Answer:
21.02986 °C
Explanation:
c = Specific heat of water at 15 °C = 4186 J/Kg°C
= Change in temperature
= Initial temperature = 21 °C
= Final temperature
m = Mass of person =
t = Time taken = 30 minutes
Heat is given by
The water temperature after half an hour is 21.02986 °C
Answer: B
Action and reaction forces act on the same object and go to the same direction
Question:
A. Engineer A: "When the sample alpha decays, it will give off alpha particles. These will
trigger a photosensor (a device that senses forms of light) that will open the lock."
B. Engineer B: "When the sample alpha decays, it only releases large amounts of heat.
This will trigger a thermometer that will open the lock."
C. Engineer C: "When the sample alpha decays, it will lose mass, This will trigger a scale
that will open the lock."
D. Engineer D: "When the sample alpha decays, it will give off negatively charged alpha
particles. Because alpha particles are electrons, they will complete a circuit that will
open the lock."
Answer:
The correct option is;
A. Engineer A: "When the sample alpha decays, it will give off alpha particles. These will trigger a photosensor ( a device that senses forms of light) that will open the lock"
Explanation:
Here, we note that in alpha decay (α-decay) there is an emission of an alpha particle or helium nucleus with the transformation of the parent element nucleus into that of a different element, having reduced mass and atomic numbers
Alpha particles are a form of ionizing radiation with a mass of 6.64 × 10⁻²⁷ kg and therefore they can be sensed by a photo sensor but will be slowly sensed by a scale.
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