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Hunter-Best [27]
3 years ago
12

The momentum of an object is determined to be 7.2 × 10-3 kg⋅m/s. Express this quantity as provided or use any equivalent unit. (

Note: 1 kg = 1000 g).
Physics
1 answer:
slavikrds [6]3 years ago
6 0

Answer:

Momentum, p = 7.2 g-m/s

Explanation:

It is given that,

The momentum of an object is p=7.2\times 10^{-3}\ kg-m/s

We need to express momentum in any equivalent units. There can be many solutions of this problem. Some of the units of mass are gram, milligram etc. units of length are meters, mm etc.

Since, 1 kg = 1000 gram

So, p=7.2\times 10^{-3}\times 10^3\ g-m/s

Therefore, the momentum of the object is 7.2 g-m/s. Hence, this is the required solution.

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Ne4ueva [31]

Speed of the projectile at its maximum height is only along horizontal direction

so at highest point

v_1 = v_x

now when he is at half of the maximum height the speed will be in x and y direction both

v_2 = \sqrt{v_y^2 + v_x^2}

here it is given that

v_1 = 0.58 v_2

v_x = 0.58\sqrt{v_x^2 + v_y^2}

2.97 v_x^2 = v_x^2 + v_y^2

1.97 v_x^2 = v_y^2

also we know that

v_y^2 = v_{iy}^2 - 2 g \frac{H}{2}

here we know that maximum height is given as

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v_y^2 = v_{iy}^2 - 2 g\frac{v_{iy}^2}{4g}

v_y^2 = \frac{v_{iy}^2}{2}

now from above

1.97 v_x^2 = \frac{v_{iy}^2}{2}

1.98 v_x = v_{iy}

also we know that angle of projection is

tan\theta = \frac{v_{iy}}{v_x}

tan\theta = \frac{1.98v_x}{v_x}

so angle is

\theta = tan^{-1} 1.98

\theta = 63.3 degree

6 0
3 years ago
On that system is<br> In an isolated system, the net __<br> zero.
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Answer:

the net work is 0

Explanation:

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