Answer: MR²
is the the moment of inertia of a hoop of radius R and mass M with respect to an axis perpendicular to the hoop and passing through its center
Explanation:
Since in the hoop , all mass elements are situated at the same distance from the centre , the following expression for the moment of inertia can be written as follows.
I = ∫ r² dm
= R²∫ dm
MR²
where M is total mass and R is radius of the hoop .
Answer:
(a) Heat transfer to the environment is: 1 MJ and (b) The efficiency of the engine is: 41.5%
Explanation:
Using the formula that relate heat and work from the thermodynamic theory as:
solving to Q_out we get:
this is the heat out of the cycle or engine, so it will be heat transfer to the environment. The thermal efficiency of a Carnot cycle gives us: 
where T_Low is the lowest cycle temperature and T_High the highest, we need to remember that a Carnot cycle depends only on the absolute temperatures, if you remember the convertion of K=°C+273.15 so T_Low=150+273.15=423.15 K and T_High=450+273.15=723.15K and replacing the values in the equation we get:
They should look for <span>a report from an independent scientific research firm,
even if they have to pay for it.
In preparing its report, the firm would have already surveyed many of the </span>
<span>citizens from several other towns that currently add fluoride to their water,
plus a lot of other relevant medical research on the subject.</span>
Answer:
Explanation:
Given:
- mass of solid disk,
- radius of disk,
- force of push applied to disk,
- distance of application of force from the center,
<em>For the condition of no slip the force of static friction must be greater than the applied force so that there is no skidding between the contact surfaces at the contact point.</em>
where:
= static frictional force
Answer:
4.2 m/s
Explanation:
Momentum is conserved.
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
(35 g) (9 m/s) + (75 g) (-7 m/s) = (35 g) (-15 m/s) + (75 g) v
315 g m/s − 525 g m/s = -525 g m/s + (75 g) v
315 g m/s = (75 g) v
v = 4.2 m/s
