Explanation:
Assuming the wall is frictionless, there are four forces acting on the ladder.
Weight pulling down at the center of the ladder (mg).
Reaction force pushing to the left at the wall (Rw).
Reaction force pushing up at the foot of the ladder (Rf).
Friction force pushing to the right at the foot of the ladder (Ff).
(a) Calculate the reaction force at the wall.
Take the sum of the moments about the foot of the ladder.
∑τ = Iα
Rw (3.0 sin 60°) − mg (1.5 cos 60°) = 0
Rw (3.0 sin 60°) = mg (1.5 cos 60°)
Rw = mg / (2 tan 60°)
Rw = (10 kg) (9.8 m/s²) / (2√3)
Rw = 28 N
(b) State the friction at the foot of the ladder.
Take the sum of the forces in the x direction.
∑F = ma
Ff − Rw = 0
Ff = Rw
Ff = 28 N
(c) State the reaction at the foot of the ladder.
Take the sum of the forces in the y direction.
∑F = ma
Rf − mg = 0
Rf = mg
Rf = 98 N
What you know:
Vi=0m/s
Vf=143.8m/s
A=-9.8m/s
d=???
Use the equation Vf^2=Vi^2+2A(d)
Rearrange to isolate d: d=Vf^2/2A
d=(143.8)^2/2(-9.8)
d=20678.4/-19.6
d=-1055m
The tank was released from a height of 1055m
Answer:
C)
Explanation:
The buoyancy and weight of the wood have to be equal for the system to be in equilibrium. The total mass (then, weight) of the wood is the same, so the total buoyancy has to be the same. Since buoyancy is the weight of the liquid displaced, the volume of liquid displaced will be the same in either case, which means that the water level will remain unchanged.
The components of the net force on the cart is determined as 67.66 N.
<h3>
Component of net force on the cart</h3>
The component of net force on the cart is determined by resolving the forces into x and y -components.
T1 = 30 N
T2 = 40 N
T1x = -30cos(0) = 30 N
T1y = 30sin(0) = 0
T2x = 40 x cos(30) = 34.64 N
T2y = 40 x sin(3) = 20 N
∑X = 30 N + 34.64 N = 64.64 N
∑Y = 0 + 20 N = 20 N
<h3>Resultant force</h3>
R = √(64.64² + 20²)
R = 67.66 N
Learn more about net force here: brainly.com/question/25239010
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When a ball is whirled using a string, it is restricted to move only in circular motion because the net force acting on the ball is towards the center of the circle. Hence, the acceleration of the ball is towards the center. But the velocity of ball is tangential to this circular path all the time. When the whirling is stropped, the string becomes slack and tension in the string becomes zero. The ball no more performs circular motion and the ball moves tangentially to the circle in straight line. Therefore, before letting go, velocity was variable. After letting go, velocity becomes constant.
