"v0" means that there are no friction forces at that speed
<span>mgsinΘ = (mv0²/r)cosΘ → the variable m cancels </span>
<span>sinΘ/cosΘ = tanΘ = v0² / gr
</span><span>Θ = arctan(v0² / gr) </span>
<span>When v > v0, friction points downslope: </span>
<span>mgsinΘ + µ(mgcosΘ + (mv²/r)sinΘ) = (mv²/r)cosΘ → m cancels: </span>
<span>gsinΘ + µ(gcosΘ + (v²/r)sinΘ) = (v²/r)cosΘ </span>
<span>µ = ((v²/r)cosΘ - gsinΘ) / (gcosΘ + (v²/r)sinΘ) </span>
<span>where Θ is defined above. </span>
<span>When v > v0, friction points upslope: </span>
<span>mgsinΘ - µ(mgcosΘ + (mv²/r)sinΘ) = (mv²/r)cosΘ → m cancels: </span>
<span>gsinΘ - µ(gcosΘ + (v²/r)sinΘ) = (v²/r)cosΘ </span>
<span>µ = (gsinΘ - (v²/r)cosΘ) / (gcosΘ + (v²/r)sinΘ) </span>
<span>where Θ is defined above. </span>
-- Although it's not explicitly stated in the question,we have to assume that
the surface is frictionless. I guess that's what "smooth" means.
-- The total mass of both blocks is (1.5 + 0.93) = 2.43 kg. Since they're
connected to each other (by the string), 2.43 kg is the mass you're pulling.
-- Your force is 6.4 N.
Acceleration = (force)/(mass) = 6.4/2.43 m/s²<em>
</em> That's about <em>2.634 m/s²</em> <em>
</em>(I'm going to keep the fraction form handy, because the acceleration has to be
used for the next part of the question, so we'll need it as accurate as possible.)
-- Both blocks accelerate at the same rate. So the force on the rear block (m₂) is
Force = (mass) x (acceleration) = (0.93) x (6.4/2.43) = <em>2.45 N</em>.
That's the force that's accelerating the little block, so that must be the tension
in the string.
Answer:
A point on the outside rim will travel 157.2 meters during 30 seconds of rotation.
Explanation:
We can find the distance with the following equation since the acceleration is cero (the disk rotates at a constant rate):
Where:
v: is the tangential speed of the disk
t: is the time = 30 s
The tangential speed can be found as follows:
Where:
ω: is the angular speed = 100 rpm
r: is the radius = 50 cm = 0.50 m
Now, the distance traveled by the disk is:
Therefore, a point on the outside rim will travel 157.2 meters during 30 seconds of rotation.
I hope it helps you!
Answer:
56
Explanation:
I just want the points to be completely honest with you.
<u>Given data</u>
Source temperature (T₁) = 177°C = 177+273 = 450 K
Sink temperature (T₂) = 27°C = 27+273 = 300 K
Energy input (Q₁) = 3600 J ,
Work done = ?
We know that, efficiency (η) = Net work done ÷ Heat supplied
η = W ÷ Q₁
W = η × Q₁
First determine the efficiency ( η ) = ?
Also, we know that ( η ) = (T₁ - T₂) ÷ (T₁)
= 33.3% = 0.333
Now, Work done is W = η × Q₁
= 0.33 × 3600
<em> W = 1188 J</em>
<em>Work done by the engine is 1188 J</em>
