In order to effectively radiate a radio signal, an antenna must be at least one-half the wavelength of transmission in length, i
1 answer:
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Solution :
Given :
Mass attached to the spring = 4 kg
Mass dropped = 6 kg
Force constant = 100 N/m
Initial amplitude = 2 m
Therefore,
a).
= 10 m/s
Final velocity, v at equilibrium position, v = 5 m/s
Now,
A' = amplitude = 1.4142 m
b).
m' = 2m
Hence,
c).
Therefore, factor
Thus, the energy will change half times as the result of the collision.
Answer:
The effective spring constant of the firing mechanism is 1808N/m.
Explanation:
First, we can use kinematics to obtain the initial velocity of the performer. Since we know the angle at which he was launched, the horizontal distance and the time in which it's traveled, we can calculate the speed by:
(This is correct because the horizontal motion has acceleration zero). Then:
Now, we can use energy to obtain the spring constant of the firing mechanism. By the conservation of mechanical energy, considering the instant in which the elastic band is at its maximum stretch as t=0, and the instant in which the performer flies free of the bands as final time, we have:
Then, plugging in the given values, we obtain:
Finally, the effective spring constant of the firing mechanism is 1808N/m.
Actually Welcome to the concept of Efficiency.
Here we can see that, the Input work is given as 2.2 x 10^7 J and the efficiency is given as 22%
The efficiency is => 22% => 22/100.
so we get as,
E = W(output) /W(input)
hence, W(output) = E x W(input)
so we get as,
W(output) = (22/100) x 2.2 x 10^7
=> W(output) = 0.22 x 2.2 x 10^7 => 0.484 x 10^7
hence, W(output) = 4.84 x 10^6 J
The useful work done on the mass is 4.84 x 10^6 J