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GaryK [48]
3 years ago
15

a shot putter accelerates a 7.3kg shot from rest to 14m/s in 1.5r seconds. what average power was developed?

Physics
1 answer:
Sedaia [141]3 years ago
3 0

Explanation:

power =  \frac{energy \: expended}{time}

m = 7.3kg

u = 0

v = 14m/s

t = 1.5sec

P = (0.5×7.3×14²) ÷ 1.5

P = 476.93

P = 477 watt

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A) Expanding. We know this because it has a similar effect with sound. When a car goes by the pitch gets deeper and deeper. It's because you're receiving less waves. Same thing for light but instead of a pitch it's light, and the farther spread the waves - the redder, the closer and more contracted - the bluer
3 0
2 years ago
Four identical masses m are evenly spaced on a frictionless 1D track. The first mass is sent at speed v toward the other three.
SpyIntel [72]

Answer:

The speed decreases 75%.

Explanation:

  • Since no friction present, assuming no external forces acting during the three collisions, total momentum must be conserved.
  • For the first collission, only mass 1 is moving before it, so we can write the following equation:

       p_{i} = p_{f} = m*v_{o}    (1)

  • Since both masses are identical, and they stick together after the collision, we can express the final momentum as follows:

       p_{f1} = 2*m*v_{1}    (2)

  • From (1) and (2) we get:
  • v₁ = v₀/2  (3)
  • Since the masses are moving on a frictionless 1D track, the speed of the set of mass 1 and 2 combined together before colliding with mass 3 is just v₁, so the initial momentum prior the second collision (p₁) can be expressed as follows:

       p_{1} = 2*m*v_{1} = 2*m*\frac{v_{o} }{2}  = m*v (4)

  • Since after the collision the three masses stick together, we can express this final momentum (p₂) as follows:

        p_{2} = 3*m*v_{2}  (5)

  • From (4) and (5) we get:
  • v₂ = v₀/3  (6)
  • Since the masses are moving on a frictionless 1D track, the speed of the set of mass 1, 2 and 3 combined together before colliding with mass 4 is just v₂, so the initial momentum prior the third collision (p₂) can be expressed as follows:

      p_{2} = 3*m*v_{2} = 3*m*\frac{v_{o} }{3}  = m*v (7)

  • Since after the collision the four masses stick together, we can express this final momentum (p₃) as follows:

       p_{3} = 4*m*v_{3}  (8)

  • From (7) and (8) we get:
  • v₃ = v₀/4
  • This means that after the last collision, the speed will have been reduced to a 25% of the initial value, so it will have been reduced in a 75% regarding the initial value of v₀.
5 0
2 years ago
Which clouds are often associated with thunder and lightning?
ehidna [41]
Your answer is cumulonimbus clouds
3 0
2 years ago
The speed of sound in air is 345 m/s. A tuning fork vibrates above the open end of a sound resonance tube. If sound waves have w
Delvig [45]

To develop this problem it is necessary to apply the concept of Frequency based on speed and wavelength.

According to the definition the frequency can be expressed as

f = \frac{v}{\lambda}

Where,

v = Velocity

\lambda = Wavelength

Our value are given by,

v = 345m/s

\lambda = 63cm

Replacing

f = \frac{345}{0.63}

f = 547.61Hz

Therefore the frequency of the tuning fork is 547.61Hz

6 0
3 years ago
The lift does 3000 J of work in 5 seconds. What is the power of the lift?
diamong [38]

Answer:

600

Explanation:

p=Work/time

3000/5=600 is power

8 0
2 years ago
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