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bonufazy [111]
3 years ago
10

HELP!!! 20 PNTS!!!! And Brainliest!!! A car, a bike, a bus, and a motorcycle are moving parallel to each other on a road. Each v

ehicle is traveling at the same speed. The ____ will have the highest kinetic energy, while the _____ will have the lowest kinetic energy.
Mathematics
2 answers:
Soloha48 [4]3 years ago
7 0
The bus will have the highest kinetic energy and the bike will have the least kinetic energy because kinetic energy =1/2 mass × velocity squared
Paraphin [41]3 years ago
3 0
In order to answer that, we would have to take the mass of each of them. The higher the mass, means there is higher the kinetic energy. We use the equation:

Ek = 1/2*m*v^2

Since velocity will be the same speed, it will be constant. The mass is what makes the difference with the bus having the maximum and the bike having the minimum kinetic energy.
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The equivalent to that is 8.
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3 years ago
Picture shown please help !!!
natulia [17]
Let's factor the top and the bottom.
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3 years ago
A manager checked production records and found that a worker produced 200 units while working 40 hours. In the previous week, th
jeka94

Answer:

a. Current: 5 units/hour. Previous: 4.4 units/hour

b. Increase

Step-by-step explanation:

a. Current period productivity is 200 / 40 = 5 units/hour

Previous period productivity is 132 / 30 = 4.4 units/hour

b. As this week's productivity = 5 units/hours which is larger than last week's productivity = 4.4 units/hour. The worker's productivity for this week has increased.

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3 years ago
A wire b units long is cut into two pieces. One piece is bent into an equilateral triangle and the other is bent into a circle.
mezya [45]
1. Divide wire b in parts x and b-x. 

2. Bend the b-x piece to form a triangle with side (b-x)/3

There are many ways to find the area of the equilateral triangle. One is by the formula A= \frac{1}{2}sin60^{o}side*side=   \frac{1}{2} \frac{ \sqrt{3} }{2}  (\frac{b-x}{3}) ^{2}= \frac{ \sqrt{3} }{36}(b-x)^{2}
A=\frac{ \sqrt{3} }{36}(b-x)^{2}=\frac{ \sqrt{3} }{36}( b^{2}-2bx+ x^{2}  )=\frac{ \sqrt{3} }{36}b^{2}-\frac{ \sqrt{3} }{18}bx+ \frac{ \sqrt{3} }{36}x^{2}

Another way is apply the formula A=1/2*base*altitude,
where the altitude can be found by applying the pythagorean theorem on the triangle with hypothenuse (b-x)/3 and side (b-x)/6

3. Let x be the circumference of the circle.

 2 \pi r=x

so r= \frac{x}{2 \pi }

Area of circle = \pi  r^{2}= \pi  ( \frac{x}{2 \pi } )^{2} = \frac{ \pi }{ 4 \pi ^{2}  }* x^{2} = \frac{1}{4 \pi } x^{2}

4. Let f(x)=\frac{ \sqrt{3} }{36}b^{2}-\frac{ \sqrt{3} }{18}bx+ \frac{ \sqrt{3} }{36}x^{2}+\frac{1}{4 \pi } x^{2}

be the function of the sum of the areas of the triangle and circle.

5. f(x) is a minimum means f'(x)=0

f'(x)=\frac{ -\sqrt{3} }{18}b+ \frac{ \sqrt{3} }{18}x+\frac{1}{2 \pi } x=0

\frac{ -\sqrt{3} }{18}b+ \frac{ \sqrt{3} }{18}x+\frac{1}{2 \pi } x=0

(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) x=\frac{ \sqrt{3} }{18}b

x= \frac{\frac{ \sqrt{3} }{18}b}{(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) }

6. So one part is \frac{\frac{ \sqrt{3} }{18}b}{(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) } and the other part is b-\frac{\frac{ \sqrt{3} }{18}b}{(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) }

4 0
3 years ago
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Xelga [282]

Answer:

3/4 cups

Step-by-step explanation:

7 0
3 years ago
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