Answer:
Vertical distance= 3.3803ft
Explanation:
First with the speed of the ball and the distance traveled horizontally we can determine the flight time to reach the plate:
Velocity= (90 mi/h) × (1 mile/5280ft) = 475200ft/h
Distance= Velocity × time⇒ time= 60.5ft / (475200ft/h) = 0.00012731h
time= 0.00012731h × (3600s/h)= 0.458316s
With this time we can determine the distance traveled vertically taking into account that its initial vertical velocity is zero and its acceleration is that of gravity, 9.81m/s²:
Vertical distance= (1/2) × 9.81 (m/s²) × (0.458316s)²=1.0303m
Vertical distance= 1.0303m × (1ft/0.3048m) = 3.3803ft
This is the vertical distance traveled by the ball from the time it is thrown by the pitcher until it reaches the plate, regardless of air resistance.
Answer:
1. -8.20 m/s²
2. 73.4 m
3. 19.4 m
Explanation:
1. Apply Newton's second law to the car in the y direction.
∑F = ma
N − mg = 0
N = mg
Apply Newton's second law to the car in the x direction.
∑F = ma
-F = ma
-Nμ = ma
-mgμ = ma
a = -gμ
Given μ = 0.837:
a = -(9.8 m/s²) (0.837)
a = -8.20 m/s²
2. Given:
v₀ = 34.7 m/s
v = 0 m/s
a = -8.20 m/s²
Find: Δx
v² = v₀² + 2aΔx
(0 m/s)² = (34.7 m/s)² + 2 (-8.20 m/s²) Δx
Δx = 73.4 m
3. Since your braking distance is the same as the car in front of you, the minimum safe following distance is the distance you travel during your reaction time.
d = v₀t
d = (34.7 m/s) (0.56 s)
d = 19.4 m
Answer:
2.5m
Explanation:
Torque is defined as the rotational effect of a force on a body.
The torque T for the maximum shear stress is given as 0.1 Nm
Frictional torque is the torque caused by a frictional force
The frictional torque F is given as 0.04 Nm/m
The maximum length of the shaft is thus given as
L = T / F
= 0.1/0.04
L= 2.5 m
