A 25N force is applied to a bar that can pivot around its end. The force is r=0.75 m away from the end of an angle at 0= 30. wha

Inertia = inactivity
The element that influences an increase in inertia is "mass." The greater the mass of an object, the more inertia it possesses. 

Answer: 339.148N

Explanation:

Given data:

Time (t) = 47s

Initial speed (U) = 0m/s

Final speed (V) = 9.5m/s

Mass of B = 540kg

Frictional force on B = 230N

Since both boats are linked, movement of A causes B to move as well.

What is the acceleration of boat A?

Applying the motion formula:

V = u + at

9.5 = 0 + a * 47

a = 9.5 / 47

a = 0.2021 m/s²

To determine the force necessary to accelerate boat B, as both boats experience the same force:

F = Mass * acceleration

F = 540 * 0.2021 = 109.14N

Given that there is a frictional force of 230N acting on boat B, the overall force (Tension) becomes:

Tension = frictional force + applied force = (109.14 + 230)N = 339.148N

Answer: a) t = 1.8 x 10^2 seconds; b) t = 54 seconds; c) t = 49 seconds. Explanation: a) To determine the time of a stationary passenger on the sidewalk, we use the position formula. Given the constant speed of the walkway, we can calculate the time taken for set distances accordingly. This calculation extends into cases where combined velocities for walking are involved in subsequent queries.

Answer:

5.7 m

Explanation:

AD = length of the ladder = L = 8 m

AB = the position of the ladder's center of mass = (0.5) L = (0.5) 8 = 4 m

AC = distance of the climber from the bottom of the ladder = x

W = weight of the ladder = 240 N

= weight of the climber = 710 N

F = force exerted by the wall on the ladder

N = normal force acting on the ladder from the ground =?

By applying force equilibrium in the vertical direction

N = + W

N = 710 + 240

N = 950 N

μ = Coefficient of static friction = 0.55

f = static friction force on the ladder

Static friction force can be expressed as

f = μ N

f = (0.55) (950)

f = 522.5 N

The equation for force along the horizontal axis reads

F = f

F = 522.5 N

using torque equilibrium around point A

F Sin50 (AD) = W Cos50 (AB) + ( Cos50 (AC))

(522.5) Sin50 (8) = (240) Cos50 (4) + (710) Cos50 (x)

x = 5.7 m