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chubhunter [2.5K]
3 years ago
14

The force exerted on a bridge pier in a river is to be tested in a 1:10 scale model using water as the working fluid. In the pro

totype the depth of water is 2.0 m, the velocity of flow is 1.5 m/s and the width of the river is 20 m. If the hydrodynamic force on the model bridge pier is 5 N, what would it be on the prototype? (hint: pressure ratio is equal to the length-scale ratio) (5 points; Ans: 5000 N)
Physics
1 answer:
kkurt [141]3 years ago
5 0

Answer:

f_p = 5000 N

Explanation:

GIVEN DATA:

pressure ratio = length ratio

force = 5 N

scale = 1:10

velocity = 1.5 m/s

B_p = 20 m

h_p = 2m

As pressure ratio = length ratio so we have

\frac{p_m}{p_p} =\frac{l_m}{l_p} =\frac{1}{10}

\frac{f_m *A_m}{f_p *A_p} ==\frac{1}{10}

\frac{f_m}{f_p} * \frac{A_p}{A_m} = \frac{1}{10}

\frac{f_m}{f_p} * \frac{b_p*h_p}{b_m*h_m} =\frac{1}{10}

5 * \frac{1}{\frac{1}{10}} *\frac{1}{\frac{1}{10}} = \frac{f_p}{10}

solving F_p

f_p = 5000 N

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IgorLugansk [536]

Answer:

Explanation:

M = 1.989 x 10^30 kg

R = 6.96 x 10^8 m

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Let the velocity is v.

v=\sqrt{\frac{2GM}{R}}

v=\sqrt{\frac{2\times 6.67\times 10^{-11}\times 1.989\times 10^{30}}{6.96\times 10^{8}}}

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3 years ago
Neon signs require about 12,000 V for their operation. Consider a neon-sign transformer that operates off 120- V lines. How many
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I can't give you the actual number of turns, because it's the RATIO
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8 0
3 years ago
A coil is wrapped with 300 turns of wire on the perimeter of a circular frame (radius = 8.0 cm). Each turn has the same area, eq
TiliK225 [7]

Answer:

Approximately 18 volts when the magnetic field strength increases from \rm 20\; mT to \rm 80\;mT at a constant rate.

Explanation:

By the Faraday's Law of Induction, the EMF \epsilon that a changing magnetic flux induces in a coil is:

\displaystyle \epsilon = N \cdot \frac{d\phi}{dt},

where

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  • \displaystyle \frac{d\phi}{dt} is the rate of change in magnetic flux through this coil.

However, for a coil the magnetic flux \phi is equal to

\phi = B \cdot A\cdot \cos{\theta},

where

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  • A\cdot \cos{\theta} is the area of the coil perpendicular to the magnetic field.

For this coil, the magnetic field is perpendicular to coil, so \theta = 0 and A\cdot \cos{\theta} = A. The area of this circular coil is equal to \pi\cdot r^{2} = \pi\times 8.0\times 10^{-2}\approx \rm 0.0201062\; m^{2}.

A\cdot \cos{\theta} = A doesn't change, so the rate of change in the magnetic flux \phi through the coil depends only on the rate of change in the magnetic field strength B. The size of the magnetic field at the instant that B = \rm 50\; mT will not matter as long as the rate of change in B is constant.

\displaystyle \begin{aligned} \frac{d\phi}{dt} &= \frac{\Delta B}{\Delta t}\times A \\&= \rm \frac{80\times 10^{-3}\; T- 20\times 10^{-3}\; T}{20\times 10^{-3}\; s}\times 0.0201062\;m^{2}\\&= \rm 0.0603186\; T\cdot m^{2}\cdot s^{-1}\end{aligned}.

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\displaystyle \epsilon = N \cdot \frac{d\phi}{dt} = \rm 300 \times 0.0603186\; T\cdot m^{2}\cdot s^{-1} \approx 18\; V.

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3 years ago
On his way to deliver presents Santa has a minor accident. If the sleigh (1200 kg) was traveling at 322 m/s and the jet(4800 kg)
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608.4m/s

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We have to find the final velocity of the two objects after the collision.

The collision is inelastic .

By using law of conservation of momentum

Mu+mu'=(m+M)v

Using the formula

1200\times 322+4800\times 680=6000v

6000v=3650400

v=\frac{3650400}{6000}

v=608.4m/s

Hence, the final  velocity of two objects after the collision=608.4m/s

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