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White raven [17]
2 years ago
8

A 0.1 kg toy contains a compressed spring. when the spring is released the toy fly 0.45 m upwards from ground level before falli

ng back down to the ground. what is the speed of the toy when it hits the ground.
Physics
1 answer:
trapecia [35]2 years ago
7 0

The speed of the toy when it hits the ground is 2.97 m/s.

The given parameters;

  • mass of the toy, m = 0.1 kg
  • the maximum height reached by the, h = 0.45 m

The speed of the toy before it hits the ground will be maximum. Apply the principle of conservation of mechanical energy to determine the maximum speed of the toy.

P.E = K.E

mgh_{max} = \frac{1}{2} mv_{max}^2\\\\gh_{max} = \frac{1}{2} v_{max}^2\\\\v_{max}^2= 2gh_{max}\\\\v_{max} = \sqrt{2gh_{max}}

Substitute the given values and solve the speed;

v_{max} = \sqrt{2\times 9.8 \times 0.45} \\\\v_{max} = 2.97 \ m/s

Thus, the speed of the toy when it hits the ground is 2.97 m/s.

Learn more here: brainly.com/question/7562874

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Answer:

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Explanation:

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A body of mass 100g moving with a velocity of 10.0m/s collides with a wall .if after the collision it moves with a velocity of 2
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Answer:

-1.2 kg - m/s

Explanation:

\pink{\frak{Given}}\begin{cases}\textsf{ A body of mass 100g moving with a velocity of 10.0m/s collides with a wall .}\\\textsf{ After the collision it moves with a velocity of 2.0m/s in the opposite direction.}\end{cases}

And we need to find out the change in momentum of the body . Here ,

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We know that momentum is defined as amount of motion contained in a body . Mathematically ,

\sf\longrightarrow momentum (p)= mass(m) * velocity(v)

Therefore change in momentum will be,

\sf\longrightarrow \triangle p = mv - mu

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\sf\longrightarrow \Delta p = 100g( -2m/s -10m/s) \\

\sf\longrightarrow \Delta p =\dfrac{100}{1000}kg ( -12m/s)  \\

\sf\longrightarrow \Delta p   = 0.1 kg * -12m/s \\

\sf\longrightarrow \boxed{\bf \Delta p = -1.2 \ kg-m/s} \\

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Four point charges are placed at the corners of a square. Each charge has the identical value +Q. The length of the diagonal of
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Answer:V_{net}=4\frac{kQ}{a}    

Explanation:

Given

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V_3=\frac{kQ}{a}

V_4=\frac{kQ}{a}

total Electric Potential At center is given by

V_{net}=V_1+V_2+V_3+V_4

V_{net}=4\times \frac{kQ}{a}

V_{net}=4\frac{kQ}{a}            

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