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3 years ago

Help Me Please!!!!!!!

Physics

2 answers:

iren2701 [21]3 years ago

5 0

B. Conductors allow electricity to go through them only. Such as copper etc.

CaHeK987 [17]3 years ago

3 0

Answer:

C. Semiconductors and conductors

Explanation:

this is because conductors let electrons flow through them and semiconductors only in some conditions

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1) A fan is to accelerate quiescent air to a velocity of 8 m/s at a rate of 9 m3/s. Determine the minimum power that must be sup

azamat

Answer:

$\dot{W} = 339.84 W$

Explanation:

given data:

flow Q = 9 m^{3}/s

velocity = 8 m/s

density of air = 1.18 kg/m^{3}

minimum power required to supplied to the fan is equal to the POWER POTENTIAL of the kinetic energy and it is given as

$\dot{W} =\dot{m}\frac{V^{2}}{2}$

here $\dot{m}$ is mass flow rate and given as

$\dot{m} = \rho*Q$

$\dot{W} =\rho*Q\frac{V^{2}}{2}$

Putting all value to get minimum power

$\dot{W} =1.18*9*\frac{8^{2}}{2}$

$\dot{W} = 339.84 W$

7 0

3 years ago

Vf=V0+at solve for a

dangina [55]

Answer:

See below

Explanation:

vf = vo + at subtract vo from both sides

vf - vo = at now divide both sides by t

(vf-vo) / t = a

6 0

1 year ago

According to the universal law of gravitation the force of attraction between two objects depends on the masses of the objects a

barxatty [35]

The distance between them
F = (GMm)/r^2

3 0

3 years ago

The only way to slow down a moving object is to apply a net force to it.

salantis [7]

Hey there,

Answer:

A, True

Hope this helps :D

~Top☺

4 0

3 years ago

Replacing an object attached to a spring with an object having 14 the original mass will change the frequency of oscillation of

Pachacha [2.7K]

Answer:

The frequency changes by a factor of 0.27.

Explanation:

The frequency of an object with mass m attached to a spring is given as

$f$ = $\frac{1}{2\pi } \sqrt{\frac{k}{m} }$

where $f$ is the frequency

k is the spring constant of the spring

m is the mass of the substance on the spring.

If the mass of the system is increased by 14 means the new frequency becomes

$f_{n}$ = $\frac{1}{2\pi } \sqrt{\frac{k}{14m} }$

simplifying, we have

$f_{n}$ = $\frac{1}{2\pi \sqrt{14} } \sqrt{\frac{k}{m} }$

$f_{n}$ = $\frac{1}{3.742*2\pi } \sqrt{\frac{k}{m} }$

if we divide this final frequency by the original frequency, we'll have

==> $\frac{1}{3.742*2\pi } \sqrt{\frac{k}{m} }$ ÷ $\frac{1}{2\pi } \sqrt{\frac{k}{m} }$

==> $\frac{1}{3.742*2\pi } \sqrt{\frac{k}{m} }$ x $2\pi \sqrt{\frac{m}{k} }$

==> 1/3.742 = 0.27

7 0

3 years ago

Help Me Please!!!!!!!​

Help Me Please!!!!!!!