Answer:
Work done, W = 6 J
Explanation:
It is given that,
Force of gravity acting on the book, weight of the book is 15 N
We need to find the work done in lifting the book straight up for a distance of 0.4 meters.
The weight of the book is acting in downward direction and the book is lifted straight up, it means angle between them is 180 degrees. Work done is given by :

So, the magnitude of work done in lifting the book is 6 joules.
Answer:
No. The protostellar cloud spins faster in the collapsing stage (stage 1) and becomes much slower in the contraction stage (stage 2)
Explanation:
Once the cloud is so dense that the heat which is being produced in its center cannot easily escape, pressure rapidly rises, and catches up with the weight, or whatever external force is causing the cloud to collapse, and the cloud becomes stable, as a protostellar cloud.
The protostellar cloud will become more dense over thousands of years. This stage of decreasing size is known as a contraction, rather than a collapse. In the contraction stage the cloud has become much slower, and because weight and pressure are more or less in balance. In the first stage of formation, the decrease of size is very rapid, and compressive forces completely overwhelm the pressure of the gas, and we say that the cloud is collapsing.
Answer:
3.39 mins.
10m/s multiplied by 6 equals to 60 seconds or 1 minute
Answer:
Quartered
Explanation:
Because you're a liberal.
To solve this problem we will apply the concepts related to the electric field such as the smelting of the Force and the load (In this case the force is equivalent to the weight). Later we will apply the ratio of the total charge as a function of the multiplication of the number of electrons and their individual charge.

Here,
m = mass
g = Acceleration due to gravity
Rearranging to find the charge,

Replacing,


Since the field is acting upwards the charge on the drop should be negative to balance it in air. The equation to find the number of electrons then is

Here,
n = Number of electrons
e = Charge of each electron

Replacing,


Therefore the number of electrons that reside on the drop is 