Answer:
Explanation:
given,
Angular speed of the tire = 32 rad/s
Displacement of the wheel = 3.5 rev
Δ θ = 3.5 x 2 π
= 7 π rad
now,
Time interval of the car to rotate 7π rad
using equation
Time taken to rotate 3.5 times is equal to 0.687 s.
Answer:
A(3.56m)
Explanation:
We have a conservation of energy problem here as well. Potential energy is being converted into linear kinetic energy and rotational kinetic energy.
We are given ω= 4.27rad/s, so v = ωr, which is 6.832 m/s. Place your coordinate system at top of the hill so E initial is 0.
Ef= Ug+Klin+Krot= -mgh+1/2mv^2+1/2Iω^2
Since it is a solid uniform disk I= 1/2MR^2, so Krot will be 1/4Mv^2(r^2ω^2= v^2).
Ef= -mgh+3/4mv^2
Since Ef=Ei=0
Mgh=3/4mv^2
gh=3/4v^2
h=0.75v^2/g
plug in givens to get h= 3.57m
False. Since the forces are pulling in equal and opposite directions, the net force is 0.
Answer:
The dose is 6 mSV
Explanation:
The absorbed dose (in gray - Gy) is the amount of energy that ionizing radiation deposits per unit mass of tissue. That is,
Absorbed dose = Energy deposited / Mass
while Dose equivalent (DE) (in Seivert -Sv) is given by
DE = Absorbed dose × RBE (Relative biological effectiveness)
First, we will determine the Absorbed dose
From the question, Energy deposited = 30mJ and Mass = 50kg
From,
Absorbed dose = Energy deposited / Mass
Absorbed dose = 30mJ/50kg
Absorbed dose = 0.6 mGy
Now, for the Dose equivalent (DE)
DE = Absorbed dose × RBE
From the question, RBE = 10
Hence,
DE = 0.6mGy × 10
DE = 6 mSv
