Answer:
Explanation:
Given that,
AC frequency of 2.3KHz
f=2.3×10³Hz
Vrms produce is
Vrms=1.5V
Current rms
Irms= 31mA
The capacitor is reconnected to a generator of frequency
f=4.8KHz =4800Hz
The current rms becomes
Irms= 85mA
Vrms=?
Solution
First genrator
The capacitive reactance is given as
Xc=Vrms/Irms
Xc=1.5/31×10^-3
Xc=48.39 ohms
Now, to know the capacitance of the capacitor
Xc=1/2πfC
Then,
C=1/2πfXc
So,
C=1/2×π×2300×48.39
C=1.43×10^-6C
C=1.43μF
Note: the capacitance of the capacitor did not change,
Now for generator two.
The reactance are given as
Xc=1/2πfC
Xc=1/2×π×4800×1.43×10^-6
Xc=23.19ohms
Then,
Vrms2=Irms2 ×Xc
Vrms2=85×10^-3×23.19ohms
Vrms2=1.97V
Vrms2=1.97Volts
The bowling ball will require more force to roll because it is more massive.
Answer:
Explanation:
Force, F = - mg j
r = - 7x i + y j
Torque is defined as the product f force and the perpendicular distance.
It is also defined as the cross product of force vector and the displacement vector.
[tex]\overrightarrow{\tau }= 7 m g x k
Here, we observe that the torque is independent of y coordinate.
Answer: The height above the release point is 2.96 meters.
Explanation:
The acceleration of the ball is the gravitational acceleration in the y axis.
A = (0, -9.8m/s^)
For the velocity we can integrate over time and get:
V(t) = (9.20m/s*cos(69°), -9.8m/s^2*t + 9.20m/s^2*sin(69°))
for the position we can integrate it again over time, but this time we do not have any integration constant because the initial position of the ball will be (0,0)
P(t) = (9.20*cos(69°)*t, -4.9m/s^2*t^2 + 9.20m/s^2*sin(69°)*t)
now, the time at wich the horizontal displacement is 4.22 m will be:
4.22m = 9.20*cos(69°)*t
t = (4.22/ 9.20*cos(69°)) = 1.28s
Now we evaluate the y-position in this time:
h = -4.9m/s^2*(1.28s)^2 + 9.20m/s^2*sin(69°)*1.28s = 2.96m
The height above the release point is 2.96 meters.
