<span>6.7 x 10-11....Have a good day with a big fat A+
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To solve this problem it is necessary to use the calorimetry principle. From the statement it asks about the remaining ice, that is, to the point where the final temperature is 0 ° C.
We will calculate the melted ice and in the end we will subtract the total initial mass to find out how much mass was left.
The amount of heat transferred is defined by
Where,
m = mass
c = Specific heat
Change in temperature
There are two states, the first is that of heat absorbed by that mass 'm' of melted ice and the second is that of heat absorbed by heat from -35 ° C until 0 ° C is reached.
Performing energy balance then we will have to
Where,
= Heat absorbed by whole ice
= Heat absorbed by mass
= Heat energy by latent heat fusion/melting
Replacing with our values we have that
Rearrange and find m,
Therefore the Ice left would be
Therefore there is 0.079kg ice in the containter when it reaches equilibrium
Answer:
All we need to know for this is SUVAT equations. There are 5 SUVAT equations you need to learn. S = distance, U = initial velocity, V = final velocity, A = acceleration, T = time V= U + AT V^2 = U^2 + 2AS S = UT + 1/2AT^2 S = VT -
Explanation:
don't now if this hellps or not
Answer:
1. The precession of the equinoxes.
2. Changes in the tilt angle of Earth’s rotational axis relative to the plane of Earth’s orbit around the Sun.
3. Variations in the eccentricity
Explanation:
These variations listed above; the precession of the equinoxes (refers, changes in the timing of the seasons of summer and winter), this occurs on a roughly about 26,000-year interval; changes in the tilt angle of Earth’s rotational axis relative to the plane of Earth’s orbit around the Sun, this occurs roughly in a 41,000-year interval; and changes in the eccentricity (that is a departure from a perfect circle) of Earth’s orbit around the Sun, occurring on a roughly 100,000-year timescale. which influences the mean annual solar radiation at the top of Earth’s atmosphere.
Supposing the runner is condensed to a point and moves upward at 2.2 m/s.
It takes a time = 2.2/g = 2.2/9.8 = 0.22 seconds to increase to max height.
Now looking at this condition in opposite - that is the runner is at max height and drops back to earth in 0.22 s (symmetry of this kind of motion).
From what height does any object take 0.22 s to fall to earth (supposing there is no air friction)?
d = 1/2gt²= (0.5)(9.8)(0.22)²= 0.24 m
