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iVinArrow [24]
3 years ago
6

The driver of a 810.0 kg car decides to double the speed from 23.6 m/s to 47.2 m/s. What effect would this have on the amount of

work required to stop the car, that is, on the kinetic energy of the car? KEi= × 105 J KEf= × 105 J times as much work must be done to stop the car.
Mathematics
1 answer:
Setler [38]3 years ago
8 0

Answer:

  • KEi = 2.256×10^5 J
  • KEf = 9.023×10^5 J
  • 4 times as much work

Step-by-step explanation:

The kinetic energy for a given mass and velocity is ...

  KE = (1/2)mv^2 . . . . . m is mass

At its initial speed, the kinetic energy of the car is ...

  KEi = (1/2)(810 kg)(23.6 m/s)^2 ≈ 2.256×10^5 J . . . . . m is meters

At its final speed, the kinetic energy of the car is ...

  KEf = (1/2)(810 kg)(47.2 m/s)^2 ≈ 9.023×10^5 J

The ratio of final to initial kinetic energy is ...

  (9.023×10^5)/(2.256×10^5) = 4

4 times as much work must be done to stop the car.

_____

You know this without computing the kinetic energy. KE is proportional to the square of speed, so when the speed doubles, the KE is multiplied by 2^2 = 4.

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A. Independent variable: number of rubber stamps n sold. Dependent variable: F(n), the total sales.

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C. F(n) = $4.95n, which is 4.95 times n.

So, plugin 5 for n

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6 0
3 years ago
Read 2 more answers
Find the value of cos 28°​cos 62°​– sin 28°​sin 62°​
Romashka [77]
<h2>Answer:</h2>

cos 28°​cos 62°​– sin 28°​sin 62°​ = 0

<h2>Step-by-step explanation:</h2>

From one of the trigonometric identities stated as follows;

<em>cos(A+B) = cosAcosB - sinAsinB             -----------------(i)</em>

We can apply such identity to solve the given expression.

<em>Given:</em>

cos 28°​cos 62°​– sin 28°​sin 62°​

<em>Comparing the given expression with the right hand side of equation (i), we see that;</em>

A = 28°

B = 62°

<em>∴ Substitute these values into equation (i) to have;</em>

<em>⇒ cos(28°+62°) = cos28°cos62° - sin28°sin62°</em>

<em />

<em>Solve the left hand side.</em>

<em>⇒ cos(90°) = cos28°cos62° - sin28°sin62°</em>

⇒ 0 = <em>cos28°cos62° - sin28°sin62°     (since cos 90° = 0)</em>

<em />

<em>Therefore, </em>

<em>cos28°cos62° - sin28°sin62° = 0</em>

<em />

<em />

8 0
3 years ago
I need help on number 59. It makes no sense to me, I cannot figure out how I would use Pythagorean identities to solve this. Ple
Readme [11.4K]
If you have never used a unit circle, I recommend making one. 
R for the following degrees are
30*= 4((3^1/2)/2)^2
60*= 8(2^1/2)(2^1/2)^2
90*= (8((3^1/2)/2))/4
5 0
3 years ago
The two triangles shown in the figure are similar. Find the distance, d, across Coyote Ravine.
Masteriza [31]

Answer:

d = 157.5 m

Step-by-step explanation:

from similar triangle:

350/400 = d/180

d = 350/400 * 180

d= 157.5 m

4 0
3 years ago
Helppppppppppppppppppppp
docker41 [41]
Answer:
1.796

Explanation:
log base 6 of 25 can be rewritten as:
log base 6 of (5^2) = 2 * log base 6 of 5 
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Hope this helps :)
3 0
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