Explanation:
It is given that,
Initial speed of sprinter, u = 0
Final speed of sprinter, v = 10 m/s
Time taken, t = 1.28 s
a. We need to find the acceleration of sprinter. It can be calculated using first equation of motion as :
b. Final speed of the sprinter, v = 36 km/h
Time, t = 0.000355 h
Acceleration, 
Hence, this is the required solution.
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
Answer:
a) v = 6.43 m/s
b) v = 15.8 m/s
Explanation:
Speed of car = 56 km/h
56 km/h = 14.4 m/s
Angle rain makes on the glass to the vertical = 66°
Thus knowing that the opposite side of the angle is the distance moved by the car, and the adjacent side is the distance traveled by the rain in the same time
both of which are directly proportional to their velocities
Then
tan(66°) = 14.44m/s ÷ x
or x = 14.44/tan(66°)
Which is the vertical raindrop velocity of the relative to earth
v = 6.43 m/s vertically towards earth
For v relative to the car is we have vector sum of both velocities
v = √(14.44^2 + 6.43^2) = 15.8 m/s which is the velocity relative to car
= 15.8 m/s
Answer:
2.23 × 10^6 g of F- must be added to the cylindrical reservoir in order to obtain a drinking water with a concentration of 0.8ppm of F-
Explanation:
Here are the steps of how to arrive at the answer:
The volume of a cylinder = ((pi)D²/4) × H
Where D = diameter of the cylindrical reservoir = 2.02 × 10^2m
H = Height of the reservoir = 87.32m
Therefore volume of cylindrical reservoir = (3.142×202²/4)m² × 87.32m = 2798740.647m³
1ppm = 1g/m³
0.8ppm = 0.8 × 1g/m³
= 0.8g/m³
Therefore to obtain drinking water of concentration 0.8g/m³ in a reservoir of volume 2798740.647m³, F- of mass = 0.8g/m³ × 2798740.647m³ = 2.23 × 10^6 g must be added to the tank.
Thank you for reading.
