Answer:
a)this graph is also a line b) in both cases we have a uniform movement
Explanation:
In this exercise we have a uniform movement
v = d / t
d = v t
in the table we give some values to make the graph
t (s) d (m)
1 10
2 20
3 30
In the attached we can see the graph that is a straight line
we have another vehicle at v = 50 me / S
t (s) d (m)
1 50
2 100
3 150
this graph is also a line
b) in both cases we have a uniform movement
Answer:
unchanged
Explanation:
Let the voltage of the battery be V
Inductance L1 = 5 mH
Inductance L2 = 10 mH
consider resistance R of the circuit (wire, battery).
V = I R + L dI/dt
where, I is the current in the circuit and t is the time.
After few seconds of connection being made, the factor dI/dt is negligible. There is no change in the current flowing through the circuit. when inductor was just attached in the circuit, a current
Answer: The power is 156 watt
Explanation:
is in the attachment
This question is incomplete, the complete question is;
A 8.45μC particle with a mass of 6.15 x 10⁻⁵ kg moves perpendicular to a 0.493-T magnetic field in a circular path of radius 34.1 m.
How much time will it take for the particle to complete one orbit?
a. 92.7 s
b. 0.0927 s
c. 9.27 s
d. 927 s
Answer:
it will take 92.7 seconds for the particle to complete one orbit.
Option a) 92.7 s is the correct option
Explanation:
Given that;
mass m = 6.15 x 10⁻⁵ kg
q = 8.45μC = 8.45 × 10⁻⁶ C
B = 0.493
we know that
Time period T = 2πr / V
where r = mv/qB
so T = 2πm/qB
we substitute
T = (2 × 3.14 × 6.15 x 10⁻⁵) / ( 8.45 × 10⁻⁶ × 0.493)
T = 0.0003862 / 0.000004165
T = 92.7 sec
Therefore it will take 92.7 seconds for the particle to complete one orbit.
Option a) 92.7 s is the correct option