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2 years ago

What is the momentum of a 1.5 * 10 ^ 3 - kil kilogram car as it travels at 30. meters per second due east for 60. seconds ?

Physics

1 answer:

serg [7]2 years ago

6 0

Answer:

45000kgm/s due east

Explanation:

Given parameters:

Mass of the car = 1.5 x 10³kg

Velocity = 30m/s

Time taken = 60s

Unknown:

Momentum = ?

Solution:

Momentum is the quantity of motion a body possess,

Momentum = mass x velocity

So;

Momentum = 1.5 x 10³ x 30 = 45000kgm/s due east

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7. A toy car of mass 1.2 kg is driving vertical circles inside a hollow cylinder of radius 2.0m. It is moving at a constant spee

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$N_{top}=9.8N\\N_{bottom}=33.4N$

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2 years ago

A projectile is thrown with velocity v at an angle θ with horizontal. When the projectile is at a height equal to half of the ma

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Let, the maximum height covered by projectile be $\sf{H_m}$

$\purple{ \longrightarrow \bf{h_m = \dfrac{ {v}^{2} \: {sin}^{2} \theta }{2g} }}$

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Velocity of projectile at a height half of the maximum height covered be $\sf{v_0}$

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$\qquad$ $\longrightarrow \sf{ \dfrac{ {v}^{2} \: {sin}^{2} \theta }{2g} \times \dfrac{1}{2} = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta }{2g} }$

$\qquad$ $\longrightarrow \sf{ \dfrac{ {v}^{2} \: {sin}^{2} \theta }{4g} = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta }{2g} }$

$\qquad$ $\longrightarrow \sf{ \dfrac{ {v}^{2} \: {sin}^{2} \theta }{2} = {v_0}^{2} \: {sin}^{2} \theta }$

$\qquad$ $\longrightarrow \sf{ \dfrac{ {v}^{2} }{2} = {v_0}^{2} }$

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$\qquad$ $\longrightarrow \bf{v_{(y)} = \dfrac{v}{ \sqrt{2} } \: sin \theta = \dfrac{v \: sin \: \theta}{ \sqrt{2} } }$

Therefore, the vertical component of velocity of projectile at this height will be–

☀️ $\qquad$ $\pink {\bf{ \dfrac{v \: sin \: \theta}{ \sqrt{2} }} }$

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