Answer:
35.55 ohm
Explanation:
C = 80 micro farad
F = 56 hertz
Reactance
Xc = 1 / (2 × pi × F × C)
Xc = 1 / ( 2 × 3.14 × 56 × 80 × 10^-6)
Xc = 35.55 ohm
It is the Jupiter from outer space. And the years from 170 light years away from the Earth.
Answer:
visible light
Explanation:
we can not see with out visibl e light
Answer:
The value is 
Explanation:
From the question we are told that
The velocity is ![a = 120 \ km/h = [tex]\frac{120 * 1000}{3600} = 33.3 \ m/s](https://tex.z-dn.net/?f=a%20%3D%20%20120%20%5C%20%20km%2Fh%20%3D%20%20%5Btex%5D%5Cfrac%7B120%20%2A%20%201000%7D%7B3600%7D%20%3D%20%2033.3%20%5C%20%20m%2Fs)
[/tex]
The time taken is 
The time taken for contact is 
The velocity of the of the car after contact is ![v_c = 71.0 \ km/h = [tex]\frac{71 *1000}{3600} = 19.7 2 \ m/s](https://tex.z-dn.net/?f=v_c%20%3D%20%2071.0%20%5C%20%20km%2Fh%20%3D%20%5Btex%5D%5Cfrac%7B71%20%2A1000%7D%7B3600%7D%20%3D%20%2019.7%202%20%5C%20%20m%2Fs)
[/tex]
From the equation of kinematics we have that
Here u = 0 \ m/s since the car is initially motionless
=> 
=>
Answer:
X(t) = 9.8 *t - 4.9 * t^2
Explanation:
We set a frame of reference with origin at the hand of the girl the moment she releases the ball. We assume her hand will be in the same position when she catches it again. The positive X axis point upwards.The ball will be subject to a constant gravitational acceleration of -9.81 m/s^2.
We use the equation for position under constant acceleration:
X(t) = X0 + V0 * t + 1/2 * a *t^2
X0 = 0 because it is at the origin of the coordinate system.
We know that at t = 2, the position will be zero.
X(2) = 0 = V0 * 2 + 1/2 * -9.81 * 2^2
0 = 2 * V0 - 4.9 * 4
2 * V0 = 19.6
V0 = 9.8 m/s
Then the position of the ball as a function of time is:
X(t) = 9.8 *t - 4.9 * t^2
