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Mashcka [7]
3 years ago
7

A 10.0-g bullet moving at 300 m/s is fired into a 1.00-kg block at rest. the bullet emerges (the bullet does not get embedded in

the block with half of its original speed. what is the velocity of the block right after the collision?
Physics
1 answer:
sesenic [268]3 years ago
7 0
Momentum before the hit:
p = mv = 0.01 * 300 + 1 * 0 

Momentum after the hit:
p = 0.01 * 150 + 1 * v

Momentum is conserved:
0.01 * 300 = 0.01 * 150 + v 
3 = 1.5 + v
v = 1.5

The velocity of the block after the collision is 1.5 m/s.
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2 years ago
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A cannon, located 60.0 m from the base of a vertical 25.0-m-tall cliff, shoots a 15-kg shell at 43.0o above the horizontal towar
Artist 52 [7]

Answer:

a)   v₀ = 32.64 m / s , b)  x = 59.68 m

Explanation:

a) This is a projectile launching exercise, we the distance and height of the cliff

         x = v₀ₓ t

         y = v_{oy} t - ½ g t²

We look for the components of speed with trigonometry

         sin 43 = v_{oy} / v₀

         cos 43 = v₀ₓ / v₀

         v_{oy} = v₀ sin 43

         v₀ₓ = v₀ cos 43

Let's look for time in the first equation and substitute in the second

         t = x / v₀ cos 43

         y = v₀ sin 43 (x / v₀ cos 43) - ½ g (x / v₀ cos 43)²

          y = x tan 43 - ½ g x² / v₀² cos² 43

          1 / v₀² = (x tan 43 - y) 2 cos² 43 / g x²

           v₀² = g x² / [(x tan 43 –y) 2 cos² 43]

Let's calculate

          v₀² = 9.8 60 2 / [(60 tan 43 - 25) 2 cos 43]

          v₀ = √ (35280 / 33.11)

          v₀ = 32.64 m / s

.b) we use the vertical distance equation with the speed found

         y = v_{oy} t - ½ g t²

         .y = v₀ sin43 t - ½ g t²

        25 = 32.64 sin 43 t - ½ 9.8 t²

        4.9 t² - 22.26 t + 25 = 0

         t² - 4.54 t + 5.10 = 0

We solve the second degree equation

         t = (4.54 ±√(4.54 2 - 4 5.1)) / 2

         t = (4.54 ± 0.46) / 2

         t₁ = 2.50 s

         t₂ = 2.04 s

The shortest time is when the cliff passes and the longest when it reaches the floor, with this time we look for the horizontal distance traveled

         x = v₀ₓ t

         x = v₀ cos 43 t

         x = 32.64 cos 43  2.50

         x = 59.68 m

8 0
3 years ago
A rocket powered sled accelerates a jet pilot in training straight forward from rest to 270 km/h in 12.1 seconds. Find:
Ilia_Sergeevich [38]

Answer:

  1. 6.198 m/s²
  2. 4.48 s
  3. 453.77 m

Explanation:

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3 years ago
11 A motor is used to lift a load of 40N
RSB [31]

Answer:

The correct answer is option D

4 0
2 years ago
The power in an electric circuit varies inversely with the resistance. If the power is 2,200 watts when the resistance is 25 ohm
valentinak56 [21]

Answer:

The value of resistance when power is 1100 watts = R_{2} = 50 ohms

Explanation:

Power P_{1} = 2200 Watts

Resistance R_{1} = 25 ohms

Power P_{2} = 1100 Watts

Resistance R_{2} = we have to calculate

Given that the power in an electric circuit varies inversely with the resistance

⇒ P ∝ \frac{1}{R}

⇒ \frac{P_{2} }{P_{1} } = \frac{R_{1} }{R_{2} }

⇒ \frac{1100}{2200} = \frac{25}{R_{2} }

⇒ R_{2} = 50 ohms

This is the value of resistance when power is 1100 watts.

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3 years ago
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