Neap tides occur at or immediately after the First Quarter
and Third Quarter moon phases. At those times, the moon
is 'to the side' of the Earth. I mean, the sun, Earth, and Moon
make a right angle, with Earth at the vertex.
Answer: A car slows from 22 m/s to 3.0 m/s at a constant rate of 2.1 m/s2. How many seconds are required before the car is a r2.1 m/s2 traveling at 3.0 m/s?
5 pages
Explanation:
Answer:
use the formula
Explanation:
F = 1/ T
where F stands for frequency and T stands for period
sub in 3 Hz into frequency
3 = 1/T
0.333 = T
or in a fraction form
1/ 3
hope this helps
1) Let's call

the speed of the southbound boat, and

the speed of the eastbound boat, which is 3 mph faster than the southbound boat. We can write the law of motion for the two boats:


2) After a time

, the two boats are

apart. Using the laws of motion written at step 1, we can write the distance the two boats covered:


The two boats travelled in perpendicular directions. Therefore, we can imagine the distance between them (45 mi) being the hypotenuse of a triangle, of which

and

are the two sides. Therefore, we can use Pythagorean theorem and write:

Solving this, we find two solutions. Discarding the negative solution, we have

, which is the speed of the southbound boat.
Answer:
Explanation:
a ) gauge pressure will be due to water column of length 15 cm .
pressure = h d g , h is height of column , d is density of column and g is gravitational acceleration .
= .15 x 10³ x 9.8
= 1470 Pa .
b )
Let due of weight of water column , mercury level in left column goes down by distance h . The level of mercury in right column will rise by the same distance ie by distance h .
So mercury column of 2h height is balancing the water column of height 15 cm
2h x 13.6 x 10³ x g = .15 x 10³ x g
h = .15 / (2 x 13.6)
= .55 x 10⁻² m
= . 55 cm
Difference of height of water column and mercury column
= 15 - 2 x .55 cm
= 13.9 cm .