Rocks, earth aging, fossils
Answer:
229,098.96 J
Explanation:
mass of water (m) = 456 g = 0.456 kg
initial temperature (T) = 25 degrees
final temperature (t) = - 10 degrees
specific heat of ice = 2090 J/kg
latent heat of fusion =33.5 x 10^(4) J/kg
specific heat of water = 4186 J/kg
for the water to be converted to ice it must undergo three stages:
- the water must cool from 25 degrees to 0 degrees, and the heat removed would be Q = m x specific heat of water x change in temp
Q = 0.456 x 4186 x (25 - (-10)) = 66808.56 J
- the water must freeze at 0 degrees, and the heat removed would be Q = m x specific heat of fusion x change in temp
Q = 0.456 x 33.5 x 10^(4) = 152760 J
- the water must cool further to -10 degrees from 0 degrees, and the heat removed would be Q = m x specific heat of ice x change in temp
Q = 0.456 x 2090 x (0 - (-10)) = 9530.4 J
The quantity of heat removed from all three stages would be added to get the total heat removed.
Q total = 66,808.56 + 152,760 + 9,530.4 = 229,098.96 J
Answer:
1. The length is 8.35m
2. The period on the moon is 14.05 secs
Explanation:
1. Data obtained from the question. This includes the following:
Period (T) = 5.8 secs
Acceleration due to gravity (g) = 9.8 m/s2
Length (L) =...?
The length can be obtained by using the formula given below:
T = 2π√(L/g)
5.8 = 2π√(L/9.8)
Take the square of both side
(5.8)^2 = 4π^2 x L/ 9.8
Cross multiply
4π^2 x L = (5.8)^2 x 9.8
Divide both side by 4π^2
L = (5.8)^2 x 9.8 / 4π^2
L= 8.35 m
2. Data obtained from the question. This includes the following:
Acceleration due to gravity (g) = 1.67 m/s2
Length (L) = 8.35m (the length remains the same)
Period (T) =?
The period can be obtained as follow:
T = 2π√(L/g)
T = 2π√(8.35/1.67)
T = 14.05 secs
Therefore, the period on the moon is 14.05 secs
Answer: 1.6Hz
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Answer:
Explanation:
We know that when we don't have air friction on a free fall the mechanical energy (I will symbololize it with ME) is equal everywhere. So we have:
where me(1) is mechanical energy while on h=10m
and me(2) is mechanical energy while on the ground
Ek(1) + DynamicE(1) = Ek(2) + DynamicE(2)
Ek(1) is equal to zero since an object that has reached its max height has a speed equal to zero.
DynamicE(2) is equal to zero since it's touching the ground
Using that info we have
we divide both sides of the equation with mass to make the math easier.
