Answer:
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Answer:
Approximately
(assuming that the melting point of ice is
.)
Explanation:
Convert the unit of mass to kilograms, so as to match the unit of the specific heat capacity of ice and of water.

The energy required comes in three parts:
- Energy required to raise the temperature of that
of ice from
to
(the melting point of ice.) - Energy required to turn
of ice into water while temperature stayed constant. - Energy required to raise the temperature of that newly-formed
of water from
to
.
The following equation gives the amount of energy
required to raise the temperature of a sample of mass
and specific heat capacity
by
:
,
where
is the specific heat capacity of the material,
is the mass of the sample, and
is the change in the temperature of this sample.
For the first part of energy input,
whereas
. Calculate the change in the temperature:
.
Calculate the energy required to achieve that temperature change:
.
Similarly, for the third part of energy input,
whereas
. Calculate the change in the temperature:
.
Calculate the energy required to achieve that temperature change:
.
The second part of energy input requires a different equation. The energy
required to melt a sample of mass
and latent heat of fusion
is:
.
Apply this equation to find the size of the second part of energy input:
.
Find the sum of these three parts of energy:
.
Answer: The distance is 723.4km
Explanation:
The velocity of the transverse waves is 8.9km/s
The velocity of the longitudinal wave is 5.1 km/s
The transverse one reaches 68 seconds before the longitudinal.
if the distance is X, we know that:
X/(9.8km/s) = T1
X/(5.1km/s) = T2
T2 = T1 + 68s
Where T1 and T2 are the time that each wave needs to reach the sesmograph.
We replace the third equation into the second and get:
X/(9.8km/s) = T1
X/(5.1km/s) = T1 + 68s
Now, we can replace T1 from the first equation into the second one:
X/(5.1km/s) = X/(9.8km/s) + 68s
Now we can solve it for X and find the distance.
X/(5.1km/s) - X/(9.8km/s) = 68s
X(1/(5.1km/s) - 1/(9.8km/s)) = X*0.094s/km= 68s
X = 68s/0.094s/km = 723.4 km
<span>A gymnast with mass m1 = 43 kg is on a balance beam that sits on (but is not attached to) two supports. The beam has a mass m2 = 115 kg and length L = 5 m. Each support is 1/3 of the way from each end. Initially the gymnast stands at the left end of the beam.
1)What is the force the left support exerts on the beam?
2)What is the force the right support exerts on the beam?
3)How much extra mass could the gymnast hold before the beam begins to tip?
Now the gymnast (not holding any additional mass) walks directly above the right support.
4)What is the force the left support exerts on the beam?
5)What is the force the right support exerts on the beam?</span>