Answer:
E = h ν energy of electromagnetic particle
(b) has the greater frequency
Answer:
a) V_f = 25.514 m/s
b) Q =53.46 degrees CCW from + x-axis
Explanation:
Given:
- Initial speed V_i = 20.5 j m/s
- Acceleration a = 0.31 i m/s^2
- Time duration for acceleration t = 49.0 s
Find:
(a) What is the magnitude of the satellite's velocity when the thruster turns off?
(b) What is the direction of the satellite's velocity when the thruster turns off? Give your answer as an angle measured counterclockwise from the +x-axis.
Solution:
- We can apply the kinematic equation of motion for our problem assuming a constant acceleration as given:
V_f = V_i + a*t
V_f = 20.5 j + 0.31 i *49
V_f = 20.5 j + 15.19 i
- The magnitude of the velocity vector is given by:
V_f = sqrt ( 20.5^2 + 15.19^2)
V_f = sqrt(650.9861)
V_f = 25.514 m/s
- The direction of the velocity vector can be computed by using x and y components of velocity found above:
tan(Q) = (V_y / V_x)
Q = arctan (20.5 / 15.19)
Q =53.46 degrees
- The velocity vector is at angle @ 53.46 degrees CCW from the positive x-axis.
Answer:
The gauge pressure is 
Explanation:
From the question we are told that
The height of the water contained is 
The height of liquid in the cylinder is 
At the bottom of the cylinder the gauge pressure is mathematically represented as
Where 
is the pressure of water which is mathematically represented as
Now 
is the density of water with a constant values of 
substituting values
While 
is the pressure of oil which is mathematically represented as
Where 
is the density of oil with a constant value
substituting values
Therefore
Answer:
C) upward
Explanation:
The problem can be solved by using the right-hand rule.
First of all, we notice at the location of the negatively charged particle (above the wire), the magnetic field produced by the wire points out of the page (because the current is to the right, so by using the right hand, putting the thumb to the right (as the current) and wrapping the other fingers around it, we see that the direction of the field above the wire is out of the page).
Now we can apply the right hand rule to the charged particle:
- index finger: velocity of the particle, to the right
- middle finger: direction of the magnetic field, out of the page
- thumb: direction of the force, downward --> however, the charge is negative, so we must reverse the direction --> upward
Therefore, the direction of the magnetic force is upward.
The rms current in the transmission lines is I = 487.18 A.
The root-imply-rectangular (rms) voltage of a sinusoidal supply of electromotive force is used to represent the source. it is the rectangular root of the time average of the voltage squared.
Alternating-present day circuits. the root-imply-square (rms) voltage of a sinusoidal source of electromotive force is used to symbolize the supply. it's far the square root of the time average of the voltage squared.
Electric power is by using present day or the waft of electric fee and voltage or the capacity of rate to deliver electricity. A given cost of power can be produced by using any combination of contemporary and voltage values
power = 38 M watt
rms voltage = 78 K v
power = IV
I = power/V
I = (38 * 1000000)/78*1000
I = 487.18 A.
Learn more about rms current here:-brainly.com/question/20913680
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