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Alchen [17]
3 years ago
6

What is the equation for an inelastic collision

Physics
1 answer:
abruzzese [7]3 years ago
7 0
M1 v1 = (m1 + m2)v2.

All of the exponents should be lowered to the bottom right of the letters.
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What is the purpose of oil used in a car's engine?
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B) to reduce friction
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Lithium (chemical symbol Li) is located in Group 1, Period 2. Which is lithium most likely to be? O A. A soft, shiny, highly rea
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A. A soft, shiny, highly reactive metal
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A Carnot heat engine has an efficiency of 0.400. If it operates between a deep lake with a constant temperature of 298.0k and a
tatuchka [14]

Answer:

496.7 K

Explanation:

The efficiency of a Carnot engine is given by the equation:

\eta = 1 - \frac{T_H}{T_L}

where:

T_H is the temperature of the hot reservoir

T_C is the temperature of the cold reservoir

For the engine in the problem, we know that

\eta = 0.400 is the efficiency

T_C = 298.0 K is the temperature of the cold reservoir

Solving for T_H, we find:

\frac{T_C}{T_H}=1-\eta\\T_H = \frac{T_C}{1-\eta} =\frac{298.0}{1-0.400}=496.7 K

6 0
3 years ago
A truck moves 100 miles to the south in 2 hours. What is the trucks velocity
ruslelena [56]

v = d/t

v = 100/2

v=50

50 mph

3 0
3 years ago
A motorcycle is following a car that is traveling at a constant speed on a straight highway. Initially, the car and the motorcyc
Artist 52 [7]

Answer:

(a) 3.807 s

(b) 145.581 m

Explanation:

Let Δt = t2 - t1 be the time it takes from the moment when the motorcycle starts to accelerate until it catches up with the car. We know that before the acceleration, both vehicles are travelling at a constant speed. So they would maintain a distance of 58 m prior to the acceleration.

The distance traveled by car after Δt (seconds) at v_c = 23m/s speed is

s_c = \Delta t v_c = 23\Delta t

The distance traveled by the motorcycle after Δt (seconds) at m_m = 23 m/s speed and acceleration of a = 8 m/s2 is

s_m = \Delta t v_m + a\Delta t^2/2

s_m = 23\Delta t + 8\Delta t^2/2 = 23 \Delta t + 4 \Delta t^2

We know that the motorcycle catches up to the car after Δt, so it must have covered the distance that the car travels, plus their initial distance:

s_m = s_c + 58

23 \Delta t + 4 \Delta t^2 = 23\Delta t + 58

4 \Delta t^2 = 58

\Delta t^2 = 14.5

\Delta t = \sqrt{14.5} = 3.807s

(b)

s_m = 23 \Delta t + 4 \Delta t^2

s_m = 23*3.807 + 58 = 145.581 m

5 0
3 years ago
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