Answer:
2a) x = 32 [mil/h]; 2b) t = 0.5[h]; 3a) t = 2.5 [h]; 3b) x = 185[mil]
Explanation:
2a)
We can solve this problem by using the kinematics equation, which relates speed to time and displacement.
![v=\frac{x}{t} \\v=velocity [\frac{mil}{h} ] = 32 [\frac{mil}{h}] \\t=time = 1 [h]\\x=v*t\\x=32[\frac{mil}{h} ]*1[h]\\x=32[mil}](https://tex.z-dn.net/?f=v%3D%5Cfrac%7Bx%7D%7Bt%7D%20%5C%5Cv%3Dvelocity%20%5B%5Cfrac%7Bmil%7D%7Bh%7D%20%5D%20%3D%2032%20%5B%5Cfrac%7Bmil%7D%7Bh%7D%5D%20%5C%5Ct%3Dtime%20%3D%201%20%5Bh%5D%5C%5Cx%3Dv%2At%5C%5Cx%3D32%5B%5Cfrac%7Bmil%7D%7Bh%7D%20%5D%2A1%5Bh%5D%5C%5Cx%3D32%5Bmil%7D)
2b)
We can solve this problem by using the kinematics equation, which relates speed to time and displacement.
![v=\frac{x}{t} \\t=\frac{x}{v} \\t=\frac{420}{840}\\ t=0.5[h]](https://tex.z-dn.net/?f=v%3D%5Cfrac%7Bx%7D%7Bt%7D%20%5C%5Ct%3D%5Cfrac%7Bx%7D%7Bv%7D%20%5C%5Ct%3D%5Cfrac%7B420%7D%7B840%7D%5C%5C%20t%3D0.5%5Bh%5D)
3a)
We can solve this problem by using the kinematics equation, which relates speed to time and displacement.
![v=\frac{x}{t} \\t=\frac{x}{v} \\t=\frac{35}{14}\\ t=2.5[h]](https://tex.z-dn.net/?f=v%3D%5Cfrac%7Bx%7D%7Bt%7D%20%5C%5Ct%3D%5Cfrac%7Bx%7D%7Bv%7D%20%5C%5Ct%3D%5Cfrac%7B35%7D%7B14%7D%5C%5C%20t%3D2.5%5Bh%5D)
3b)
We can solve this problem by using the kinematics equation, which relates speed to time and displacement.
![v=\frac{x}{t} \\v=velocity [\frac{mil}{h} ] = 74 [\frac{mil}{h}] \\t=time = 2.5 [h]\\x=v*t\\x=74[\frac{mil}{h} ]*2.5[h]\\x=185[mil}](https://tex.z-dn.net/?f=v%3D%5Cfrac%7Bx%7D%7Bt%7D%20%5C%5Cv%3Dvelocity%20%5B%5Cfrac%7Bmil%7D%7Bh%7D%20%5D%20%3D%2074%20%5B%5Cfrac%7Bmil%7D%7Bh%7D%5D%20%5C%5Ct%3Dtime%20%3D%202.5%20%5Bh%5D%5C%5Cx%3Dv%2At%5C%5Cx%3D74%5B%5Cfrac%7Bmil%7D%7Bh%7D%20%5D%2A2.5%5Bh%5D%5C%5Cx%3D185%5Bmil%7D)
Answer:
a) 
b) This value of specific heat is close to the specific heat of ice at -40° C and the specific heat of peat (a variety of coal).
c) The material is peat, possibly.
d) The material cannot be ice because ice doesn't exists at a temperature of 100°C.
Explanation:
Given:
- mass of aluminium,

- mass of water,

- initial temperature of the system,

- mass of copper block,

- temperature of copper block,

- mass of the other block,

- temperature of the other block,

- final equilibrium temperature,

We have,
specific heat of aluminium, 
specific heat of copper, 
specific heat of water, 
Using the heat energy conservation equation.
The heat absorbed by the system of the calorie-meter to reach the final temperature.



The heat released by the blocks when dipped into water:

where
specific heat of the unknown material
For the conservation of energy : 
so,


b)
This value of specific heat is close to the specific heat of ice at -40° C and the specific heat of peat (a variety of coal).
c)
The material is peat, possibly.
d)
The material cannot be ice because ice doesn't exists at a temperature of 100°C.
Answer: 3
Explanation:
Given
One cloud is traveling at rate of 
combined velocity of the two is 
Suppose the masses of the clouds be 
Conserving momentum

Answer:
(a) 0.613 m
(b) 0.385 m
(c) vₓ = 1.10 m/s, vᵧ = 3.50 m/s
v = 3.68 m/s², θ = 72.6° below the horizontal
Explanation:
(a) Take down to be positive.
Given in the y direction:
v₀ = 0 m/s
a = 10 m/s²
t = 0.350 s
Find: Δy
Δy = v₀ t + ½ at²
Δy = (0 m/s) (0.350 s) + ½ (10 m/s²) (0.350 s)²
Δy = 0.613 m
(b) Given in the x direction:
v₀ = 1.10 m/s
a = 0 m/s²
t = 0.350 s
Find: Δx
Δx = v₀ t + ½ at²
Δx = (1.10 m/s) (0.350 s) + ½ (0 m/s²) (0.350 s)²
Δx = 0.385 m
(c) Find: vₓ and vᵧ
vₓ = aₓt + v₀ₓ
vₓ = (0 m/s²) (0.350 s) + 1.10 m/s
vₓ = 1.10 m/s
vᵧ = aᵧt + v₀ᵧ
vᵧ = (10 m/s²) (0.350 s) + 0 m/s
vᵧ = 3.50 m/s
The magnitude is:
v² = vₓ² + vᵧ²
v = 3.68 m/s²
The direction is:
θ = atan(vᵧ / vₓ)
θ = 72.6° below the horizontal
RDA stands for Recommended Daily Allowance. To determine the amount needed of a certain adult per day, we simply multiply the mass of the adult to the value of RDA. For this case, we do as follows:
Daily requirement = 12 mg/kg ( 78 kg ) = 936 mg of lysine = 936000 g