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Vlada [557]
3 years ago
14

Three mountain climbers set out to climb a mountain from the same altitude and all arrive at the same location at the top. Mount

ain climber A took a long gradual slope to the top, B went a steeper but shorter path, and C tackled the sheer straight side to the top. Assume all three climbers weigh the same. Which did the most work?
A
B
C
all did the same amount of work
Physics
2 answers:
vampirchik [111]3 years ago
6 0
C did the most effort, but A walked the longest, so a reasonable answer is that they all did the same amount of work, as B simply mediates.
ipn [44]3 years ago
6 0

Answer: all did the same amount of work

Explanation:

The work done by each mountain climber to climb the mountain is equal to how much gravitational potential energy they gained in the climb, and this is given by

W=\Delta U=mg\Delta h

where m is the mass of the climber, g is the gravitational acceleration, and \Delta h is the gap in altitude.

Since the three climbers have same weight, they also have same mass, so the term "m" in the formula is the same. Also, they started from the same altitude and arrive at the same altitude, so the gap \Delta h is the same for all of them. Therefore, they gain the same gravitational potential energy, and so they did the same amount of work. So, the work does not depend on the path taken.

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The heat capacity of object B is twice that of object A. Initially A is at 300 K and B at 450 K. They are placed in thermal cont
ivann1987 [24]

Answer:

The final temperature of both objects is 400 K

Explanation:

The quantity of heat transferred per unit mass is given by;

Q = cΔT

where;

c is the specific heat capacity

ΔT is the change in temperature

The heat transferred by the  object A per unit mass is given by;

Q(A) = caΔT

where;

ca is the specific heat capacity of object A

The heat transferred by the  object B per unit mass is given by;

Q(B) = cbΔT

where;

cb is the specific heat capacity of object B

The heat lost by object B is equal to heat gained by object A

Q(A) = -Q(B)

But heat capacity of object B is twice that of object A

The final temperature of the two objects is given by

T_2 = \frac{C_aT_a + C_bT_b}{C_a + C_b}

But heat capacity of object B is twice that of object A

T_2 = \frac{C_aT_a + C_bT_b}{C_a + C_b} \\\\T_2 = \frac{C_aT_a + 2C_aT_b}{C_a + 2C_a}\\\\T_2 = \frac{c_a(T_a + 2T_b)}{3C_a} \\\\T_2 = \frac{T_a + 2T_b}{3}\\\\T_2 = \frac{300 + (2*450)}{3}\\\\T_2 = 400 \ K

Therefore, the final temperature of both objects is 400 K.

4 0
3 years ago
A box is being pulled to the right. What is the magnitude of the Kinect frictional force?
Anna35 [415]
The answer to this question is A - 25 N
3 0
3 years ago
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A motor produces less mechanical energy than the energy it uses because the motor?
In-s [12.5K]
<span>A motor produces less mechanical energy than the energy it uses because the motor looses some energy to heat.</span>
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3 years ago
If a 15 kg mass accelerates at a rate of 4 m/s2, what net force acts on it?
jarptica [38.1K]
<h2>Answer: D 60N</h2>

<h3>Explanation:</h3>

Mass(M)=15 kg

Acceleration(A)=4 m/s2

Force=?

Now,

Force(F)=M×A

F=15×4

F=60N Ans

5 0
3 years ago
A car was moving at 14 m/s After 30 s, its speed increased to 20 m/s. What was the acceleration during this time ( need help fas
Arada [10]

Answer:let initial velocity u=14m/s

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Explanation:

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7 0
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