Answer:
4°C
Explanation:
Water is densest at 4°C. Since dense water sinks, the bottom of the lake will be 4°C.
Answer:

t'=1.1897 μs
Explanation:
First we will calculate the velocity of micrometeorite relative to spaceship.
Formula:

where:
v is the velocity of spaceship relative to certain frame of reference = -0.82c (Negative sign is due to antiparallel track).
u is the velocity of micrometeorite relative to same frame of reference as spaceship = .82c (Negative sign is due to antiparallel track)
u' is the relative velocity of micrometeorite with respect to spaceship.
In order to find u' , we can rewrite the above expression as:


u'=0.9806c
Time for micrometeorite to pass spaceship can be calculated as:

(c = 3*10^8 m/s)


t'=1.1897 μs
The answer is to increase energy. Hope this helps!
Answer:
it shows the products of a chemical reaction to the right of the reaction arrow
Answer:
The answer cannot be determined.
Explanation:
The energy of the diver when he hits the pool will be equal to its potential energy
, and for the temperature of the pool to rise up, this energy has to be converted into the heat energy of the pool.
The change in temperature
then will be

Where m is the mass of water in the pool, c is the specific heat capacity of water, and
is the added heat which in this case is the energy of the diver.
Since we do not know the mass of the water in the pool, we cannot make this calculation.