Answer:
Yes
Explanation:
The given parameters are;
The speed with which the fastball is hit, u = 49.1 m/s (109.9 mph)
The angle in which the fastball is hit, θ = 22°
The distance of the field = 96 m (315 ft)
The range of the projectile motion of the fastball is given by the following formula
Where;
g = The acceleration due to gravity = 9.81 m/s², we have;
Yes, given that the ball's range is larger than the extent of the field, the batter is able to safely reach home.
Answer:
r = 2161.9 m
Explanation:
Aerodynamic lift(L) is perpendicular to the wing, which is tilted 40 degrees to the horizontal.
Since the plane is moving in a horizontal circle, the vertical component of the lift must cancel the weight W of the airplane, but the horizontal component is the centripetal force that keeps it in a circle.
L is perpendicular to wing at angle θ with respect to horizontal
Thus,
Vertical component of lift is:
L cosθ = W = mg
Thus, m = L cosθ / g - - - - (eq1)
Horizontal component of lift is:
L sinθ = centripetal force = mv² / r - - - - (eq2)
Combining equations 1 and 2,we have;
L sinθ = (L cosθ / g)(v² / r)
L cancels out on both sides to give;
tanθ = v²/ rg
r = v² / (g tanθ)
We are given;
velocity; v = 480 km/hr = 480 x 10/36 = 133.33 m/s
r = 133.33²/[(9.8) tan(40)] = 2161.9 m
Answer:
13,750 N
Yes
Explanation:
Given:
v₀ = 90 km/h = 25 m/s
v = 0 m/s
t = 4 s
Find: a and Δx
a = Δv / Δt
a = (0 m/s − 25 m/s) / (4 s)
a = -6.25 m/s²
F = ma
F = (2200 kg) (-6.25 m/s²)
F = -13,750 N
Δx = ½ (v + v₀) t
Δx = ½ (0 m/s + 25 m/s) (4 s)
Δx = 50 m
Answer:
Explanation:
In order to measure the coefficient of friction , we apply external force to move the body . When external force comes in motion , we adjust the external force so that it moves with zero acceleration or uniform velocity . In this case external force becomes equal to kinetic frictional force and then net force becomes zero because
net force = mass x acceleration = m x 0 = 0
Now frictional force = μ mg where μ is coefficient of kinetic friction
so F = μ mg where F is external force applied
μ = F / mg
Hence , to make external force equal to frictional force , it is necessary to make acceleration of body zero .