A 0.5 kg object is whirled on the end of a string that is 1.2 m long at a speed of 7.5 m/s. Calculate the angular momentum of th
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The equilibrant force of the two given forces is 14.14 N.
<h3 /><h3 /><h3>What is equilibrant force?</h3>- This is a single force that balances other given forces.
The given parameters:
- First force, F₁ = 10 N
- Second force, F₂ = 10 N
- Angle between the forces, θ = 90⁰
The equilibrant force of the two given forces is calculated as follows;
Thus, the equilibrant force of the two given forces is 14.14 N.
Learn more about equilibrant force here: brainly.com/question/8045102
Refer to the diagram shown below.
m = the mass of the object
x = the distance of the object from the equilibrium position at time t.
v = the velocity of the object at time t
a = the acceleration of the object at time t
A = the amplitude ( the maximum distance) of the mass from the equilibrium
position
The oscillatory motion of the object (without damping) is given by
x(t) = A sin(ωt)
where
ω = the circular frequency of the motion
T = the period of the motion so that ω = (2π)/T
The velocity and acceleration are respectively
v(t) = ωA cos(ωt)
a(t) = -ω²A sin(ωt)
In the equilibrium position,
x is zero;
v is maximum;
a is zero.
At the farthest distance (A) from the equilibrium position,
x is maximum;
v is zero;
a is zero.
In the graphs shown, it is assumed (for illustrative purposes) that
A = 1 and T = 1.
m = the mass of the object
x = the distance of the object from the equilibrium position at time t.
v = the velocity of the object at time t
a = the acceleration of the object at time t
A = the amplitude ( the maximum distance) of the mass from the equilibrium
position
The oscillatory motion of the object (without damping) is given by
x(t) = A sin(ωt)
where
ω = the circular frequency of the motion
T = the period of the motion so that ω = (2π)/T
The velocity and acceleration are respectively
v(t) = ωA cos(ωt)
a(t) = -ω²A sin(ωt)
In the equilibrium position,
x is zero;
v is maximum;
a is zero.
At the farthest distance (A) from the equilibrium position,
x is maximum;
v is zero;
a is zero.
In the graphs shown, it is assumed (for illustrative purposes) that
A = 1 and T = 1.
Answer:
The air resistance on the skydiver is 68.6 N
Explanation:
When the skydiver is falling down, there are two forces acting on him:
- The force of gravity, of magnitude , in the downward direction (where m is the mass of the skydiver and g is the acceleration due to gravity)
- The air resistance, , in the upward direction
So the net force on the skydiver is:
where
m = 7.0 kg is the mass
According to Newton's second law of motion, the net force on a body is equal to the product between its mass and its acceleration (a):
In this problem, however, the skydiver is moving with constant velocity, so his acceleration is zero:
Therefore the net force is zero:
And so, we have:
And so we can find the magnitude of the air resistance, which is equal to the force of gravity: