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hjlf
3 years ago
11

Technician A says mechanical efficiency is a comparison between brake horsepower and indicated horsepower. Technician B says vol

umetric efficiency is a measurement of the amount of air-fuel mixture that actually enters the combustion chamber compared to the amount that could be drawn in. Who is correct?
Physics
1 answer:
Reil [10]3 years ago
8 0

Answer:Both are correct

Explanation:

Both are correct because

Mechanical efficiency is the dimensionless term which is the ratio of brake horsepower to the Indicated horse Power

Where brake power is the Power obtained at the crankshaft  and  

Indicated horsepower is the power obtained in the combustion chamber and this power is the loss in the form of friction.    

Volumetric efficiency is the ratio of actual fuel intake to the maximum air fuel that could be taken.  

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Which one is having lesser ressistance. a 60w bulb or a 40w bulb?
Ivan
Power dissipation = (voltage across the component)² / (resistance of the component)

Since the resistance is in the denominator of the fraction in this formula for the
quantity of power dissipated, you can see that when the supply voltage is constant,
the smaller resistance dissipates more power.

So the <u>60w bulb</u> has lower resistance than the 40w bulb.
8 0
3 years ago
A lead ball is dropped into a lake from a diving board 5.0 m above the water. After entering the water, it sinks to the bottom w
nirvana33 [79]

Answer:

|D_{depth} |=19.697m

Explanation:

To find Depth D of lake we must need to find the time taken to hit the water.So we use equation of simple motion as:

Δx=vit+(1/2)at²

x_{f}-x_{i}=v_{i}t+(1/2)at^{2}\\  -5.0m=(o)t+(1/2)(-9.8m/s^{2} )t^{2}\\ -4.9t^{2}=-5.0\\ t^{2}=5/4.9\\t=\sqrt{1.02} \\t=1.01s

As we have find the time taken now we need to find the final velocity vf from below equation as

v_{f}=v_{i}+at\\v_{f}=0+(-9.8m/s^{2} )(1.01s) \\v_{f}=-9.898m/s

So the depth of lake is given by:

first we need to find total time as

t=3.0-1.01 =1.99 s

|D_{depth} |=|vt|\\|D_{depth} |=|(-9.898m/s)(1.99s)|\\|D_{depth} |=19.697m

6 0
3 years ago
How to atoms behave in non-magnetic items?
Anastaziya [24]

Answer:

By altering the quantum interactions of the electrons in the atoms of a metal's atoms, scientists from the University of Leeds have generated magnetism in metals that aren’t normally magnetic.

Explanation:

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What is the difference between the inner and outer planets?
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3 0
3 years ago
Read 2 more answers
A bug is 12 cm from the center of a turntable that is rotating with a frequency of 45 rev/min . What minimum coefficient frictio
Agata [3.3K]

Answer:

The minimum coefficient of friction is 0.27.

Explanation:

To solve this problem, start with identifying the forces at play here. First, the bug staying on the rotating turntable will be subject to the centripetal force constantly acting toward the center of the turntable (in absence of which the bug would leave the turntable in a straight line). Second, there is the force of friction due to which the bug can stick to the table. The friction force acts as an intermediary to enable the centripetal acceleration to happen.

Centripetal force is written as

F_c = m\frac{v^2}{r}

with v the linear velocity and r the radius of the turntable. We are not given v, but we can write it as

v = r\omega

with ω denoting the angular velocity, which we are given. With that, the above becomes:

F_c = m\frac{v^2}{r}=m\omega^2 r

Now, the friction force must be at least as much (in magnitude) as Fc. The coefficient (static) of friction μ must be large enough. How large?

F_r=\mu mg \geq m\omega^2 r = F_c\implies\\\mu \geq \frac{\omega^2 r}{g}

Let's plug in the numbers. The angular velocity should be in radians per second. We are given rev/min, which can be easily transformed by a factor 2pi/60:

\frac{1 rev}{1 min}\cdot\frac{\frac{2\pi rad}{rev}}{\frac{60s}{1 min}}=\frac{2\pi}{60}\frac{rad}{s}

and so 45 rev/min = 4.71 rad/s.

\mu \geq \frac{\omega^2 r}{g}=\frac{4.71^2\frac{1}{s^2}\cdot 0.12m}{9.8\frac{m}{s^2}}=0.27

A static coefficient of friction of at least be 0.27 must be present for the bug to continue enjoying the ride on the turntable.



3 0
3 years ago
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