Answer:
The mass of the another block is 60 kg.
Explanation:
Given that,
Mass of block M= 100 kg
Height = 1.0 m
Time = 0.90 s
Let the mass of the other block is m.
We need to calculate the acceleration of each block
Using equation of motion
![s=ut+\dfrac{1}{2}at^2](https://tex.z-dn.net/?f=s%3Dut%2B%5Cdfrac%7B1%7D%7B2%7Dat%5E2)
Put the value into the formula
![1.0=0+\dfrac{1}{2}\times a\times(0.90)^2](https://tex.z-dn.net/?f=1.0%3D0%2B%5Cdfrac%7B1%7D%7B2%7D%5Ctimes%20a%5Ctimes%280.90%29%5E2)
![a=\dfrac{2\times1.0}{(0.90)^2}](https://tex.z-dn.net/?f=a%3D%5Cdfrac%7B2%5Ctimes1.0%7D%7B%280.90%29%5E2%7D)
![a=2.46\ m/s^2](https://tex.z-dn.net/?f=a%3D2.46%5C%20m%2Fs%5E2)
We need to calculate the mass of the other block
Using newton's second law
The net force of the block M
![Ma=Mg-T](https://tex.z-dn.net/?f=Ma%3DMg-T)
....(I)
The net force of the block m
![ma=T-mg](https://tex.z-dn.net/?f=ma%3DT-mg)
Put the value of T from equation (I)
![ma=Mg-Ma-mg](https://tex.z-dn.net/?f=ma%3DMg-Ma-mg)
![m(a+g)=M(g-a)](https://tex.z-dn.net/?f=m%28a%2Bg%29%3DM%28g-a%29)
![m=\dfrac{M(g-a)}{(a+g)}](https://tex.z-dn.net/?f=m%3D%5Cdfrac%7BM%28g-a%29%7D%7B%28a%2Bg%29%7D)
Put the value into the formula
![m=\dfrac{100(9.8-2.46)}{2.46+9.8}](https://tex.z-dn.net/?f=m%3D%5Cdfrac%7B100%289.8-2.46%29%7D%7B2.46%2B9.8%7D)
![m=59.8\ \approx60\ kg](https://tex.z-dn.net/?f=m%3D59.8%5C%20%5Capprox60%5C%20kg)
Hence, The mass of the another block is 60 kg.
Answer:
Capacitive Reactance is 4 times of resistance
Solution:
As per the question:
R = ![X_{L} = j\omega L = 2\pi fL](https://tex.z-dn.net/?f=X_%7BL%7D%20%3D%20j%5Comega%20L%20%3D%202%5Cpi%20fL)
where
R = resistance
![X_{L} = Inductive Reactance](https://tex.z-dn.net/?f=X_%7BL%7D%20%3D%20Inductive%20Reactance)
f = fixed frequency
Now,
For a parallel plate capacitor, capacitance, C:
![C = \frac{\epsilon_{o}A}{x}](https://tex.z-dn.net/?f=C%20%3D%20%5Cfrac%7B%5Cepsilon_%7Bo%7DA%7D%7Bx%7D)
where
x = separation between the parallel plates
Thus
C ∝ ![\frac{1}{x}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bx%7D)
Now, if the distance reduces to one-third:
Capacitance becomes 3 times of the initial capacitace, i.e., x' = 3x, then C' = 3C and hence Current, I becomes 3I.
Also,
![Z = \sqrt{R^{2} + (X_{L} - X_{C})^{2}}](https://tex.z-dn.net/?f=Z%20%3D%20%5Csqrt%7BR%5E%7B2%7D%20%2B%20%28X_%7BL%7D%20-%20X_%7BC%7D%29%5E%7B2%7D%7D)
Also,
Z ∝ I
Therefore,
![\frac{Z}{I} = \frac{Z'}{I'}](https://tex.z-dn.net/?f=%5Cfrac%7BZ%7D%7BI%7D%20%3D%20%5Cfrac%7BZ%27%7D%7BI%27%7D)
![\frac{\sqrt{R^{2} + (R - X_{C})^{2}}}{3I} = \frac{\sqrt{R^{2} + (R - \frac{X_{C}}{3})^{2}}}{I}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%7BR%5E%7B2%7D%20%2B%20%28R%20-%20X_%7BC%7D%29%5E%7B2%7D%7D%7D%7B3I%7D%20%3D%20%5Cfrac%7B%5Csqrt%7BR%5E%7B2%7D%20%2B%20%28R%20-%20%5Cfrac%7BX_%7BC%7D%7D%7B3%7D%29%5E%7B2%7D%7D%7D%7BI%7D)
![{R^{2} + (R - X_{C})^{2}} = 9({R^{2} + (R - \frac{X_{C}}{3})^{2}})](https://tex.z-dn.net/?f=%7BR%5E%7B2%7D%20%2B%20%28R%20-%20X_%7BC%7D%29%5E%7B2%7D%7D%20%3D%209%28%7BR%5E%7B2%7D%20%2B%20%28R%20-%20%5Cfrac%7BX_%7BC%7D%7D%7B3%7D%29%5E%7B2%7D%7D%29)
![{R^{2} + R^{2} + X_{C}^{2} - 2RX_{C} = 9({R^{2} + R^{2} + \frac{X_{C}^{2}}{9} - 2RX_{C})](https://tex.z-dn.net/?f=%7BR%5E%7B2%7D%20%2B%20R%5E%7B2%7D%20%2B%20X_%7BC%7D%5E%7B2%7D%20-%202RX_%7BC%7D%20%3D%209%28%7BR%5E%7B2%7D%20%2B%20R%5E%7B2%7D%20%2B%20%5Cfrac%7BX_%7BC%7D%5E%7B2%7D%7D%7B9%7D%20-%202RX_%7BC%7D%29)
Solving the above eqn:
![X_{C} = 4R](https://tex.z-dn.net/?f=X_%7BC%7D%20%3D%204R)
when the ball hits the floor and bounces back the momentum of the ball changes.
the rate of change of momentum is the force exerted by the floor on it.
the equation for the force exerted is
f = rate of change of momentum
![f = \frac{mv - mu}{t}](https://tex.z-dn.net/?f=f%20%3D%20%5Cfrac%7Bmv%20-%20mu%7D%7Bt%7D)
v is the final velocity which is - 3.85 m/s
u is initial velocity - 4.23 m/s
m = 0.622 kg
time is the impact time of the ball in contact with the floor - 0.0266 s
substituting the values
![f = \frac{0.622 kg (3.85 m/s - (-)4.23 m/s)}{0.0266}](https://tex.z-dn.net/?f=f%20%3D%20%5Cfrac%7B0.622%20kg%20%283.85%20m%2Fs%20-%20%28-%294.23%20m%2Fs%29%7D%7B0.0266%7D)
since the ball is going down, we take that as negative and ball going upwards as positive.
f = 189 N
the force exerted from the floor is 189 N
Answer:
at resonance impedence is equal to resistance and quality factor is dependent on R L AND C all
Explanation:
we know that for series RLC circuit impedance is given by
![Z=\sqrt{R^2+\left ( X_L-X_C \}right )^2](https://tex.z-dn.net/?f=Z%3D%5Csqrt%7BR%5E2%2B%5Cleft%20%28%20X_L-X_C%20%5C%7Dright%20%29%5E2)
but we know that at resonance
putting
in impedance formula , impedance will become
Z=R so at resonance impedance of series RLC is equal to resistance only
now quality factor of series resonance is given by
so from given expression it is clear that quality factor depends on R L and C