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miskamm [114]
3 years ago
14

A 0.11 kg bullet traveling at speed hits a 18.3 kg block of wood and stays in the wood. The block with the bullet imbedded in it

moves forward with a velocity of 8.8 m/s. What was the velocity (speed) of the bullet immediately before it hit the block (in m/s)?
Physics
1 answer:
arlik [135]3 years ago
5 0

Explanation:

The energy of the system before the collision must equal the energy after the collision.

After the collision the bullet and the block have a total mass of 18.41 kg and they move at a speed of 8.8 m/s. The kinetic energy after the collision is

\frac{18.41 kg (8.8 m/s)^2}{2} = 713 J

Before the collision only the bullet has kinetic energy.

So we can now determine the speed of the bullet using

\frac{0.11kg (v^2)}{2} = 713 J\\v = 114 m/s

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Aleks04 [339]

Answer:

Kinetic energy = (1/2) (mass) (speed²)

Original KE = (1/2) (1430 kg) (7.5 m/s)²  =  40,218.75 joules

Final KE  =  (1/2) (1430 kg) (11.0 m/s)²  =   86,515 joules

Work done during the acceleration = (40218.75 - 86515) = 46,296.25 joules

Power = work/time = 46,296.25 joules / 9.3 sec  =  4,978.1 watts .

Explanation:

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3 0
3 years ago
A car initially traveling 7 m/s speeds up uniformly at a rate of 3 m/s2 until it reaches a velocity of 22 m/s. How much time did
dezoksy [38]

Answer:

t = 5 s

Explanation:

Data:

  • Initial Velocity (Vo) = 7 m/s
  • Acceleration (a) = 3 m/s²
  • Final Velocity (Vf) = 22 m/s
  • Time (t) = ?

Use formula:

  • \boxed{t=\frac{Vf - Vo}{a}}

Replace:

  • \boxed{t=\frac{22\frac{m}{s} -7\frac{m}{s}}{3\frac{m}{s^{2}}}}

Solve the subtraction of the numerator:

  • \boxed{t=\frac{15\frac{m}{s}}{3\frac{m}{s^{2}}}}

It divides:

  • \boxed{t=5\ s}

How much time did it take the car to reach this final velocity?

It took a time of <u>5 seconds.</u>

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3 years ago
Blood ki kami se konsa rog hota hai?​
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Answer:

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If a system has 475 kcal of work done to it, and releases 5.00 × 102 kJ of heat into its surroundings, what is the change in int
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First, we convert kcal to joules:
1 kcal = 4.184 kJ
475 kcal = 1987.4 kJ

Now, calculating the change in internal energy:

ΔU = Q + W; where Q is the heat supplied to the system and W is the work done on the system.

ΔU = -500 + 1987.4
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4 0
4 years ago
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Answer:

opposite

Explanation:

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