Answer:
7,700
Step-by-step explanation:
-175 x 6 = -1,050
-1,050 + 8,750 = 7,700
Answer:
<h2>
-29.61m/s</h2>
Step-by-step explanation:
Given the distance of fall of the student in term of the time t expressed by the equation s(t) = −16t² + 8√t, to get the average speed of fall of the pencil after 2.8 secs, we will need to differentiate the given function first since Velocity is the change in distance of a body with respect to time i.e
V = d(s(t))/dt
s(t) = −16t² + 8t^1/2
V = -32t+1/2(8)t^(1/2 - 1)
V = -32t+4t^-1/2
The average speed of the fall Using the fact that the pencil hit the ground in exactly 2.8 seconds, will be gotten by substituting t = 2.8 into the resulting equation.
V = -32t+4(2.8)^-1/2
V = -32t+4/√2.8
V = -32+4/1.6733
V = -32+2.391
v = -29.61m/s
<em>Hence the average speed of the fall is -29.61m/s</em>
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]
