a) The angular acceleration of a disk is 2.0 rad/s².
b) The average angular velocity of the disk is 5 rad/s.
What is angular acceleration?The temporal rate at which angular velocity changes is known as angular acceleration. The standard unit of measurement is radians per second per second.
Rotational acceleration is another name for angular acceleration. It is a numerical representation of the variation in angular velocity over time.
A pseudoscalar, angular acceleration, exists. If the angular speed rises counterclockwise, the sign of angular acceleration is taken to be positive; if it grows clockwise, it is taken to be negative. Angular acceleration is used to study the motion of rotating objects like the wheel, fan, and earth.
Assume that the sense of rotation is positive.
This means that w, α, and other quantities will then be positive also.
a) we know,
25 rad = 1/2 α 5²
α = 2.0 rad/s²
∴ the angular acceleration of a disk is 2.0 rad/s².
b) w avg = Δθ/Δt
w avg = 25rad/5s
w avg = 5rad/s
∴ the average angular velocity of the disk is 5 rad/s.
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Gravity obeys an inverse square relation. This statement implies that the force due to gravity betweentwo massesa.will increase as the distance between the two masses increases.*b.will decrease as the square of the distance between the two masses increases.c.will cause the two masses to move away from each other.d.will cause the two masses to move in a straight line.e.will cause the two masses to orbit each other.
The inverse square relation implies that force resulting from gravity between two masses will decrease as the square of the distance between the two masses increases.
The force of gravity between any two objects is given by,
F = GM₁M₂/R²
Where,
M₁ and M₂ are the masses of the bodies.
R is the distance between them.
We know, if the masses are constant. then the force of gravity will be inversely proportional to the square of the distance between them.
This is what inverse square relation means here.
As the force of gravity and the square of the distance are inversely proportional to each other,
So, when the square of the distance between the masses increases, the force of gravity between the decreases.
Hence of all the options provided, b is the correc0,t which says, the force will decrease on increasing the inverse square of the distance.
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a car speeds up from 22 m/s to 26 m/s in 2 seconds. what is the average acceleration of the car
Answer:
2 m/s^2
Explanation:
Acceleration is change in velocity over change in time
(26-22) m/s / 2 s = 2 m/s^2
a gyroscope flywheel of radius 2.83 cm is accelerated from rest at 14.2 rad/s2 until its angular speed is 2760 rev/min. (a) what is the tangential acceleration of a point on the rim of the flywheel during this spin-up process? (b) what is the radial acceleration of this point when the flywheel is spinning at full speed? (c) through what distance does a point on the rim move during the spin-up?
The tangential acceleration of the gyroscope flywheel is 40.186 x [tex]10^{-2[/tex] m/[tex]s^2[/tex], Radial acceleration of gyroscope flywheel is 2364.08 [tex]m/s^2[/tex] and During the spin up process the distance moved by a point on rim is 83.24 m.
What is meant by tangential acceleration?The centripetal force causes an acceleration known as radial acceleration, which is measured in radians per square second. Radial acceleration is directed toward the center.Tangential acceleration occurs when a body or object moves at a speed that is not uniform.a gyroscope flywheel of radius, R = 2.83 cm =2.83 x [tex]10^{-2}[/tex] m
[tex]\omega_0 = 0[/tex]
[tex]\omega = 2760 X \frac{2\pi}{60} rad/sec[/tex]
[tex]\alpha = 14.2 rad/ sec^2[/tex]
The tangential acceleration :
[tex]a_t = R\alpha[/tex] =2.83 x [tex]10^{-2}[/tex] m x 14.2 [tex]rad/s^2[/tex] =40.186 x [tex]10^{-2[/tex] m/[tex]s^2[/tex]
The radial Acceleration :
[tex]a_r = R\omega^2\\=(2.83 X 10^-2) (2760 X \frac{\pi}{30} )^2[/tex] =2364.08 [tex]m/s^2[/tex]
During the spin up process the distance moved by a point on rim
d = [tex]R\theta[/tex]
where [tex]\theta =\frac{ \omega^2 - \omega_0^{2}}{2\alpha}[/tex]
= [tex]\frac{(2760 X \pi/30)^2}{2 X 14.2}[/tex] =2941.42 rad
d = 2.83 x [tex]10^{-2}[/tex] x 2941.42 =83.24.m
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What is the potential energy of a 25 kg bicycle resting at the top of a hill 3 m high? (Formula: PE = mgh)
The potential energy of a 25 kg bicycle resting at the top of a hill 3 m high will be 735 J
potential energy = m*g*h
where
m = mass
g = acceleration due to gravity
h = height
potential energy = 25 * 9.8 * 3 = 735 J
The potential energy of a 25 kg bicycle resting at the top of a hill 3 m high will be 735 J
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