Answer:
D, C, B, A
Explanation:
The derivative from a velocity-time graph provides the acceleration value.
Segment A
Segment B
Segment C
Segment D
Sorted from the lowest to the highest acceleration:
D, C, B, A
Heat is allowed to flow from the heat source of a heat engine at 425 K to a cold sink at 313 K. What is the efficiency of the he
2 answers:
I presume that the engine operates under a Carnot cycle, which allows for maximum efficiency.
Assuming this, the Carnot cycle's efficiency can be expressed as
where
represents the temperature of the cold sink
and
signifies the temperature of the heat source.
In this scenario, the cold sink temperature is 313 K, while the heat source temperature is 425 K; thus, we can calculate the engine's efficiency as
[[TAG_15]]
Assuming this, the Carnot cycle's efficiency can be expressed as
where
signifies the temperature of the heat source.
[[TAG_15]]
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Response:
A)
B)
C)
Clarification:
The question is not fully provided. The complete question was:
A skateboarder with a mass of ms = 54 kg is at the top of a ramp with a height of hy = 3.3 m. He then jumps on his skateboard and goes down the ramp. His speed at the base is vf = 6.2 m/s.
Part (a) Formulate an expression for the work, Wf, done by the frictional force on the skateboarder in terms of the variables listed in the problem.
Part (b) The ramp is at an angle θ with the ground, where θ = 30°. Formulate an expression for the frictional force's magnitude, fr, between the skateboarder and the ramp.
Part (c) Upon reaching the bottom, the skateboarder continues with speed vf onto a grass-covered flat surface. The friction between the grass and the skateboarder brings him to a halt after 5.00 m. Determine the frictional force, Fgrass in newtons, between the skateboarder and the grass.
For part A), we assess the energy balance to determine the work done by the friction:
For part B), we utilize the previously calculated work:
Rearranging for friction force:
For part C), we first ascertain the acceleration through kinematic equations and subsequently find the frictional force using dynamic methods:
Rearranging for 'a':
Now, using dynamics:
Response:
Explanation:
Let T denote the tension.
By employing Newton's second law to analyze the bucket's downward motion, we have:
mg - T = ma
A torque, TR, acts on the drum, inducing an angular acceleration α in it. If I refers to the moment of inertia of the drum, then:
TR = Iα
Rearranging gives: TR = Ia/R
This leads to T = Ia/R²
Substituting this expression for T back into the previous equation yields:
mg - T = ma
mg - Ia/R² = ma
Consequently, we find that mg = Ia/R² + ma
Therefore, a (I/R² + m) = mg
This results in: a = mg / (I/R² + m)
Next, we aim to express T as:
mg - T = ma
which simplifies to mg - ma = T
Rearranging gives mg - m²g / (I/R² + m) = T
Thus, we arrive at: mg - mg / (1 + I / m R²) = T
For part (b), T = Ia/R²
and for part (c), the moment of inertia of a hollow cylinder calculates to:
I = 1/2 M (R² - (R² / 4))
This simplifies to 3/4 x 1/2 MR², yielding 3/8 MR²
Thus, I / R² = 3/8 M
When we substitute, we find a = mg / (3/8 M + m)
and subsequently T = Ia/R²
= 3/8 MR² × mg / (3/8 M + m) × 1/R²
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