Answer:
B. The current increases.
Explanation:
As we know that rate of flow of charge through the conductor is known as electric current
So we have
here we know that charge Q flowing through the conductor is constant while the time in which it passes through it is decreased
so we can say that the ratio of charge and time will increase
so here we have
So correct answer will be
B. The current increases.
We know that:
d=vt
d=32mph*5h
d=160mi
"60 kg" is not a weight. It's a mass, and it's always the same
no matter where the object goes.
The weight of the object is
(mass) x (gravity in the place where the object is) .
On the surface of the Earth,
Weight = (60 kg) x (9.8 m/s²)
= 588 Newtons.
Now, the force of gravity varies as the inverse of the square of the distance from the center of the Earth.
On the surface, the distance from the center of the Earth is 1R.
So if you move out to 5R from the center, the gravity out there is
(1R/5R)² = (1/5)² = 1/25 = 0.04 of its value on the surface.
The object's weight would also be 0.04 of its weight on the surface.
(0.04) x (588 Newtons) = 23.52 Newtons.
Again, the object's mass is still 60 kg out there.
___________________________________________
If you have a textbook, or handout material, or a lesson DVD,
or a teacher, or an on-line unit, that says the object "weighs"
60 kilograms, then you should be raising a holy stink.
You are being planted with sloppy, inaccurate, misleading
information, and it's going to be YOUR problem to UN-learn it later.
They owe you better material.
Answer:
Wnet, in, = 133.33J
Explanation:
Given that
Pump heat QH = 1000J
Warm temperature TH= 300K
Cold temperature TL= 260K
Since the heat pump is completely reversible, the combination of coefficient of performance expression is given as,
From first law of thermodynamics,
COP(HP, rev) = 1/(1-TL/TH)
COP(HP, rev) = 1/(1-260/300)
COP(HP, rev) = 1/(1-0.867)
COP(HP, rev) = 1/0.133
COP(HP, rev) = 7.5
The power required to drive the the heat pump is given as
Wnet, in= QH/COP(HP, rev)
Wnet, in = 1000/7.5
Wnet, in = 133.333J. QED
So the 133.33J was the amount heat that was originally work consumed in the transfer.
Extra....
According to the first law, the rate at which heat is removed from the low temperature reservoir is given as
QL=QH-Wnet, in
QL=1000-133.333
QL=866.67J
