Answer:
a) W₁ = 78400 [J]
b)Wt = 82320 [J]
Step-by-step explanation:
a) W = ∫ f*dl general expression for work
If we have a chain with density of 10 Kg/m, distributed weight would be
9.8 m/s² * 10 kg = mg
Total length of th chain is 40 m, and the function of y at any time is
f(y) = (40 - y ) mg where ( 40 - y ) is te length of chain to be winded
At the beggining we have to wind 40 meters y = 0 at the end of the proccess y = 40 and there is nothing to wind then:
f(y) = mg* (40 - y )
W₁ = ∫f(y) * dy ⇒ W₁ = ∫₀⁴⁰ mg* (40 - y ) dy ⇒ W₁ = mg [ ∫₀⁴⁰ 40dy - ∫₀⁴⁰ ydy
W₁ = mg [ 40*y |₀⁴⁰ - 1/2 * y² |₀⁴⁰ ⇒ W₁ = mg* [ 40*40 - 1/2 (40)² ]
W₁ = mg * [1/2] W₁ = 10*9,8* ( 800 )
W₁ = 78400 [J]
b) Now we can calculate work to do if we have a 25 block and the chain is weightless
W₂ = ∫ mg* dy ⇒ W₂ = ∫₀⁴⁰ mg*dy ⇒ W₂ = mg y |₀⁴⁰
W₂ = mg* 40 = 10*9.8* 40
W₂ = 3920 [J]
Total work
Wt = W₁ + W₂ ⇒ Wt = 78400 + 3920
Wt = 82320 [J]
Step 1: divide: 14-3+3*4^2
Step 2: evaluate the power: 14-3+3*16
Step 3: multiply: 14-3+48
Step 4: add and subtract: 59
Answer:59
It means about 15 centimeters i believe thats right
x = 2y
1/x + 1/y = 3/10
Since we have a value for x, let's plug it into the second equation.
1/2y + 1/y = 3/10
Now, let's make the denominators equal.
Multiply the second term by 2.
1/2y + 2/2y = 3/10
Multiply the final term by 0.2y
1/2y + 2/2y = 0.6y/2y
Compare numerators after adding.
3 = 0.6y
Divide both sides by 0.6
<h3>y = 5</h3>
Now that we have the value of the second integer, we can find the first.
x = 2y
x = 2(5)
<h3>x = 10</h3>
Let's plug in these values in our equations to verify.
10 = 2(5) √ this is true
1/10 + 1/5 = 3/10 √ this is true
<h3>The first integer is equal to 10, and the second is equal to 5.</h3>
