A brick is resting on a rough incline as shown in the figure. The friction force acting on the brick, along the incline, is
2 answers:
Answer:
D) less than the gravitational force of the brick.
Explanation:
When a brick is at rest on the inclined plane
then we will have all forces balanced on the brick
so we will have force balance along the inclined plane given as
on the other side we can find force balance perpendicular to the inclined plane
so from above equation we can say that friction force is the component of the weight of the object along the inclined plane
so correct answer is
D) less than the gravitational force of the brick.
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Remember your kinematic equations for constant acceleration. One of the equations is 
, where 
= final position, 
= initial position, 
= initial velocity, t = time, and a = acceleration.
Your initial position is where you initially were before you braked. That means = 100m. You final position is where you ended up after t seconds passed, so = 350m. The time it took you to go from 100m to 350m was t = 8.3s. You initial velocity at the initial position before you braked was = 60.0 m/s. Knowing these values, plug them into the equation and solve for a, your acceleration:
Your acceleration is approximately
.
Your initial position is where you initially were before you braked. That means
Your acceleration is approximately
Centripetal Force (Fcp) = ?
His arm length = Radius (R) = 0.75 m
Discus velocity = Linear Velocity (V) = 5 m/s
Discus mass (m) = 2 kg
Centripetal Acceleration (Acp) = V^2/R or W^2 x R
In this case i will use the V^2/R formula, because it uses the discus velocity (V).
Answer: Last option, 66 N.
His arm length = Radius (R) = 0.75 m
Discus velocity = Linear Velocity (V) = 5 m/s
Discus mass (m) = 2 kg
Centripetal Acceleration (Acp) = V^2/R or W^2 x R
In this case i will use the V^2/R formula, because it uses the discus velocity (V).
Answer: Last option, 66 N.