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Inga [223]
3 years ago
9

13. Under what condition (if any) does a moving body experience no energy even though there

Physics
1 answer:
djyliett [7]3 years ago
6 0

Answer:

When the Net Forces are equal to 0

Explanation:

Momentum of a body can be defined as product of mass and velocity. It is in the same direction as in velocity. When the momentum of a body doesn't change, it is said to be conserved. If the momentum of a body is constant, the the net forces acting on a body becomes zero. When net forces acting on a body is zero, it means that no kinetic energy is being lost or gained, hence the kinetic energy is also conserved. If no energy is being gained or lost, it means that the body will experience no energy.

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Clear cutting has a similar effect as controlled burning of a forest. <br><br> True<br><br> False
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True </span>
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Block 1, of mass m₁ = 1.30 kg , moves along a frictionless air track with speed v₁ = 29.0 m/s. It collides with block 2, of mass
Alecsey [184]

Answer:

a. 37.7 kgm/s b. 0.94 m/s c. -528.85 J

Explanation:

a. The initial momentum of block 1 of m₁ = 1.30 kg with speed v₁ = 29.0 m/s is p₁ = m₁v₁ = 1.30 kg × 29.0 m/s = 37.7 kgm/s

The initial momentum of block 2 of m₁ = 39.0 kg with speed v₂ = 0 m/s since it is initially at rest is p₁ = m₁v₁ = 39.0 kg × 0 m/s = 0 kgm/s

So, the magnitude of the total initial momentum of the two-block system = (37.7 + 0) kgm/s = 37.7 kgm/s

b. Since the blocks stick together after the collision, their final momentum is p₂ = (m₁ + m₂)v where v is the final speed of the two-block system.

p₂ = (1.3 + 39.0)v = 40.3v

From the principle of conservation of momentum,

p₁ = p₂

37.7 kgm/s = 40.3v

v = 37.7/40.3 = 0.94 m/s

So the final velocity of the two-block system is 0.94 m/s

c. The change in kinetic energy of the two-block system is ΔK = K₂ - K₁ where K₂ = final kinetic energy of the two-block system = 1/2(m₁ + m₂)v² and K₁ = final kinetic energy of the two-block system = 1/2m₁v₁²

So, ΔK = K₂ - K₁ = 1/2(m₁ + m₂)v² - 1/2m₁v₁² = 1/2(1.3 + 39.0) × 0.94² - 1/2 × 1.3 × 29.0² = 17.805 J - 546.65 J = -528.845 J ≅ -528.85 J

7 0
3 years ago
_________ liquids and gases exert a buoyant force.
Eduardwww [97]

Answer:

C. Both

Explanation:

5 0
3 years ago
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A hockey puck has a coefficient of kinetic friction of μk = .35. If the puck feels a normal force (FN) of 5 N, what is the frict
alina1380 [7]

Answer:

The frictional force is  F_f =  1.75 \  N

Explanation:

From the question we are told that

     The coefficient of kinetic force is  μk = 0.35

     The normal force felt by the puck is  F_N  =  5 \  N

Generally the frictional force that acts on the puck is mathematically represented as

          F_f =  \mu_k  *  F_N

=>       F_f =  0.35  *  5

=>       F_f =  1.75 \  N

3 0
3 years ago
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