An object is moving east, and its velocity changes from 65 m/s to 25 m/s in 10 seconds Which describes the acceleration?
1 answer:
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Answer:
h = 3.5 m
Explanation:
First, we will calculate the final speed of the ball when it collides with a seesaw. Using the third equation of motion:
where,
g = acceleration due to gravity = 9.81 m/s²
h = height = 3.5 m
vf = final speed = ?
vi = initial speed = 0 m/s
Therefore,
Now, we will apply the law of conservation of momentum:
where,
m₁ = mass of colliding ball = 3.6 kg
m₂ = mass of ball on the other end = 3.6 kg
v₁ = vf = final velocity of ball while collision = 8.3 m/s
v₂ = vi = initial velocity of other end ball = ?
Therefore,
Now, we again use the third equation of motion for the upward motion of the ball:
where,
g = acceleration due to gravity = -9.81 m/s² (negative for upward motion)
h = height = ?
vf = final speed = 0 m/s
vi = initial speed = 8.3 m/s
Therefore,
<u>h = 3.5 m</u>
Answer:
Amplitude = 8 Volts
Frequency = 0.067 kHz
Explanation:
Note: The missing picture in question is attached for your review.
Given:
Volts/Div = 2 V/div
Time/Div = 5 msec/div
Finding Amplitude:
Now, as you can see in the attached picture, there are 4 division between two peaks of the waveform, so,
(Multiplying by 2 V/div because oscilloscope dial is set at 2 V/div)
Finding Frequency:
As can be seen in attached picture, 3 division are there for one complete cycle of waveform,so,
Since,
Answer:
The Entropy generated by the steam = 2.821 kJ/K
Explanation:
Total volume of container = 5m³
Heat transfer does not exist between system and surrounding, dQ = 0
At the first chamber, temperature of water at saturated liquid is 300°C
From the steam table:
Specific enthalpy of saturated liquid at 300°C ,
Specific internal energy of saturated liquid at 300°C, 1332.7 kJ/kg
For closed system, the first law of thermodynamics state that:
dQ = dw + dU..................(1)
work done for free expansion, dw =0
0 = 0 + dU
dU = 0 , i.e. U₁ = U₂
At the second chamber,
The final pressure, P₂ = 50 kPa
From the steam table, at P₂ = 50 kPa, = 340.49 kJ/kg
Let the dryness fraction at the second chamber = x
Since U₁ = U₂
1332.7 = 340.49 + x2140.7
Dryness fraction, x = 0.463
From steam table, the specific volume is,
V₂ = 5 m³
1.5 = 5/m₂
m₂ = 3.33 kg
At 300°C
From the steam table,
Therefore the entropy generated will be :
Entropy = mass* (S₂ - S₁)
Entropy = 3.33* (4.102 - 3.2548)
Entropy = 2.821 kJ/K