Answer:i=300 mA
Explanation:
Given
inductance(L)=40 mH
Resistor(R)=
Voltage(V)=15 V
Time constant()=
current
Current as a function of time is given by
i= 299.95 mA
Two identical metal spheres A and B are in contact. Both are initially neutral. 1.0× 10 12 electrons are added to sphere A, then
1 answer:
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Answer:
a) 0.500 s
b) greater than 0.500 s
c) greater than 0.500 s
Explanation:
The time period of an oscillating spring-mass system is given by:
where, m is the mass and k is the spring constant.
a) As the period of oscillation does not depend on the distance by which the mass is pulled, the period would remain same as 0.500 s for the given system.
b) As the period varies inversely with the square root of spring constant, so with the decrease in the spring constant, the period would increase. So, the new period would be greater than 0.500 s.
c) As the period varies directly with the square root of mass, so with the increase in mass, the period will also increase. The new period will be greater than 0.500 s.
Answer:
42.11 years old
Explanation:
Given that:
In 2000, a 20-year-old astronaut left Earth to explore the galaxy; her spaceship travels at 2.5 x 10^8 m/s. She returns in 2040
To find her age we use:
Δtm is time interval for the observer stationary relative to the sequence of
events = 2040 - 2000 = 40 years
Δts is is the time interval for an observer moving with a speed v relative to the sequence of event
v = velocity = 2.5 x 10^8 m/s
c = speed of light = 3 x 10^8 m/s
Here age in 2000 is 20 year, therefore when she appear she would be 20 year + 22.11 year = 42.11 years old
Remain the same
Explanation:
If the force exerted by the intern is doubled and the distance is halved, the work done by the intern remains the same.
Work done is the force applied to move a body through a distance.
Work done = F x d
where F is the applied force
d is the distance moved
Now;
if:
f = 2f
d = d
Input the parameter:
Work done = fxd = 2f x d = fd
The work done will still remain the same
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