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3 years ago

Three Carnot engines operate between the following temperature limits.

Physics

1 answer:

Kipish [7]3 years ago

6 0

Answer:c>a>b

Explanation:

Given

All the engine extracts same amount of heat(Q) from High-temperature reservoir

For a) 400 and 500 K

$\eta _{engine}=1-\frac{T_L}{T_H}$

$\eta _{engine}=\frac{work\ supplied}{heat\ supplied}$

$1-\frac{400}{500}=\frac{W_a}{Q}$

$W_a=\frac{Q}{5}$

For b)500 K and 600K

$1-\frac{500}{600}=\frac{W_b}{Q}$

$W-b=\frac{Q}{6}$

For c) 400 K and 600 K

$1-\frac{400}{600}=\frac{W_c}{Q}$

$W_c=\frac{2Q}{3}$

So c will give the highest amount of work

c>a>b

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Calculate the average speed of a gazelle that runs 140 meters in 5.0 seconds

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140 m / 5 s = 28 meters per second

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The correct choice is A. 0 meters.

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3 years ago

Read 2 more answers

At high speeds, a particular automobile is capable of an acceleration of about 0.540 m/s^2. At this rate, how long (in seconds)

Mama L [17]

Answer:

t = 6.68 seconds

Explanation:

The acceleration of the automobile, $a=0.54\ m/s^2$

Initial speed of the automobile, u = 91 km/hr = 25.27 m/s

Final speed of the automobile, v = 104 km/hr = 28.88 m/s

Let t is the time taken to accelerate from u to v. It can be calculated as the following formula as :

$t=\dfrac{v-u}{a}$

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t = 6.68 seconds

So, the time taken by the automobile to accelerate from u to v is 6.68 seconds. Hence, this is the required solution.

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3 years ago

A 124-kg balloon carrying a 22-kg basket is descending with a constant downward velocity of 20.0 m/ s. A I.O-kg stone is thrown

nadya68 [22]

(a) 296.6 m

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$v_x = 15.0 m/s$

and with an initial vertical velocity of

$v_{y0} = -20.0 m/s$

where we have put a negative sign to indicate that the direction is downward.

The vertical position of the stone at time t is given by

$y(t) = h + v_{0y} t + \frac{1}{2}gt^2$ (1)

where

h is the initial height

g = -9.81 m/s^2 is the acceleration due to gravity

The stone hits the ground after a time t = 6.00 s, so at this time the vertical position is zero:

y(6.00 s) = 0

Substituting into eq.(1), we can solve to find the initial height of the stone, h:

$0 = h + v_{0y} y + \frac{1}{2}gt^2\\h = -v_{0y} y - \frac{1}{2}gt^2=-(-20.0 m/s)(6.00 s) - \frac{1}{2}(9.81 m/s^2)(6.00 s)^2=296.6 m$

(b) 176.6 m

The balloon is moving downward with a constant vertical speed of

$v_y = -20 m/s$

So the vertical position of the balloon after a time t is

$y(t) = h + v_y t$

and substituting t = 6.0 s and h = 296.6 m, we find the height of the balloon when the rock hits the ground:

$y(t) = 296.6 m + (-20.0 m)(6.00 s)=176.6 m$

(c) 198.2 m

In order to find how far is the rock from the balloon when it hits the ground, we need to find the horizontal distance covered by the rock during the time of the fall.

The horizontal speed of the rock is

$v_x = 15.0 m/s$

So the horizontal distance travelled in t = 6.00 s is

$d_x = v_x t = (15.0 m/s)(6.00 s)=90 m$

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$d_y = 176.6 m$

The distance between the balloon and the rock can be found by using Pythagorean theorem:

$d=\sqrt{(90 m)^2+(176.6 m)^2}=198.2 m$

(di) 15.0 m/s, -58.8 m/s

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Instead, the vertical velocity of the rock for an observer at rest in the basket is

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Substituting time t=6.00 s, we find

$v_y = (-9.8 m/s)(6.00 s)=-58.8 m/s$

(dii) 15.0 m/s, -78.8 m/s

For an observer at rest on the ground, the rock is still moving horizontally with a velocity of

$v_x = 15.0 m/s$

Instead, the vertical velocity of the rock for an observer on the ground is now given by

$v_y (t) = v_{0y} + gt$

Substituting time t=6.00 s, we find

$v_y = (-20.0 m/s)+(-9.8 m/s)(6.00 s)=-78.8 m/s$

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3 years ago

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Zigmanuir [339]

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3 years ago