Answer:
1. 625,000 J
2. 100 J
4. 5 kg
5. √5 ≈ 2.236 m/s
Step-by-step explanation:
You should be aware that the SI derived units of Joules are equivalent to kg·m²/s².
To reduce confusion between <em>m</em> for mass and m for meters, we'll use an <em>italic m</em> for mass.
In each case, the "find" variable is what's left after we put the numbers into the formula. It is what the question is asking for. The "given" values are the ones in the problem statement and are the values we put into the formula. The formula is the same in every case.
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1. KE = (1/2)<em>m</em>v² = (1/2)(2000 kg)(25 m/s)² = 625,000 kg·m²/s² = 625,000 J
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2. KE = (1/2)<em>m</em>v² = (1/2)(0.5 kg)(20 m/s)² = 100 kg·m²/s² = 100 J
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4. KE = (1/2)<em>m</em>v²
250 J = (1/2)<em>m</em>(10 m/s)² = 50 m²/s²
(250 kg·m²/s²)/(50 m²/s²) = <em>m</em> = 5 kg
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5. KE = (1/2)<em>m</em>v²
2000 kg·m²/s² = (1/2)(800 kg)v²
(2000 kg·m²/s²)/(400 kg) = v² = 5 m²/s²
v = √5 m/s ≈ 2.236 m/s
The equation is -2x-4. u can use y2-y1/x2-x1.
If the surface area of the cylinder is 12π square meters. Then the radius of the sphere will be 1.7 meters.
<h3>What is a cylinder?</h3>
A cylinder is a closed solid that has two parallel circular bases connected by a curved surface.
A cylinder has a height of 4 meters and a radius of 1.5 meters.
Then the approximate radius of a sphere that has the same surface area as the cylinder.
We know that the Surface area of the cylinder will be is given as
Surface area = 2πrh
Surface area = 2 x π x 1.5 x 4
Surface area = 12π square meters
Then we have
Surface area of sphere = surface area of cylinder
4πr² = 12π
r² = 3
r = 1.7 meters
More about the cylinder link is given below.
brainly.com/question/3692256
#SPJ1
First find the side lengths of the box by factoring each area:
12: 1, 2, 3, 4, 6, 12
15: 1, 3, 5, 15
20: 1, 2, 4, 5, 10, 20
Find 3 numbers that both have 2 sides have in common and can multiply to make the areas of the faces:
12: 3 * 4
15: 3 * 5
20: 4 * 5
Sides: 3, 4, and 5
Then multiply to get the volume:
3 * 4 * 5 = 60 units^2
The answer is A
hope you found this usefull
