Answer: Option (b) is the correct answer.
Explanation:
The given data is as follows.
F =
N
g = 9.8 m/s
radius =
=
= 15 cm = 0.15 m (as 1 m = 100 cm)
Formula to calculate depth is as follows.
F = 
or, h =
h =
= 751 m
Thus, we can conclude that the maximum depth in a lake to which the submarine can go without damaging the window is closest 750 m.
Answer:
The distance from Witless to Machmer is 438.63 m.
Explanation:
Given that,
Machmer Hall is 400 m North and 180 m West of Witless.
We need to calculate the distance
Using Pythagorean theorem

Where,
=distance of Machmer Hall
=distance of Witless
Put the value into the formula


Hence, The distance from Witless to Machmer is 438.63 m.
I don't know what the exact word is, but I do know that the bigger an objects mass is the more it will attract other objects toward it, mainly smaller objects with less mass. it might be gravity or something around those lines....is it a multiple choice question?
Answer:
The fraction of its energy that it radiates every second is
.
Explanation:
Suppose Electromagnetic radiation is emitted by accelerating charges. The rate at which energy is emitted from an accelerating charge that has charge q and acceleration a is given by

Given that,
Kinetic energy = 6.2 MeV
Radius = 0.500 m
We need to calculate the acceleration
Using formula of acceleration

Put the value into the formula

Put the value into the formula


We need to calculate the rate at which it emits energy because of its acceleration is

Put the value into the formula


The energy in ev/s


We need to calculate the fraction of its energy that it radiates every second


Hence, The fraction of its energy that it radiates every second is
.
Metamorphic rock
Coal
Changes in pressure.
the matching will be in comments