Answer:
649kg/m^3
Explanation:
Let p be the density of this particular object.
Formula for density:

We can substitute the givenmass and volume to find density of the object.

Therefore the density of this object is 649kg/m^3.
Answer: 71.7 KJ
Explanation:
The rotational kinetic energy of a rotating body can be written as follows:
Krot = ½ I ω2
Now, any point on the rim of the flywheel, is acted by a centripetal force, according to Newton’s 2nd Law, as follows:
Fc = m. ac
It can be showed that the centripetal acceleration, is related with the angular velocity and the radius, as follows:
ac = ω2 r
We know that this acceleration has a limit value, so , we can take this limit to obtain a maximum value for the angular velocity also.
As the flywheel is a solid disk, the rotational inertia I is just ½ m r2.
Replacing in the expression for the Krot, we have:
Krot= ½ (1/2 mr2.ac/r) = ¼ mr ac = ¼ 67.0 Kg. 1.22 m . 3,510 m/s2 = 71. 7 KJ
Answer:
The vector magnitudes F and r are always postive, so the sign o W is determined entirely by the angle e between the force and the displacement.Submit Figure 1 off 1 part C
Answer:
vDP = 21.7454 m/s
θ = 200.3693°
Explanation:
Given
vDE = 7.5 m/s
vPE = 20.2 m/s
Required: vDP
Assume that
vDE to be in direction of - j
vPE to be in direction of i
According to relative motion concept the velocity vDP is given by
vDP = vDE - vPE (I)
Substitute in (I) to get that
vDP = - 7.5 j - 20.2 i
The magnitude of vDP is given by
vDP = √((- 7.5)²+(- 20.2)²) m/s = 21.7454 m/s
θ = Arctan (- 7.5/- 20.2) = 20.3693°
θ is in 3rd quadrant so add 180°
θ = 20.3693° + 180° = 200.3693°