The force of friction depends upon
2 answers:
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<h3>
</h3><h3>Given</h3>
v = 20m\s
a = 3m\s^2
t = 4sec
Firstly we have to find u
a =
3m\s =
12m\s = 20 - u
20 - u = 12m\s
- u = -8
u = 8
Now we can easily find distance by using second equation of motion
s = ut + 1\2 at^2
s = 8(4) + 1\2(3)(16)
s = 32 + 24
s = 56
So distance is 56 m\s hope it helps
The brightness of the lamp is proportional to the current flowing through the lamp: the larger the current, the brighter the lamp.
The current flowing through the lamp is given by Ohm's law:
where
V is the potential difference across the lamp, which is equal to the emf of the battery, and R is the resistance of the lamp.
The problem says that the battery is replaced with one with lower emf. Looking at the formula, this means that V decreases: if we want to keep the same brightness, we need to keep I constant, therefore we need to decrease R, the resistance of the lamp.
The correct answer is "As the distance from the earth increases, the gravitational pull on the spaceship would decrease."
In fact, the gravitational force (attractive) exerted by the Earth on the spaceship is given by
where G is the gravitational constant, M the Earth's mass, m the mass of the spaceship and d the distance of the spaceship from the Earth. As we can see from the formula, as the distance d between the spaceship and the Earth increases, the gravitational force F decreases, so answer D) is the correct one.
In fact, the gravitational force (attractive) exerted by the Earth on the spaceship is given by
where G is the gravitational constant, M the Earth's mass, m the mass of the spaceship and d the distance of the spaceship from the Earth. As we can see from the formula, as the distance d between the spaceship and the Earth increases, the gravitational force F decreases, so answer D) is the correct one.