Answer:
W = 1.06 MJ
Explanation:
- We will use differential calculus to solve this problem.
- Make a differential volume of water in the tank with thickness dx. We see as we traverse up or down the differential volume of water the side length is always constant, hence, its always 8.
- As for the width of the part w we see that it varies as we move up and down the differential element. We will draw a rectangle whose base axis is x and vertical axis is y. we will find the equation of the slant line that comes out to be y = 0.5*x. And the width spans towards both of the sides its going to be 2*y = x.
- Now develop and expression of Force required:
F = p*V*g
F = 1000*(2*0.5*x*8*dx)*g
F = 78480*x*dx
- Now, the work done is given by:
W = F.s
- Where, s is the distance from top of hose to the differential volume:
s = (5 - x)
- We have the work as follows:
dW = 78400*x*(5-x)dx
- Now integrate the following express from 0 to 3 till the tank is empty:
W = 78400*(2.5*x^2 - (1/3)*x^3)
W = 78400*(2.5*3^2 - (1/3)*3^3)
W = 78400*13.5 = 1058400 J
Answer: Cross-cutting features are always younger than the surrounding rock.
When material erodes before sediment is deposited on it, a geologic gap results.
Explanation:
The options include:
1. An unconformity is created when lava pours out on Earth’s surface.
2. Faults are the result of volcanic activity.
3. Intrusions and extrusions are sedimentary formations.
4. Cross-cutting features are always younger than the surrounding rock.
5. When material erodes before sediment is deposited on it, a geologic gap results.
The law of superposition simply states that when there is a layers of rocks, we would see that the younger layers will lie and be on top of the layers that are older.
Other tools that can help scientist with relative dating are:
• Cross-cutting features are always younger than the surrounding rock.
• When material erodes before sediment is deposited on it, a geologic gap results.
Jack------------ force of 92.5 n eastward-------Fjack(X)=92.5 n Fjack(Y)=0
<span>jill ------------------------------- force of 89.9 n northeast
Fjill(X)=cos45*89.9=63.57 n
</span>Fjill(Y)=sin45*89.9=63.57 n<span>
</span>jane -----------------------------force of 163 n southeast
Fjane(X)=cos45*163=115.26 n
Fjane(X)=-sin45*163=-115.26 n
Ftotal (X)=92.5+63.57+115.26=271.33 n
Ftotal (Y)=0+63.57-115.26=-51.69 n
Fotal=((271.33)^2+(-115.26)^2) ^0.5=294.80 n southeast
the magnitude of the net force the people exert on the donkey. is 294.80 n southeast
<h2>K.E/P.E = m/k tan²φ x ω²</h2>
Explanation:
The given position of block x = x₀ cos(ωt + φ)
The velocity of block v = dx/dt = - x₀ sin(ωt + φ) x ω
The kinetic energy = 1/2 mv² = 1/2 m x₀² sin²(ωt + φ) x ω²
The potential energy of spring = 1/2 k x² , where k is the spring constant
Thus P.E = 1/2 x k x x₀² cos²(ωt + φ)
When t = 0
K.E = 1/2 m x₀²sin²φ x ω²
P.E = 1/2 k x₀² cos²φ
Dividing these , we have
K.E/P.E = m/k tan²φ x ω²
Answer:
the magnitude of Vpg = 493.711 km/h
Explanation:
given data
speed Vpg = 560 km/h
speed Vwg = 80 km/h
solution
we get here magnitude of the plane velocity w.r.t. ground is
we know that the Vpg = Vpw + Vwg .....................1
writing the component of the velocity that is
Vpw = (0 km/h î - 560 km/h j )
Vwg = (80 cos 45 km/h î + 80 sin 45 km/h j)
adding these
Vpg = (0+80 cos 45 km/h ) î + ( -560 + 80 sin 45 km/h j)i
Vpg = (42.025 ) î (-491.92 km/h)j
now we take magnitude
the magnitude of Vpg = 
the magnitude of Vpg = 493.711 km/h
