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3 years ago

Need help with a physics question.

Physics

1 answer:

Verdich [7]3 years ago

8 0

QUESTION:

WHAT IS THE MAGNITUDE OF THE MAGNETIC FIELD AT RIGHT ANGLES TO THE PROTON'S PATH?

ANSWER:

=☑ 2.4 T

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A building is 512 m high. What must be the minimum water pressure in a pipe at ground level in order to get water out of a tap i

vivado [14]

Answer:

Pressure = 5 x 10⁶ Pa

Explanation:

Given:

Height of building = 512 m

Find:

Pressure

Computation:

P2 = P1+dgh

P2 = 1 + (1000)(9.8)(512)

P2 = 51.2 atm

Pressure = 5 x 10⁶ Pa

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3 years ago

An engineering team has come to the stage in the engineering design

SIZIF [17.4K]

Answer: D

Explanation: read the answers above and compare to your test

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3 years ago

A sealed tank containing seawater to a height of 12.8 m also contains air above the water at a gauge pressure of 2.90 atm . Wate

exis [7]

Answer:

The velocity of water at the bottom, $v_{b} = 28.63 m/s$

Given:

Height of water in the tank, h = 12.8 m

Gauge pressure of water, $P_{gauge} = 2.90 atm$

Solution:

Now,

Atmospheric pressue, $P_{atm} = 1 atm = 1.01\tiems 10^{5} Pa$

At the top, the absolute pressure, $P_{t} = P_{gauge} + P_{atm} = 2.90 + 1 = 3.90 atm = 3.94\times 10^{5} Pa$

Now, the pressure at the bottom will be equal to the atmopheric pressure, $P_{b} = 1 atm = 1.01\times 10^{5} Pa$

The velocity at the top, $v_{top} = 0 m/s$ , l;et the bottom velocity, be $v_{b}$ .

Now, by Bernoulli's eqn:

$P_{t} + \frac{1}{2}\rho v_{t}^{2} + \rho g h_{t} = P_{b} + \frac{1}{2}\rho v_{b}^{2} + \rho g h_{b}$

where

$h_{t} - h_{b} = 12.8 m$

Density of sea water, $\rho = 1030 kg/m^{3}$

$\sqrt{\frac{2(P_{t} - P_{b} + \rho g(h_{t} - h_{b}))}{\rho}} = v_{b}$

$\sqrt{\frac{2(3.94\times 10^{5} - 1.01\times 10^{5} + 1030\times 9.8\times 12.8}{1030}} = v_{b}$

$v_{b} = 28.63 m/s$

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3 years ago

The water line from the street to my house is 1 inch diameter and made of PVC (i.e. smooth). The line is roughly 450 ft long. Th

jekas [21]

Answer:

The right solution is "126 Psi".

Explanation:

The given values are:

P₁ = 130 psig

i.e.,

= $130\times 6.894$

= $896.22 \ Kpa$

or,

= $896.22\times 10^3 \ Pa$

Z₂ = 10ft

= 3.05 m

$\delta$ = 1000 kg/m³

According to the question,

Z₁ = 0

V₁ = V₂

As we know,

⇒ $\frac{P_1}{\delta_g} +\frac{V_1^2}{2g} +Z_1=\frac{P_2}{\delta_g} +\frac{V_2^2}{2g} +Z_2$

On substituting the values, we get

⇒ $\frac{P_1}{\delta_g} +0+0=\frac{P_2}{\delta_g} +0+Z_2$

⇒ $\frac{896.22\times 10^3}{1000\times 9.8} =\frac{P_2}{1000\times 9.8} +3.05$

⇒ $P_2=866330 \ P_a$

i.e.,

⇒ $=866330\times 0.000145$

⇒ $=126 \ Psi$

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2 years ago

N which device is chemical energy transformed into electrical energy?

densk [106]

The answer is batteries

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3 years ago

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