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KIM [24]
3 years ago
5

Using the results of Question 1 that would apply if the collision were inelastic, compute (using Excel in the yellow highlighted

box) the fractional kinetic energy loss during the collision. Express the "loss" as a percent. What became of the "lost" energy?
Physics
1 answer:
ch4aika [34]3 years ago
4 0

Answer:

The fractional kinetic energy will be lost if the collision is inelastic. In inelastic collision, the kinetic energy is converted into other forms of energy.

The lost energy became heat and sound energy.

Explanation:

During inelastic collision, the kinetic energy of a moving object does not conserve. It changes into another form of energy such as sound energy and heat energy etc.

For example, when a moving car hit another car or wall etc, the kinetic energy is converted into sound and heat energy. This type of collision is inelastic collision.

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At high noon, the sun delivers 1150 w to each square meter of a blacktop road. if the hot asphalt loses energy only by radiation
Pie
Stephen`s Law:
P = (Sigma) · A · e · T^4
P in = P out
e = 1 for blacktop;
1150 W = (Sigma) · T^4
(Sigma) = 5.669 · 10 ^(-8) W/m²K^4
T^4 = 1150 : ( 5.669 · 10^(-8) )
T^4 = 202.875 · 10^8
T =  \sqrt[4]{202.857 * 10 ^{8} }
T = 3.774 · 10² = 377.4 K
Answer: Equilibrium temperature is 377.4 K. 
3 0
3 years ago
A garbage truck has a mass of 2100 kg. It starts from rest and travels 75 meters in 15 seconds. The truck uniformly accelerates
Ivanshal [37]

Answer: F=(2100)(0.33)=693N

Explanation:

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3 years ago
active and passive secruity measures are employed to identify, detect, classify and analyze possible threats inside of which zon
Juli2301 [7.4K]

Answer:

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8 0
3 years ago
A man does 4,475 J of work in the process of pushing his 2.50 103 kg truck from rest to a speed of v, over a distance of 26.0 m.
Tcecarenko [31]

Answer:

a) 1.89 m/s  b) 172.1 N

Explanation:

a)

  • Applying the work-energy theorem, if we can neglect the friction between truck and road, the total change in kinetic energy must be equal to the work done by the external forces.
  • This work, is just 4,475 J.
  • So we can write the following equation:

        \Delta K = \frac{1}{2} * m*v^{2} = 4,475 J

  • where m= mass of the truck = 2.5*10³ kg.
  • So, we can find the speed v, as follows:

        v =\sqrt{\frac{2*W}{m}} =\sqrt{\frac{2*4,475J}{2.5e3kg} }  = 1.89 m/s

b)

  • The work done by the man, is just the horizontal force applied, times the displacement produced by the force horizontally:

        W = F*d

  • We can solve for F, as follows:

        F = \frac{W}{d} = \frac{4,475 J}{26.0m} =  172.1 N

4 0
3 years ago
What do u put on the last blank?
Nookie1986 [14]

The greater the MASS of a moving object, the more kinetic energy it has. <3

8 0
3 years ago
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