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Sergio039 [100]
3 years ago
9

given the perimeter of a rectangle and either the length or width, find the unknown measurement using the appropriate formula. T

he perimeter is 54cm, and the length is 15cm.
Physics
1 answer:
LekaFEV [45]3 years ago
3 0
54-(15x2) divided by 2 = 12
54 - 15 = 39
39-15=24
24 divided by 2=12

Length:15
Width:12
Perimeter:54
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Five race cars speed toward the finish line at the Jasper County Speedway. The table lists each car’s speed in meters/second. If
nekit [7.7K]
We have: v = d/t
From the expression, we conclude speed is indirectly proportional to speed.
So, the car which will take longer time must have the smallest speed. Among all the options Car C has the smallest speed. So, it would be your answer.

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Hope this helps!
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3 years ago
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Help!! Or I get an F!!! <br> What planet could you jump the highest?
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You're driving down the highway late one night at 20 m/s when a deer steps onto the road 49 m in front of you. You reaction time
Leno4ka [110]

Answer:

v = 26.7 m/s

Explanation:

Given,

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distance between the car and the deer = 49 m

time taken to press the brake = 0.50 s

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Now,

distance travel by the car in 0.5 s = u x t = 20 x 0.5 = 10 m

distance travel by the car after the break is pressed

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v² = u² + 2 a s

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s = 20 m

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b. Maximum speed a car could have

Distance travel by the car in reaction time = v' x 0.5

v' is the maximum speed of the car.

maximum distance car can cover = 49 - 0.5 v'

Now, Using equation of motion

v² = u² + 2 a s

0² =v'² - 2 x 10 x (49- 0.5 x v')

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v = 26.7 m/s

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4 0
3 years ago
A Carnot engine operates between two heat reservoirs at temperatures THTH and TCTC. An inventor proposes to increase the efficie
Marat540 [252]

Answer:

e_12=1-Tc/Th

This is same as the original Carnot engine.  

Explanation:

For original Carnot engine, its efficiency is given by

e = 1-Tc/Th

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e_12=(W_1+W_2)/Q_H1

where Q_H1 is the heat input to the first engine, W_1 s the work done by the first engine and W_2 is the work done by the second engine.  

But the work done can be written as  

W= Q_H + Q_C with Q_H as the heat input and Q_C as the heat emitted to the cold reservoir. So.  

e_12=(Q_H1+Q_C1+Q_H2+Q_C2)/Q_H1

But Q_H2 = -Q_C1 so the second and third terms in the numerator cancel  

each other.

 e_12=1+Q_C2/Q_H1

but, Q_C2/Q_H2= -T_C/T'

⇒ Q_C2 = -Q_H2(T_C/T')

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But, Q_C1 = -Q_H1(T'/T_H)

so, Q_C2 =  -Q_H1(T'/T_H)(T_C/T') = Q_H1(T_C/T_H) So the efficiency of the composite engine is given by  

e_12=1-Tc/Th

This is same as the original Carnot engine.  

7 0
3 years ago
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Ksenya-84 [330]

Answer:

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Explanation:

5 0
4 years ago
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