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bixtya [17]
2 years ago
3

The viewing portion of the rectangular glass window in a fish tank is 63 inches wide and runs from 1.5 inches below the water's

surface to 40.5 inches below the surface. Find the fluid force against this portion of the window. The weight-density of seawater is 64 lb/ft What is the fluid force against the window? lb te
Physics
1 answer:
AlekseyPX2 years ago
3 0

Answer:

fluid force = 1309 lb

Explanation:

Given data

wide = 63 inches

distance below water surface = 1.5 inches

distance below surface = 40.5 inches

weight-density of seawater W =  64 lb/ft

to find out

fluid force against the window

solution

we consider here A(y) is the depth of fluid and B(y) length of portion at y

so we can say y1 = 0 so y2 will be 33.5 - 0.5 = 33 inches = 2.75 feet

so  A(y) = 33.5/12 - y ) feet

and B(y) = 63 inches = 5.25 feet

fluid force = W \int_{y1}^{y2} A(y) B(y) dy

so take limit y = 0 to y = 2.77

fluid force = 64 \int_{0}^{2.75} ((33.5/12) -y ) (5.25) dy

fluid force = 64×5.25/12 \int_{0}^{2.75} ((33.5  -12y ) dy

fluid force = 28 (33.5y - 6y^2)^{2.75} _0

fluid force = 28 × 46.75 - 0

so fluid force = 1309 lb

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two people, each with a mass of 70 kg, are wearing inline skates and are holding opposite ends of a 15m rope. One person pulls f
Zina [86]

Answer:

7.75 s

Explanation:

Newton's second law:

∑F = ma

35 N = (70 kg) a

a = 0.5 m/s²

Given v₀ = 0 m/s and Δx = 15 m:

Δx = v₀ t + ½ at²

(15 m) = (0 m/s) t + ½ (0.5 m/s²) t²

t = 7.75 s

5 0
2 years ago
A celebrating student throws a water balloon horizontally from a dormitory window that is 50 m above the ground. It hits the gro
Contact [7]

Answer:

a) The horizontal velocity of the balloon just before it hits the ground is 6 m/s

b) The magnitude of the vertical velocity of the balloon just before it hits the ground is 98 m/s.

Explanation:

Hi there!

The velocity and position vectors of the water balloon are given by the following equations:

r =(x0 + v0x · t, y0 + v0y · t + 1/2 · g · t²)

v =(v0x, v0y + g · t)

where:

r = position vector at time t.

x0 = initial horizontal position.

v0x = initial horizontal velocity.

t = time.

y0 = initial vertical position.

v0y = initial vertical position.

g = acceleration due to gravity (-9.8 m/s² considering the upward position as positive) .

v = velocity vector at time t.

a) Please, see the attached figure for a graphic description of the problem.

Considering the origin of the frame of reference as the point of launch, notice that the position vector when the balloon hits the ground is

r1 = (60, -50) m

Then:

r1x = 60 m = v0x · t

r1y = -50 m = 1/2 · (-9.8 m/s²) · t²

(notice that the initial vertical velocity is zero, see figure).

Solving r1y for t:

(-50 m · 2) / -9.8 m/s² = t²

t = 10 s

Now, let´s replace t in the r1x equation and solve it for the horizontal component of the velocity:

60 m = v0x · 10 s

v0x = 60 m / 10 s

v0x = 6 m/s

The initial horizontal component of the velocity is 6 m/s. This velocity is constant because there is no air resistance. Then, just before the balloon hits the ground, it will have a horizontal velocity of 6 m/s.

b) To calculate the vertical component of the velocity when the balloon hits the ground, let´s use the equation of the vertical component of the velocity:

v1y = v0y + g · t

Since v0y = 0

v1y = -9.8 m/s² · (10 s) = -98 m/s

The magnitude of the vertical velocity of the balloon when it hits the ground is 98 m/s.

4 0
2 years ago
The chemical formula for glucose is C6H12O6. Therefore, four molecules of glucose will have carbon atoms, hydrogen atoms, and ox
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If there are 4 molecules of glucose, there will be 24 carbon, 48 hydrogen, and 24 oxygen.
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2 years ago
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Isabella drops a pen off her balcony by accident while celebrating the successful completion of a physics problem. assuming air
expeople1 [14]
U = 0, initial vertical velocity

Neglect air resistance, and g = 9.8 m/s².

The time, t, required for the pen to attain a vertical velocity of 19.62 m/s is given by
19.62 m/s = 0 + (9.8 m/s²)*(t s)
t = 19.62/9.8 = 2.00 s

Answer:  2.0 s
8 0
2 years ago
A solid cylindrical bar conducts heat at a rate of 25 W from a hot to a cold reservoir under steady state conditions. If both th
expeople1 [14]

Answer:

Using the new cylinder the heat rate between the reservoirs would be 50 W

Explanation:

  1. Conduction could be described by the Law of Fourierin the form: Q=kA\frac{T_1-T_2}{L} where Q is the rate of heat transferred  by conduction, k is the thermal conductivity of the material, T_1 and T_2 are the temperatures of each heat deposit, A is the cross area to the flow of heat, and {L} is the distance that the flow of heat has to go.
  2. For the original cylinder the Fourier's law would be: kA_1\frac{T_1-T_2}{L_1}=25W, and if A_1=\frac{\pi D_{1}^{2}}{4}, then the expression would be:k\frac{\pi D_1^{2}}{4} \frac{T_1-T_2}{L_1}=25W where D_1 is the diameter of the original cylinder, and {L_1} is the length of the original cylinder.
  3. For the new cylinder, in the same fashion that for the first, Fourier's Law would be: Q_2=k\frac{\pi D_2^2}{4}\frac{T_1-T_2}{L_2},where Q_2 is the heat rate in the second case, D_2 and {L_2 are the new diameter and length.
  4. But, D_2=2D_1 and L_2=2L_1, substituting in the expression for Q_2: Q_2=k\frac{\pi (2D_1)^2}{4}\frac{T_1-T_2}{2L_1}.
  5. Rearranging: Q_2=\frac{2^2}{2}(k\frac{\pi D_1^2}{4}\frac{T_1-T_2}{L_1}).
  6. In the last declaration of  Q_2, it could be noted that the expressión inside the parenthesis is actually  Q_1, then:  Q_2=\frac{2^2}{2}(25W)=50W.
  7. <u>It should be noted, that the temperatures in the hot and cold reservoirs never change.</u>
7 0
2 years ago
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