Answer:
new atmospheric pressure is 0.9838 ×
Pa
Explanation:
given data
height = 21.6 mm = 0.0216 m
Normal atmospheric pressure = 1.013 ✕ 10^5 Pa
density of mercury = 13.6 g/cm³
to find out
atmospheric pressure
solution
we find first height of mercury when normal pressure that is
pressure p = ρ×g×h
put here value
1.013 ×
= 13.6 × 10³ × 9.81 × h
h = 0.759 m
so change in height Δh = 0.759 - 0.0216
new height H = 0.7374 m
so new pressure = ρ×g×H
put here value
new pressure = 13.6 × 10³ × 9.81 × 0.7374
atmospheric pressure = 98380.9584
so new atmospheric pressure is 0.9838 ×
Pa
The magnet will be a useful tool to pick the paper clips because the magnet can attract the paper clips.
<h3>
Attraction of magnets</h3>
The like poles of magnets repel while unlike poles of magnets attracts. Magnets attracts irons or metallic materials.
The paper clips are mettalic or made of iron and hence the magnet will attract them.
There we can conclude that the magnet will be a useful tool to pick the paper clips because the magnet can attract the paper clips and will help to gather them together for easy picking.
Learn more about magnets here: brainly.com/question/14997726
The frequency of rotation of Mars is 0.0000113 Hertz.
<u>Given the following data:</u>
- Period = 1 day and 37 minutes.
To find the frequency of rotation in Hertz:
First of all, we would convert the the value of period in days and minutes to seconds because the period of oscillation of a physical object is measured in seconds.
<u>Conversion:</u>
1 day = 24 hours
24 hours to minutes =
×
=
minutes

1 minute = 60 seconds
1477 minute = X seconds
Cross-multiplying, we have:
× 
X = 88620 seconds
Now, we can find the frequency of rotation of Mars by using the formula:

<em>Frequency </em><em>of rotation</em> = <em>0.0000113 Hertz</em>
Therefore, the frequency of rotation of Mars is 0.0000113 Hertz.
Read more: brainly.com/question/14708169
Answer:
ρ = 830.32 kg/m³
Explanation:
Given that
Oil head = 12.2 m
h= 12.2 m
Pressure P = 1.013 x 10⁵ Pa
Lets take density of the liquid =ρ
The pressure due to liquid P given as
P = ρ g h
Now by putting the all values in the above equation
1.013 x 10⁵ Pa = ρ x 10 x 12.2 ( take g =10 m/s²)
ρ = 830.32 kg/m³
Therefore the density of oil is 830.32 kg/m³
Answer:
The magnitude of the rate of change of the child's momentum is 794.11 N.
Explanation:
Given that,
Mass of child = 27 kg
Speed of child in horizontal = 10 m/s
Length = 3.40 m
There is a rate of change of the perpendicular component of momentum.
Centripetal force acts always towards the center.
We need to calculate the magnitude of the rate of change of the child's momentum
Using formula of momentum


Put the value into the formula


Hence, The magnitude of the rate of change of the child's momentum is 794.11 N.