A ball is shot straight up from the surface of the earth with an initial speed of 19.6 m/s. Neglect any effects due to air resis
1 answer:
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13.0m/s
1.2m/s
Explanation:
Given parameters:
Initial speed of the body = 7.1m/s
time taken = 2.23s
Acceleration = 2.64m/s²
Unknown:
Final speed = ?
Solution:
Acceleration is the rate of change of velocity with time.
a =
a = acceleration
V = final speed
U = initial speed
T = time taken
Input the variables and solve for V;
2.64 =
V - 7.1 = 5.9 expression 1
V = 5.9 + 7.1 = 13.0m/s
B
Using the same parameters, the speed after a uniform deceleration of -2.64m/s², the negative sign implies deceleration;
from expression 1;
V - 7.1 = -5.9
V = -5.9 + 7.1 = 1.2m/s
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Answer: Add an incline or grade to the road track.
Explanation:
Refer to the figure shown below.
When a vehicle travels on a level road in a circular path of radius r, a centrifugal force, F, tends to make the vehicle skid away from the center of the circular path.
The magnitude of the force is
F = mv²/r
where
m = mass of the vehicle
v = linear (tangential) velocity to the circular path.
The force that resists the skidding of the vehicle is provided by tractional frictional force at the tires, of magnitude
μN = μW = μmg
where
μ = dynamic coefficient of friction.
At high speeds, the frictional force will not overcome the centrifugal force, and the vehicle will skid.
When an incline of θ degrees is added to the road track, the frictional force is augmented by the component of the weight of the vehicle along the incline.
Therefore the force that opposes the centrifugal force becomes
μN + Wsinθ = W(sinθ + μ cosθ).
Explanation:
Refer to the figure shown below.
When a vehicle travels on a level road in a circular path of radius r, a centrifugal force, F, tends to make the vehicle skid away from the center of the circular path.
The magnitude of the force is
F = mv²/r
where
m = mass of the vehicle
v = linear (tangential) velocity to the circular path.
The force that resists the skidding of the vehicle is provided by tractional frictional force at the tires, of magnitude
μN = μW = μmg
where
μ = dynamic coefficient of friction.
At high speeds, the frictional force will not overcome the centrifugal force, and the vehicle will skid.
When an incline of θ degrees is added to the road track, the frictional force is augmented by the component of the weight of the vehicle along the incline.
Therefore the force that opposes the centrifugal force becomes
μN + Wsinθ = W(sinθ + μ cosθ).