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gladu [14]
3 years ago
12

Two racecars are driving at constant speeds around a circular track. both cars are the same distance away from the center of the

track, but car 2 is driving twice as fast as car 1. the acceleration of car 2 is ___________ the acceleration of car 1.
Physics
1 answer:
emmainna [20.7K]3 years ago
8 0
If the track has circumference c miles, then and the car's speeds are x and y mi/s, then since time = distance/speed, 

<span>In the same direction, the slow car has a "lead" of 1 mile which the faster car has to make up. </span>

<span>1/(y-x) = 120 </span>
<span>1/(x+y) = 30 </span>

<span>120(y-x) = 30(x+y) </span>
<span>120y-120x = 30x+30y </span>
<span>90y = 150x </span>
<span>y = 5/3 x </span>


<span>1/(x + 5/3 x) = 30 </span>
<span>8/3 x * 30 = 1 </span>
<span>80x = 1 </span>
<span>x = 1/80 mi/sec = 45 mph </span>
<span>y = 75 mph</span>
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exis [7]

Answer:

Cant say for sure, its been a while sense ive done this, but im almost certain its 70

Explanation:

Take the zero off of your 10 s

Then go 7 m/s x 1 s

Which equals 7

Add the zero back to the end of your answer

10 -- 0 = 1 x 7 = 7 ++ 0 = 70

(PS, two of the same sign is just adding a number to the end of your original answer, that is not what it actually stands for in mathematical terms but that is what i'm using to make it clearer as to whats happening)

I'm not too good at explaining and formulas but i hope this helped

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3 years ago
Near the poles, more energy is reflected back into space than is absorbed. near the poles, more energy is reflected back into sp
VladimirAG [237]
Correct answer: a) True.

In fact, the soil at the poles consists mainly of snow. Snow has an albedo of 90%: albedo is the fraction of light (and so, of energy) that is reflected back to the space. This means that at the poles, about 90% of the light is reflected back to the space, while only 10% is absorbed.
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2 years ago
A ray diagram without the produced image is shown.
Mashutka [201]
The answer is real and smaller than the object. The image point of the top of the object is the point where the two refracted rays intersect. Tracing the entire image having the same distance from the mirror as the image of the top of the object and with the bottom on the principal axis. Hence, a real inverted image will be formed for an object outside the focal point. 
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gizmo_the_mogwai [7]
The answer your looking for is conduction.
8 0
2 years ago
An offshore oil well is 2 kilometers off the coast. The refinery is 4 kilometers down the coast. Laying pipe in the ocean is twi
shusha [124]

Answer:

Rectangular path

Solution:

As per the question:

Length, a = 4 km

Height, h = 2 km

In order to minimize the cost let us denote the side of the square bottom be 'a'

Thus the area of the bottom of the square, A = a^{2}

Let the height of the bin be 'h'

Therefore the total area, A_{t} = 4ah

The cost is:

C = 2sh

Volume of the box, V = a^{2}h = 4^{2}\times 2 = 128            (1)

Total cost, C_{t} = 2a^{2} + 2ah            (2)

From eqn (1):

h = \frac{128}{a^{2}}

Using the above value in eqn (1):

C(a) = 2a^{2} + 2a\frac{128}{a^{2}} = 2a^{2} + \frac{256}{a}

C(a) = 2a^{2} + \frac{256}{a}

Differentiating the above eqn w.r.t 'a':

C'(a) = 4a - \frac{256}{a^{2}} = \frac{4a^{3} - 256}{a^{2}}

For the required solution equating the above eqn to zero:

\frac{4a^{3} - 256}{a^{2}} = 0

\frac{4a^{3} - 256}{a^{2}} = 0

a = 4

Also

h = \frac{128}{4^{2}} = 8

The path in order to minimize the cost must be a rectangle.

8 0
3 years ago
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