The answer to your question is A, 11/24
Answer:
11 & 5 are the factors of 55
Step-by-step explanation:
Factors of 55
11 × 5 = 55
Thus, 11 & 5 are the factors of 55
<u>-TheUnknownScientist</u><u> 72</u>
The cinematic equation is:
h (t) = (1/2) * a * t ^ 2 + vo * t + h0
Where,
a: acceleration
vo: initial speed
h0: initial height
Substituting values:
h (t) = (1/2) * (- 32) * t ^ 2 + (0) * t + 9
h (t) = - 16t ^ 2 + 9
For t = 0.2 we have:
h (0.2) = - 16 * (0.2) ^ 2 + 9
h (0.2) = 8.36 feet
To touch the ground we have:
-16t ^ 2 + 9 = 0
16t ^ 2 = 9
t = root (9/16)
t = 0.75 s
Answer:
The height of the cherry after 0.2 seconds is:
h (0.2) = 8.36 feet
the cherry hits the ground at:
t = 0.75 s
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Answer: 500$
Explanation:
I hope this helped!
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- Zack Slocum
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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]
