25.82 m/s
Explanation:
Given:
Force applied by the baseball player; F = 100 N
Distance the ball travels; d = 0.5 m
Mass of the ball; m = 0.15 kg
To find the velocity at which the ball is released, we will equate the work done with the kinetic energy involved.
It's important to recognize that work done reflects the energy the baseball player has used. Thus, the relationship can be represented as follows:
F × d = ½mv²
100 × 0.5 = ½ × 0.15 × v²
Solving gives:
v² = (2 × 100 × 0.5) / 0.15
v² = 666.67
v = √666.67
v = 25.82 m/s.
Answer:
Acceleration(a) = 0.75 m/s²
Explanation:
Given:
Force(F) = 3 N
Mass of object(m) = 4 kg
Find:
Acceleration(a)
Computation:
Force(F) = ma
3 = (4)(a)
Acceleration(a) = 3/4
Acceleration(a) = 0.75 m/s²
Conclusion:
The total net force acting on the objects is 16 N, directed towards the right.
Clarification:
It is stated that,
The force exerted by the dog, 
(to the right)
The force exerted by Simone, 
(backward)
Here, assume the backward direction is negative and the right direction is positive.
The net force will move in the direction where the larger force is present. The net force can be calculated as:
F = 16 N
Thus, the net force amounts to 16 N, acting towards the right.
Answer:
A rock weighing 50kg should be positioned at a distance of 0.5m from the pivot of the seesaw.
Explanation:
τchild=τrock
We will utilize the formula for torque:
(F)child(d)child)=(F)rock(d)rock)
The gravitational force acts equally on both objects.
(m)childg(d)child)=(m)rockg(d)rock)
We can eliminate gravity from both sides of the equation for simplification.
(m)child(d)child)=(m)rock(d)rock)
Now employing the given masses for the rock and child. The seesaw's total length is 2 meters, with the child sitting at one end, placing them 1 meter from the center of the seesaw.
(25kg)(1m)=(50kg)drock
Solve for the distance where the rock should be positioned in relation to the seesaw's center.
drock=25kg⋅m50kg
drock=0.5m
