Answer:
570 N
Explanation:
Draw a free body diagram on the rider. There are three forces: tension force 15° below the horizontal, drag force 30° above the horizontal, and weight downwards.
The rider is moving at constant speed, so acceleration is 0.
Sum of the forces in the x direction:
∑F = ma
F cos 30° - T cos 15° = 0
F = T cos 15° / cos 30°
Sum of the forces in the y direction:
∑F = ma
F sin 30° - W - T sin 15° = 0
W = F sin 30° - T sin 15°
Substituting:
W = (T cos 15° / cos 30°) sin 30° - T sin 15°
W = T cos 15° tan 30° - T sin 15°
W = T (cos 15° tan 30° - sin 15°)
Given T = 1900 N:
W = 1900 (cos 15° tan 30° - sin 15°)
W = 570 N
The rider weighs 570 N (which is about the same as 130 lb).
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.
Answer:
The speed of the 270g cart after the collision is 0.68m/s
Explanation:
Mass of air track cart (m1) = 320g
Initial velocity (u1) = 1.25m/s
Mass of stationary cart (m2) = 270g
Velocity after collision (V) = m1u1/(m1+m2) = 320×1.25/(320+270) = 400/590 = 0.68m/s
Answer:
76.73 ft/s
Explanation:
Let the final velocity is v.
initial velocity, u = 96 ft/s
g = 32 ft/s²
height, h = 52 feet
use third equation of motion
v² = u² - 2 gh
v² = 96 x 96 - 2 x 32 x 52
v = 76.73 ft/s
Thus, the speed of the ball as it reaches the ground is 76.73 ft/s.
