The compression curve would be theoretically given for a system of bodies in which the spring applies the force (Although in the same way the following process can be extrapolated to any system, depending on the type of Force to consider) For a spring mass system, the strength is given by Hooke's law as
Where,
K = Spring constant
x = Displacement
If we integrate based on distance we would have
This integral represents the area under the Force Curve based on each distance segment traveled.
This is the same formula that represents the elastic potential energy of a body. Therefore the correct answer is D.
The distance between the two cities is 513.24 km.
<h3>Time of motion when the two trains meet</h3>
The time spent on the journey when the two trains meet is calculated as follows;
(Va - Vb)t = d
where;
- d is the distance between the trains before meeting
(76 - 65)t = 40
11t = 40
t = 40/11
t = 3.64 hr
<h3>Distance traveled by the fast train</h3>
d1 = 76 km/h x 3.64 h
d1 = 276.64 km
<h3>Distance traveled by the slow train</h3>
d2 = 65 km/h x 3.64 h
d2 = 236.6 km
The distance between the two cities = 276.64 km + 236.6 km
= 513.24 km
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The internal energy of the gas is 49,200 J
Explanation:
The internal energy of a diatomic gas, such as , is given by
where
n is the number of moles
R is the gas constant
T is the absolute temperature of the gas
For the gas in this problem, we have:
n = 4.50 (number of moles)
R = 8.31 J/(mol·K) (gas constant)
(absolute temperature)
Substituting, we find:
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Answer:
option (D)
Explanation:
Here initial rotation speed is given, final rotation speed is given and asking for time.
If we use
A) θ=θ0+ω0t+(1/2)αt2
For this equation, we don't have any information about the value of angular displacement and angular acceleration, so it is not useful.
B) ω=ω0+αt
For this equation, we don't have any information about angular acceleration, so it is not useful.
C) ω2=ω02+2α(θ−θ0)
In this equation, time is not included, so it is not useful.
D) So, more information is needed.
Thus, option (D) is true.
Answer:
Part a)
the tension force is equal to the weight of the crate
Part b)
tension force is more than the weight of the crate while accelerating upwards
tension force is less than the weight of crate if it is accelerating downwards
Explanation:
Part a)
When large crate is suspended at rest or moving with uniform speed then it is given as
here since speed is constant or it is at rest
so we will have
so the tension force is equal to the weight of the crate
Part b)
Now let say the crate is accelerating upwards
now we can say
so tension force is more than the weight of the crate
Now if the crate is accelerating downwards
so tension force is less than the weight of crate if it is accelerating downwards