Answer:
x = 11.23 m
Explanation:
For this interesting exercise, we must use angular kinematics, linear kinematics and the relationship between angular and linear quantities.
Let's reduce to SI system units
θ = 155 rev (2pi rad / rev) = 310π rad
α = 2.00rev / s2 (2pi rad / 1 rev) = 4π rad / s²
Let's look for the angular velocity at the time the piece is released, with starting from rest the initial angular velocity is zero (wo = 0)
w² = w₀² + 2 α θ
w =√ 2 α θ
w = √(2 4pi 310pi)
w = 156.45 rad / s
The relationship between angular and linear velocity
v = w r
v = 156.45 0.175
v = 27.38 m / s
In this part we have the linear speed and the height that it travels to reach the floor, so with the projectile launch equations we can find the time it takes to arrive
y = 
t - ½ g t²
As it leaves the highest point its speed is horizontal
y = 0 - ½ g t²
t = √ (-2y / g)
t = √ (-2 (-0.820) /9.8)
t = 0.41 s
With this time we calculate the horizontal distance, because the constant horizontal speed
x = vox t
x = 27.38 0.41
x = 11.23 m
3.5 seconds is 3500 milliseconds.
The tension in the string B) It quadruples.
Explanation:
The ball is in uniform circular motion in a horizontal circle, so the tension in the string is providing the centripetal force that keeps the ball in circular motion. So we can write:
where:
T is the tension in the string
m is the mass of the ball
v is the speed of the ball
r is the radius of the circle (the lenght of the string)
In this problem, we are told that the speed of the ball is doubled, so
v' = 2v
Substituting into the previous equation, we find the new tension in the string:
Therefore, the tension in the string will quadruple.
Learn more about circular motion:
brainly.com/question/2562955
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The work done on the box by the applied force is zero.
The work done by the force of gravity is 75.95 J
The work done on the box by the normal force is 75.95 J.
<h3>The given parameters:</h3>
- Mass of the box, m = 3.1 kg
- Distance moved by the box, d = 2.5 m
- Coefficient of friction, = 0.35
- Inclination of the force, θ = 30⁰
<h3>What is work - done?</h3>
- Work is said to be done when the applied force moves an object to a certain distance
The work done on the box by the applied force is calculated as;
where;
a is the acceleration of the box
The acceleration of the box is zero since the box moved at a constant speed.
The work done by the force of gravity is calculated as follows;
The work done on the box by the normal force is calculated as follows;
Learn more about work done here: brainly.com/question/8119756
Answer:
7.7 km 26°
Explanation:
The total x component is:
x = 2.5 cos(35°) + 5.2 cos(22°) = 6.87
The total y component is:
y = 2.5 sin(35°) + 5.2 sin(22°) = 3.38
The magnitude is:
d = √(x² + y²)
d = 7.7 km
The direction is:
θ = atan(y/x)
θ = 26°
