The angular speed is decreasing and direction of rotation clockwise of the rod immediately after time t.
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</h3><h3>What is angular speed ?</h3>
The rate of change of angular displacement is defined as angular speed. It is stated as follows:
ω = θ t
Where,
θ is the angle of rotation,
t is the time
ω is the angular velocity
The torque is found as;l
If the force is acting on the rod from the three point is the same, the value of the torque is depends upon the radius or the perpendicular distance.
The perpendicular distance of the right force is grater. Hence, the force acting on the right side is more, and the rod will rotate clockwise.
Both the forces are acting downwards. Thus, the resultant force is the less due to which the speed is increasing.
Hence, the angular speed is decreasing and direction of rotation clockwise of the rod immediately after time t.
To learn more about the angular speed, refer to the link;
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Answer:
Steven has to row at a speed to reach the same horizontal spot at the other side of the river is, V = 6 m/s
Explanation:
Given data,
The river flowing south at the rate, v = 3 m/s
To reach the other side directly across the river, he aims the raft, Ф = 30°
The speed of his raft across the river is given by the formula,
V = v / Sin Ф
= 3 / Sin 30°
= 6 m/s
Steven has to row at a speed to reach the same horizontal spot at the other side of the river is, V = 6 m/s
Answer:
Explanation:
Given that,
Initial angular velocity is 0
ωo=0rad/s
It has angular velocity of 11rev/sec
ωi=11rev/sec
1rev=2πrad
Then, wi=11rev/sec ×2πrad
wi=22πrad/sec
And after 30 revolution
θ=30revolution
θ=30×2πrad
θ=60πrad
Final angular velocity is
ωf=18rev/sec
ωf=18×2πrad/sec
ωf=36πrad/sec
a. Angular acceleration(α)
Then, angular acceleration is given as
wf²=wi²+2αθ
(36π)²=(22π)²+2α×60π
(36π)²-(22π)²=120πα
Then, 120πα = 8014.119
α=8014.119/120π
α=21.26 rad/s²
Let. convert to revolution /sec²
α=21.26/2π
α=3.38rev/sec
b. Time Taken to complete 30revolution
θ=60πrad
∆θ= ½(wf+wi)•t
60π=½(36π+22π)t
60π×2=58πt
Then, t=120π/58π
t=2.07seconds
c. Time to reach 11rev/sec
wf=wo+αt
22π=0+21.26t
22π=21.26t
Then, t=22π/21.26
t=3.251seconds
d. Number of revolution to get to 11rev/s
∆θ= ½(wf+wo)•t
∆θ= ½(0+11)•3.251
∆θ= ½(11)•3.251
∆θ= 17.88rev.
The object's speed will not change.
In fact, after the astronaut throws the object, no additional forces will act on it (since the object is in free space). According to Newton's second law:
where the first term is the resultant of the forces acting on the body, m is the mass of the object and a its acceleration, we see that if no forces act on the object, then the acceleration is zero. Therefore, the acceleration of the object is zero, and its velocity remains constant.
There’s no picture so how r we supposed to answer it
