Answer: A
Explanation:
Lunar eclipses work the same way in a different order: Moon, Earth and Sun all on a line. In this case the Earth's shadow hides the Moon from view.
Answer:
9.6 m
Explanation:
This is a case of motion under variable acceleration . So no law of motion formula will be applicable here. We shall have to integrate the given equation .
a = 3.6 t + 5.6
d²x / dt² = 3.6 t + 5.6
Integrating on both sides
dx /dt = 3.6 t² / 2 + 5.6 t + c
where c is a constant.
dx /dt = 1.8 t² + 5.6 t + c
when t = 0 , velocity dx /dt is zero
Putting these values in the equation above
0 = 0 +0 + c
c = 0
dx /dt = 1.8 t² + 5.6 t
Again integrating on both sides
x = 1.8 t³ / 3 + 5.6 x t² /2 + c₁
x = 0.6 t³ + 2.8 t² + c₁
when t =0, x = 0
c₁ = 0
x = 0.6 t³ + 2.8 t²
when t = 1.6
x = .6 x 1.6³ + 2.8 x 1.6²
= 2.4576 + 7.168
= 9.6256
9.6 m
To solve this problem we will apply the concepts related to the linear kinematic movement. We will start by finding the speed of the body from time and the acceleration given.
Through the position equations we will calculate the distance traveled.
Finally, using this same position relationship and considering the previously found speed, we can determine the time to reach your goal.
For time (t) and acceleration (a) we have to,

The velocity would be,

Now the position is,



Now with the initial speed and position found we will have the time is,



Solving the polynomian we have,

Therefore the rocket will take to hit the ground around to 4.56min
<span>Resistance of automobile light is equal to 12 ohms.
</span>Resistance = Voltage / Current
where
Voltage = 12 volts
Current = 1.0 Amperes
Resistance = 12 volts/ 1.0 amperes
Resistance = 12 ohms.
Automobile light has 12 ohms resistance when it is connected to 12volts battery with 1.0 ampere current.
Answer: Option (c) is the correct answer
Explanation:
It is known that when a neutral object comes in contact with a charged object then an opposite charge develops on the neutral object. But this development of charge occurs only when both the objects come in contact and if they are not in contact with each other then there occurs no charge on the neutral object.
As in the given situation, negatively-charged glass rod is brought near, but does not touch the sphere. So, there will occur no charge on the sphere.
Thus, we can conclude that final charge on the sphere is neutral.