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suter [353]
3 years ago
15

An aircraft engine takes in 9000 j of heat and discards 6400 j each cycle. (a) what is the mechanical work output of the engine

during one cycle? (b) what is the thermal efficiency of the engine?
Physics
1 answer:
charle [14.2K]3 years ago
4 0
Work done = heat in - heat out 
<span>= 9000 -6400 = 3600 J per cycle. </span>

<span>And the efficiency = work out / heat in </span>
<span>= 3600/9000 </span>
<span>= 0.4 or 40%

Sorry if this does not help.</span>
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Round the following off to required number of significant figures.
Stells [14]

Answer:

140

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Explanation:

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A particle of ink in a ink-jet printer carries a charge of -8x 10^-13 C and is deflected onto paper force of 3.2x10^-4. Find the
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A tennis player hits a ball 2.0 m above the ground. The ball leaves his racquet with a speed of 20 m/s at an angle 5.0 ∘ above t
goldfiish [28.3K]

Answer:

Explanation:

initial height, yo = 2 m

initial velocity, u = 20 m/s

angle of projection,θ = 5 degree

distance of net = 7 m

height of net = 1 m

Let it covers a vertical distance y in time t .

Use Second equation of motion for vertical motion

y=y_{0}+uSin\theta t-1/2 gt^{2}

y=2+20Sin5 t-4.9t^{2}

As it hits the ground in time t, so put y = 0

0=2+1.74 t-4.9t^{2}

4.9t^{2}-1.74t-2=0

t= \frac{1.74\pm\sqrt{1.74^{2}+4\times\2\times4.9}}{9.8}

Taking positive sign, t = 0.84 s

The ball travels a horizontal distance x in time t

X = 20 Cos5 x t

X =  16.76 m

As this distance is more than the distance of net, so it clears the net.

Let t' be the time taken to travel a horizontal distance equal to the distance of net

7 = 20 cos5 x t'

t' = 0.35 s

Let the vertical distance traveled by the ball in time t' is y'.

So,

y'=y_{0}+uSin\theta t'-1/2 gt'^{2}

y'=2+20Sin5 t-4.9\times0.35^{2}

y' = 2.008 m

So, it clears the net which is 1 m high.

It clears the net by a vertical distance of 2.008 - 1 = 1.008 m and horizontal distance 16.76 - 7 = 9.76 m

3 0
3 years ago
A gymnast of mass 63.0 kg hangs from a vertical rope attached to the ceiling. You can ignore the weight of the rope and assume t
Sergio [31]

Answer:

Explanation:

A ) When gymnast is motionless , he is in equilibrium

T = mg

= 63 x 9.81

= 618.03 N

B )

When gymnast climbs up at a constant rate , he is still in equilibrium ie net force acting on it is zero as acceleration is zero.

T = mg

= 618.03 N

C ) If the gymnast climbs up the rope with an upward acceleration of magnitude 0.600 m/s2

Net force on it = T - mg   , acting in upward direction

T - mg = m a

T =  mg + m a

= m ( g + a )

= 63 ( 9.81 + .6)

= 655.83 N

D )  If the gymnast slides down the rope with a downward acceleration of magnitude 0.600 m/s2

Net force acting in downward direction

mg - T = ma

T = m ( g - a )

= 63 x ( 9.81 - .6 )

= 580.23 N

6 0
3 years ago
If a town was 90 miles away and you travel at 45 mph how long would it take to get there
notka56 [123]

Answer:

t = 2 hours

Explanation:

Given that,

Distance of the town, d = 90 miles

Speed, v = 45 mph

We need to find the time to get there. The speed of an object is given by :

v=\dfrac{d}{t}

Where

t is time

t=\dfrac{d}{v}\\\\t=\dfrac{90}{45}\\\\t=2\ h

So, the required time is 2 hours.

5 0
3 years ago
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