

time period

T = \dfrac{1}{f}=2\pi\sqrt{LC}

after time T/4 current reach maximum

 t = \dfrac{T}{4}

 t = \dfrac{2\pi\sqrt{LC}}{4}

 t = \dfrac{2\pi\sqrt{0.08 \times 3.4 \times 10^{-6}}}{4}

        t = 8.2 x 10⁻⁴ s

        t = 0.82 ms

b) using law of conservation

  \dfrac{1}{2}CV^2=\dfrac{1}{2}LI^2

  I^2 = \dfrac{CV^2}{L}

  I^2 = \dfrac{C}{L}\dfrac{Q^2}{C^2}

  I =\sqrt{\dfrac{Q^2}{CL}}

  I =\sqrt{\dfrac{(5.4 \times 10^{-6})^2}{0.08 \times 3.4 \times 10^{-6}}}

       I = 0.010 A

       I = 10 mA

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3 years ago
Look at a photograph of a fault. notice how the right side appears lower than the left side. this happens when pieces of crust a
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I believe it’s divergent boundary but I might be wrong
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3 years ago
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Hydrogen is the second most abundant gas in the atmosphere? True or false?
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7 0
3 years ago
A 20.00 kg lead sphere is hanging from a hook by a thin, massless wire 2.80 m long and is free to swing in a complete circle. Tr
Tema [17]

The minimum initial speed of the dart so that the combination makes a complete circular loop after the collision is 58.5 m/s.

<h3>Minimum speed for the object not fall out of the circle</h3>

The minimum speed if given by tension in the wire;

T + mg = ma

T + mg = m(v²)/R

tension must be zero for the object not fall

0 + mg = mv²/R

v = √(Rg)

<h3>Final speed of the two mass after collision</h3>

Use the principle of conservation of energy

K.Ef  = K.Ei + P.E

¹/₂mvf² = ¹/₂mv² + mg(2R)

¹/₂vf² = ¹/₂v² + g(2R)

¹/₂vf² = ¹/₂(Rg) + g(2R)

vf² = Rg + 4Rg

vf² = 5Rg

vf = √(5Rg)

vf = √(5 x 2.8 x 9.8)

vf = 11.7 m/s

<h3>Initial speed of the dart</h3>

Apply principle of conservation of linear momentum for inelastic collision;

5v = vf(20 + 5)

5v = 11.7(25)

5v = 292.5

v = 58.5 m/s

Learn  more about linear momentum here: brainly.com/question/7538238

#SPJ1

3 0
2 years ago
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