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rosijanka [135]
3 years ago
14

What is the momentum of a 950 kg car moving at 10.0 m/s?

Physics
2 answers:
pentagon [3]3 years ago
8 0

Answer:

<h3>The answer is 9500 kgm/s</h3>

Explanation:

The momentum of an object can be found by using the formula

<h3>momentum = mass × velocity</h3>

From the question

mass = 950 kg

velocity = 10.0 m/s

We have

momentum = 950 × 10

We have the final answer as

<h3>9500 kgm/s</h3>

Hope this helps you

Nesterboy [21]3 years ago
8 0
<h2><u>Given</u> :</h2>
  • A 950 kg car moving at 10.0 ms
<h2><u>To Find</u> :</h2>
  • Momentum
<h2><u>Solution</u> :</h2>

<u>We have,</u>

  • Mass = 950 kg
  • Velocity = 10.0 m/s = 10 m/s

<u>We know that,</u>

<h3>→ Momentum = Mass × Velocity</h3>

<u>Substituting the values we get,</u>

\impliesMomentum = 950 × 10

\impliesMomentum = 9500 kgm/s

∴ The momentum is 9500 kgm/s.

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The rectangular plates in a parallel-plate capacitor are 0.063 m × 5.4 m. A distance of 3.5 × 10–5 m separates the plates. The p
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Explanation:

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distance between the plate d = 3.5 × 10–5 m

dielectric value for Teflon K = 2.1

capacitance of capacitor = ?

Formula for capacitance of parallel plate is as follows ,

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An explosive projectile is launched straight upward to a maximum height h. At its peak, it explodes, scattering particles in all
Varvara68 [4.7K]

Answer:

θ = tan⁻¹ (\frac{19.6 \ h}{v})

Explanation:

This problem must be solved using projectile launch ratios. Let's analyze the situation, the projectile explodes at the highest point, therefore we fear the height (i = h), the speed at this point is the same, but the direction changes, we are asked to find the smallest angle of the speed in the point of arrival with respect to the x-axis.

The speed at the arrival point (y = 0)

           v² = vₓ² + v_y²

Let's see how this angle changes, for two extreme values:

* The particle that falls from the point of explosion, in this case the speed is vertical

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the angle with the horizontal is 90º

* The particle leaves horizontally from the point of the explosion, the initial velocity is horizontal

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the final velocity for y = 0

         v_f = vₓ² + v_y²

therefore the angle has a value greater than zero and less than 90º

As they ask for the smallest angle, we can see that we must solve the last case

the output velocity is horizontal vₓ = v

Let's find the velocity when it hits the ground y = 0, with y₀ = h

            v_{y}^2 = v_{oy}^2 - 2 g (y-y₀)

            v_{y}^2 = - 2g (0- y₀)

let's calculate

           v_{y}^2 = 2 9.8 h

         

we use trigonometry to find the angle

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let's calculate

         θ = tan⁻¹ (\frac{19.6 \ h}{v})

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