When the termination is a terminal block, care must be taken to ensure a good electrical connection without damaging the conduct
1 answer:
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We have that the spring constant is mathematically given as
Generally, the equation for angular velocity is mathematically given by
Where
k=spring constant
And
Therefore
Hence giving spring constant k
Generally
Mass of earth
Period for on complete resolution of Earth around the Sun
Therefore
In conclusion
The effective spring constant of this simple harmonic motion is
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A) The net force on the motorbike is 205.9 N
B) The acceleration of the motorbike is
C) The final speed is 5.2 m/s
D) The elapsed time is 3.80 s
Explanation:
A)
There are two forces acting on the motorbike:
- The applied force, F = 250 N, forward
- The frictional force, , backward
The frictional force can be written as
where
is the coefficient of kinetic friction
is the mass of the motorbike
is the acceleration of gravity
Therefore the net force is given by
And substituting, we find
2)
The acceleration of the motorbike can be found by using Newton's second law, which states that the net force is equal to the product between mass and acceleration:
where
m is the mass
a is the acceleration
In this problem, we have
is the net force
m = 150 kg is the mass
Solving for a, we find the acceleration:
C)
Since the motion of the motorbike is a uniformly accelerated motion, we can use the following suvat equation:
where
v is the final velocity
u is the initial velocity
a is the acceleration
s is the distance covered
For this motorbike, we have:
u = 0 (it starts from rest)
s = 350 m
Solving for v,
4)
For this part of the problem, we can use the following suvat equation:
where
v is the final velocity
u is the initial velocity
a is the acceleration
t is the elapsed time
Here we have:
v = 5.2 m/s
u = 0
Solving for t, we find
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<span>its kinetic energy is 7350kJ
</span>
Kinetic energy is given as =
Now, m = 12 gms = 0.012 kg
And, velocity = 35 kilometers/second = 35000 m/sec
Kinetic energy is given as =![\frac{1}{2} 0.012 kg * 35000*35000 m/[tex] s^{2}](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B2%7D%200.012%20kg%20%2A%2035000%2A35000%20m%2F%5Btex%5D%20s%5E%7B2%7D%20)
= 6
×1225 ×
m/
= 7350 kJ
</span>
Kinetic energy is given as =
Now, m = 12 gms = 0.012 kg
And, velocity = 35 kilometers/second = 35000 m/sec
Kinetic energy is given as =
= 6
= 7350 kJ