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Dominik [7]
2 years ago
9

A 1/40 scale model is to be used in a towing tank to study the water motion near the bottom of a shallow channel as a large barg

e passes over . Assume that the model is operated in accordance with the Froude number criteria for dynamic similitude. The prototype barge moves at a typical speed of 13 knots.
(a) At what speed (in ft/s) should the model be towed?
(b) Near the bottom of the model channel a small particle is found to move 0.12 ft in one second so that the fluid velocity at that point is approximately 0.12 ft/s. Determine the velocity at the corresponding point in the prototype channel.
Physics
1 answer:
nadezda [96]2 years ago
5 0

Answer:

I t hi ink the answer is a

Explanation:

I hoped this helpes

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We can model a pine tree in the forest as having a compact canopy at the top of a relatively bare trunk. Wind blowing on the top
notka56 [123]

Answer:

A,)FD= 114.1N

B)Torque=798.5Nm

Explanation:

We can model a pine tree in the forest as having a compact canopy at the top of a relatively bare trunk. Wind blowing on the top of the tree exerts a horizontal force, and thus a torque that can topple the tree if there is no opposing torque. Suppose a tree's canopy presents an area of 9.0 m^2 to the wind centered at a height of 7.0 m above the ground. (These are reasonable values for forest trees.)

If the wind blows at 6.5 m/s, what is the magnitude of the drag force of the wind on the canopy? Assume a drag coefficient of 0.50 and the density of air of 1.2 kg/m^3

B)What torque does this force exert on the tree, measured about the point where the trunk meets the ground?

A)The equation of Drag force equation can be expressed below,

FD =[ CD × A × ρ × (v^2/ 2)]

Where CD= Drag coefficient for cone-shape = 0.5

ρ = Density

Area of of the tree canopy = 9.0 m^2

density of air of = 1.2 kg/m^3

V= wind velocity= 6.5 m/s,

If we substitute those values to the equation, we have;

FD =[ CD × A × ρ × (v^2/ 2)]

F= [ 0.5 × 9.0 m^2 × 1.2 kg/m^3 ( 6.5 m/s/ 2)]

FD= 114.1N

B) the torque can be calculated using below formula below

Torque= (Force × distance)

= 114.1 × 7

= 798.5Nm

8 0
3 years ago
A particular 12 V car battery can send a total charge of 110 A·h (ampere-hours) through a circuit, from one terminal to the othe
DiKsa [7]
<h2>Answer:</h2>

(a) 3.96 x 10⁵C

(b) 4.752 x 10⁶ J

<h2>Explanation:</h2>

(a) The given charge (Q) is 110 A·h (ampere hour)

Converting this to A·s (ampere second) gives the number of coulombs the charge represents. This is done as follows;

=> Q = 110A·h

=> Q = 110 x 1A x 1h          [1 hour = 3600 seconds]

=> Q = 110 x A x 3600s

=> Q = 396000A·s

=> Q = 3.96 x 10⁵A·s = 3.96 x 10⁵C

Therefore, the number of coulombs of charge is 3.96 x 10⁵C

(b) The energy (E) involved in the process is given by;

E = Q x V           -----------------(i)

Where;

Q = magnitude of the charge = 3.96 x 10⁵C

V = electric potential = 12V

Substitute these values into equation (i) as follows;

E = 3.96 x 10⁵ x 12

E = 47.52 x 10⁵ J

E = 4.752 x 10⁶ J

Therefore, the amount of energy involved is 4.752 x 10⁶ J

8 0
3 years ago
It takes 52,000 Joules to heat a cup of coffee to boiling from room temperature. How long a piece of 20 cm wide Aluminum foil wo
vovikov84 [41]

Answer:

L = 1.11 x 10^{6} m, is the length of piece of 20 cm wide Aluminum foil to make capacitor large enough to hold 52000 J of energy.

Explanation:

Solution:

Data Given:

Heat Energy = 52000 J

Dielectric Constant of the plastic Bag = 3.7 = K

Thickness = 2.6 x 10^{5} m =d

V = 610 volts

A = width x Length

width = 20 cm = 20 x 10^{-2} m

Length = ?

So,

we know that,

U = 1/2 C Δv^{2}

U = 52000 J

C = ?

V = 610 volts'

So,

U = 1/2 C Δv^{2}  

52000 J = (0.5) x (C) x (610^{2})

C = 0.28 F

And we also know that,

C = \frac{K*E*A}{d}

E = 8.85 x 10 ^{-12}

K = 3.7

A = 0.20 x L

d = 2.6 x 10^{5} m

Plugging in the values into the formula, we get:

0.28 = \frac{3.7 * 8.85 .10^{-12} * (0.20 . L) }{2.6 . 10^{5} }

Solving for L, we get:

L = 1.11 x 10^{6} m,

is the length of piece of 20 cm wide Aluminum foil to make capacitor large enough to hold 52000 J of energy.  

7 0
3 years ago
A bug splats against the windshield of a car traveling at high speeds down a backcountry road. Which statement correctly compare
zvonat [6]

Answer:

C. The bug's change in momentum is equal to the car's change in momentum.

Explanation:

As we know by Newton's 2nd law

F = \frac{\Delta P}{\Delta t}

here we have also know that when car hits the bug then force applied by wind shield on the bug is same as the force applied by the bug on the car's wind shield as per Newton's III law

F_{12} = F_{21}

so we know that

\frac{\Delta P_{12}}{\Delta t} = \frac{\Delta P_{21}}{\Delta t}

so we have

\Delta P_{12} = \Delta P_{21}

so correct answer will be

C. The bug's change in momentum is equal to the car's change in momentum.

6 0
3 years ago
A resistor with R = 300 Ω and an inductor are connected in series across an ac source that has voltage amplitude 500 V. The rate
schepotkina [342]

Answer:

Explanation:

impedance z=(XL^2+R^2)^1/2

power across te resistor ==i^2r

286/300

I=.976

3 0
3 years ago
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