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Vladimir79 [104]
3 years ago
12

What would be the mass of an object that is moving at 5 m/s, with a total momentum of 1500 kg*m/s southeast?

Physics
1 answer:
USPshnik [31]3 years ago
7 0

Answer:

The mass of the object is 300 kg

Explanation:

The given parameters are;

The velocity of the object = 5 m/s southeast

The total momentum of the object = 1500 kg·m/s

The equation for linear momentum, is given as follows;

Momentum = Mass × Velocity

Therefore, the mass of the object can be found as follows;

The \ mass \ of \ the  \ object= \dfrac{The \ momentum \ of \ the  \ object}{The \ velocity  \ of \ the  \ object}

Therefore, we have;

The \ mass \ of \ the  \ object= \dfrac{1500 \ kg \cdot m/s}{5 \ m/s} = 300 \ kg

The mass of the object = 300 kg.

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Rick shoots a basketball at an angle of 35' from the horizontal. It leaves his hands 7 feet from the ground with a velocity of 2
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Given:

The angle of projection of the basketball, θ=35°

The height at which the ball leaves the hand, h=7 ft

The initial velocity of the basketball, v=20 ft/s

To find:

The parametric equations describing the shot.

Explanation:

The range, x of the basketball is given by,

x=v\cos\theta t

On substituting the known values,

\begin{gathered} x=20\times\cos35\degree\times t \\ \implies x=16.4t \end{gathered}

The change in the height, y of the basketball is given by,

y=-v\sin\theta t+\frac{1}{2}gt^2

Where g is the acceleration due to gravity.

On substituting the known values,

\begin{gathered} y=-20\times\sin35\degree\times t+\frac{1}{2}\times32\times t^2 \\ \implies y=-11.5t+16t^2 \end{gathered}

Final answer:

The parametric equations describing the shot are

\begin{gathered} \begin{equation*} x=16.4t \end{equation*} \\ \begin{equation*} y=-11.5t+16t^2 \end{equation*} \end{gathered}

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1 year ago
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Answer:

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Applying,

Q = vA.................... Equation 1

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From the question,

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Substitute these values into equation 1

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