Answer:
a) y= 3.5 10³ m, b) t = 64 s
Explanation:
a) For this exercise we use the vertical launch kinematics equation
Stage 1
y₁ = y₀ + v₀ t + ½ a t²
y₁ = 0 + 0 + ½ a₁ t²
Let's calculate
y₁ = ½ 16 10²
y₁ = 800 m
At the end of this stage it has a speed
v₁ = vo + a₁ t₁
v₁ = 0 + 16 10
v₁ = 160 m / s
Stage 2
y₂ = y₁ + v₁ (t-t₀) + ½ a₂ (t-t₀)²
y₂ = 800 + 150 5 + ½ 11 5²
y₂ = 1092.5 m
Speed is
v₂ = v₁ + a₂ t
v₂ = 160 + 11 5
v₂ = 215 m / s
The rocket continues to follow until the speed reaches zero (v₃ = 0)
v₃² = v₂² - 2 g y₃
0 = v₂² - 2g y₃
y₃ = v₂² / 2g
y₃ = 215²/2 9.8
y₃ = 2358.4 m
The total height is
y = y₃ + y₂
y = 2358.4 + 1092.5
y = 3450.9 m
y= 3.5 10³ m
b) Flight time is the time to go up plus the time to go down
Let's look for the time of stage 3
v₃ = v₂ - g t₃
v₃ = 0
t₃ = v₂ / g
t₃ = 215 / 9.8
t₃ = 21.94 s
The time to climb is
= t₁ + t₂ + t₃
t_{s} = 10+ 5+ 21.94
t_{s} = 36.94 s
The time to descend from the maximum height is
y = v₀ t - ½ g t²
When it starts to slow down it's zero
y = - ½ g t_{b}²
t_{b} = √-2y / g
t_{b} = √(- 2 (-3450.9) /9.8)
t_{b} = 26.54 s
Flight time is the rise time plus the descent date
t = t_{s} + t_{b}
t = 36.94 + 26.54
t =63.84 s
t = 64 s
Answer:
Time, ![t=1.07\times 10^{-23}\ s](https://tex.z-dn.net/?f=t%3D1.07%5Ctimes%2010%5E%7B-23%7D%5C%20s)
Explanation:
Given that,
When a high-energy proton or pion traveling near the speed of light collides with a nucleus, it may travel
before interacting.
Let t is the time interval required for the strong interaction to occur. It will move with the speed of light. So,
![t=\dfrac{d}{c}\\\\t=\dfrac{3.2\times 10^{-15}}{3\times 10^8}\\\\t=1.07\times 10^{-23}\ s](https://tex.z-dn.net/?f=t%3D%5Cdfrac%7Bd%7D%7Bc%7D%5C%5C%5C%5Ct%3D%5Cdfrac%7B3.2%5Ctimes%2010%5E%7B-15%7D%7D%7B3%5Ctimes%2010%5E8%7D%5C%5C%5C%5Ct%3D1.07%5Ctimes%2010%5E%7B-23%7D%5C%20s)
So, the time interval is ![1.07\times 10^{-23}\ s](https://tex.z-dn.net/?f=1.07%5Ctimes%2010%5E%7B-23%7D%5C%20s)
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.
![\begin{aligned}m&= 100\; \rm g \times \frac{1\; \rm kg}{1000\; \rm g} \\ &= 0.100\; \rm kg\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7Dm%26%3D%20100%5C%3B%20%5Crm%20g%20%5Ctimes%20%5Cfrac%7B1%5C%3B%20%5Crm%20kg%7D%7B1000%5C%3B%20%5Crm%20g%7D%20%5C%5C%20%26%3D%200.100%5C%3B%20%5Crm%20kg%5Cend%7Baligned%7D)
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:
D. The tea loses heat to the spoon causing the spoon to become warmer
Explanation:
When the silver spoon at a lower temperature than the tea, is added to the tea, it makes thermal contact. Hence, the heat transfer starts between the two until the equilibrium is reached. We know that the heat transfer takes place from the body with a higher temperature to a body with a lower temperature. As a result, the body with higher temperature loses heat and its temperature lowers down. While the body with a lower temperature gains heat and its temperature rises.
Therefore, the correct option is:
<u>D. The tea loses heat to the spoon causing the spoon to become warmer</u>
epicycles were orbits within orbits used to explain discrepancies between expected and observed planetary movement, including the appearance of planets slowing down, speeding up, and moving backward.