Explanation:
hope it helps.
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How does increasing the speed of a motorbike change the amount of energy in its kinetic energy store?
2 answers:
Answer:
If the speed of a motorbike increases, its kinetic energy will also increase.
Explanation:
Kinetic energy is the energy an object has due to its motion, and it is a function of the mass and velocity of the object:
From the expression, it is easy do identify a quadratic relation between kinetic energy and velocity. Therefore, if the speed of a motorbike increases, its kinetic energy will also increase.
For example, if you increase the speed of a motorbike by 10 times, and since mass is a constant value, the kinetic energy will increase by 100 times.
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Answer:
Final temperature is equal to 1291.63°R
Explanation:
given,
p₁ = 100 lb f/in², v₁ = 3.704 ft³/lb, and T₁ = 1000 °R
p₂ = 30 lb f/in² n = 1.4
Δ u = 0.171(T₂ - T₁)
we know for poly tropic process
p vⁿ = constant
p₁ v₁ⁿ = p₂ v₂ⁿ
100 × 3.704¹°⁴ = 30 × v₂¹°⁴
v₂ = 8.753 ft³/lb
work done for poly tropic process
W =
=
= 269.525 lbf/in².ft³
W = Btu/lb
= 49.87 Btu/lb
in the piston cylinder arrangement air is expanding acrobatically
Δ q = Δu + w
Δ u = - w
0.171(T₂ - T₁) = -49.87
0.171(T₁ - T₂) = -49.87
0.171 T₂ = 0.171 × 1000 + 49.87
T₂ = 1291.63 °R
Final temperature is equal to 1291.63°R
Answer:
a)
a = 2 [m/s^2]
b)
a = 1.6 [m/s^2]
c)
xt = 2100 [m]
Explanation:
In order to solve this problem we must use kinematics equations. But first we must identify what kind of movement is being studied.
a)
When the car moves from rest to 40 [m/s] by 20 [s], it has a uniformly accelerated movement, in this way we can calculate the acceleration by means of the following equation:
where:
Vf = final velocity = 40 [m/s]
Vi = initial velocity = 0 (starting from rest)
a = acceleration [m/s^2]
t = time = 20 [s]
40 = 0 + (a*20)
a = 2 [m/s^2]
The distance can be calculates as follows:
where:
x1 = distance [m]
40^2 = 0 + (2*2*x1)
x1 = 400 [m]
Now the car maintains its speed of 40 [m/s] for 30 seconds, we must calculate the distance x2 by means of the following equation, it is important to emphasize that this movement is at a constant speed.
v = x2/t2
where:
x2 = distance [m]
t2 = 30 [s]
x2 = 40*30
x2 = 1200 [m]
b)
Immediately after a change of speed occurs, such that the previous final speed becomes the initial speed, the new Final speed corresponds to zero, since the car stops completely.
Note: the negative sign of the equation means that the car is stopping, i.e. slowing down.
0 = 40 - (a *25)
a = 40/25
a = 1.6 [m/s^2]
The distance can be calculates as follows:
0 = (40^2) - (2*1.6*x3)
x3 = 500 [m]
c)
Now we sum all the distances calculated:
xt = x1 + x2 + x3
xt = 400 + 1200 + 500
xt = 2100 [m]