Answer:
Explanation:
Atmospheric pressure = 7 x 10⁴ Pa
force on a disk-shaped region 2.00 m in radius at the surface of the ocean due to atmosphere = pressure x area
= 7 x 10⁴ x 3.14 x 2 x 2
= 87.92 x 10⁴ N
b )
weight, on this exoplanet, of a 10.0 m deep cylindrical column of methane with radius 2.00 m
Pressure x area
height x density x acceleration of gravity x π r²
= 10 x 415 x 6.2 x 3.14 x 2 x 2
=323168.8 N
c ) Pressure at a depth of 10 m
atmospheric pressure + pressure due to liquid column
= 7 x 10⁴ + 10 x 415 x 6.2 ( hρg)
= 7 x 10⁴ + 10 x 415 x 6.2
(7 + 2.57 )x 10⁴ Pa
9.57 x 10⁴ Pa
Answer:
To find the circumference (orbit) of an object, you use Pi x Diameter.
As you have the circumference of B, you divide it by Pi to get the Diameter.
So 120 divided by 3.141592654 = 38.2 minutes for the Diameter.
As' radius and Diameter will be 3x greater than B.
38.2 x 3 = 114.6
To get back to the orbital period, times 114.6 by Pi, and you will get 360 minutes
HOPE THIS HELPS AND PLS MARK AS BRAINLIEST
THNXX :)
Answer:
B. 17.15 watts
Explanation:
Given that
Time = 10 seconds
height = distance = 0.7 meters
weight of sack = mg = F = 245 newtons
Power = work done/ time taken
Where work done = force × distance
Substituting the given parameters into the formula
Work done = 245 newton × 0.7 meters
Work done = 171.5 J
Recall,
Power = work done/time
Power = 171.5 J ÷ 10
Power = 17.15 watts
Hence the power expended is B. 17.15 watts
Mass of gold m₁ = 47 g
Initial temperature of gold T₁ = 99 C
Specific heat of gold C₁ = 0.129 J/gC
final temperature T₂ = 38 C
Heat needed by the gold to cool down
Q =m₁ * C₁* ( T₁ - T₂)
Q = (47)(0.129)(99-38)
Q = 369.843 J
This heat will be given by the water
we need to find out mass of water m₂
and initial temperature of water is T₃ = 25 C
Specific heat of water C₂ = 4.184 J/gC
Q = m₂*C₂*(T₂ - T₃)
369.843 = m₂(4.184)(38-25)
m₂ = 6.8 g
