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Tju [1.3M]
2 years ago
11

A graph titled velocity versus time has horizontal axis time (seconds) and vertical axis velocity (meters per second). A line ru

ns from 0 seconds 5 meters per second to 5 seconds 25 meters per second. What is the initial velocity of the object represented by the graph? m/s.
Physics
1 answer:
erma4kov [3.2K]2 years ago
5 0

"Initial" means "as our story begins", or "when we arrive", or "when we start watching it", or "at the beginning of time".

You just told us that on the graph, the speed was <em>5 m/s</em> at zero seconds.     So that's the initial speed.

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A 3.53-g lead bullet traveling at 428 m/s strikes a target, converting its kinetic energy into thermal energy. Its initial tempe
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Complete question:

A 3.53-g lead bullet traveling at 428 m/s strikes a target, converting its kinetic energy into thermal energy. Its initial temperature is 40.0°C. The specific heat is 128 J/(kg · °C), latent heat of fusion is 24.5 kJ/kg, and the melting point of lead is 327°C.

(a) Find the available kinetic energy of the bullet. J

(b) Find the heat required to melt the bullet. J

Answer:

Part (a) the available kinetic energy of the bullet is 323.32 J

Part (b) the heat required to melt the bullet is 216.17 J

Explanation:

Given;

mass of the bullet = 3.53 g = 0.00353 kg

velocity of the bullet = 428 m/s

initial temperature of the bullet = 40.0°C

final temperature of the bullet =  327°C

specific heat capacity, c= 128 J/(kg · °C)

latent heat of fusion, Hf  = 24.5 kJ/kg

Part (a) the available kinetic energy of the bullet. J

KE = ¹/₂ × mv²

KE = ¹/₂ × 0.00353 × 428²

     = 323.32 J

Part (b) the heat required to melt the bullet. J

This is the thermal energy required to increase the temperature of the bullet and the heat energy required to melt the bullet.

Quantity of heat required to raise the temperature of the bullet:

Q = mcΔT

   = 0.00353 × 128 × (327-40)

   = 0.00353 × 128 × 287

   = 129.68 J

Quantity of heat required to melt the bullet:

Q = mH_f

Q = 0.00353 × 24500

   = 86.49 J

TOTAL energy required to melt the bullet = 129.68 J + 86.49 J

                                                                      = 216.17 J

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