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2 years ago

Consider this statement: Air is matter. Which facts best support the statement?

Physics

2 answers:

Zanzabum2 years ago

6 0

A. and C. support that assertion.

yawa3891 [41]2 years ago

4 0

C. air has mass

all matter has mass

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If a person looking at a poster sees green instead of yellow and doesn't see red at all, this person most likely has color blind

katrin [286]

Red - sensitive so answer is c

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3 years ago

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Two solenoids A and B, spaced close to each other and sharing the same cylindrical axis, have 430 and 610 turns, respectively. A

KIM [24]

Answer

given,

Two solenoids A and B

Number of turn

Na = 430 turns Nb = 610 turns

Current = 2.80 A

Average flux through A = 300 μWb

Average of flux through B = 90.0 μ Wb

a) $L = \dfrac{N \phi}{I}$

$L = \dfrac{610\times 90 \times 10^{-6}}{2.80}$

$L =19.6 mH$

b) inductance of A

$L = \dfrac{N_A \phi_A}{I_A}$

$L = \dfrac{430\times 300 \times 10^{-6}}{2.80}$

$L =46 mH$

c) magnitude of the emf

$\epsilon_B = -L_B\dfrac{dI}{dT}$

$\epsilon_B = -(19.6\times 10^{-3})(0.5)$

$\epsilon_B = -9.8\times 10^{-3}\ V$

$\epsilon_B = -9.8\ mV$

7 0

3 years ago

Ipaliwanag ang paggalang

cestrela7 [59]

paraan ito ng pagrerespeto

5 0

2 years ago

The gravitational force of a star on an orbiting planet 1 is f1. planet 2, which is three times as massive as planet 1 and orbit

Margaret [11]

Let us consider two bodies having masses m and m' respectively.

Let they are separated by a distance of r from each other.

As per the Newtons law of gravitation ,the gravitational force between two bodies is given as - $F = G\frac{mm'}{r^{2} }$ where G is the gravitational force constant.

From the above we see that F ∝ mm' and $F\alpha \frac{1}{r^{2} }$

Let the orbital radius of planet A is $r_{1}$ = r and mass of planet is $m_{1}$ .

Let the mass of central star is m .

Hence the gravitational force for planet A is $f_{1} =G \frac{m_{1}*m }{r^{2} }$

For planet B the orbital radius $r_{2} =2r_{1}$ and mass $m_{2} = 3 m_{1}$

Hence the gravitational force $f_{2} =G\frac{m m_{2} }{r^{2} }$

$f_{2} =G\frac{m*3m_{1} }{[2r_{1}] ^{2} }$

$= \frac{3}{4} G\frac{mm_{1} }{r_{1} ^{2} }$

Hence the ratio is $\frac{f_{2} }{f_{1} } = \frac{\frac{3}{4}G mm_{1/r_{1} ^2} }{Gmm_{1}/r_{1} ^2 }$

$=\frac{3}{4}$ [ ans]

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3 years ago

Read 2 more answers

One of the checks that you could do for problem 1) would be to check the output resistance of the Wheatstone bridge to make sure

Ray Of Light [21]

Answer:

Explanation:

Let the four resistances of th wheat stone bridge is

P, Q, R and S and the value of each is 350 ohm.

Here, P and Q are in series.

R' = P + Q = 350 + 350 = 700 ohm

Then R and S are in series

R' = R + S = 350 + 350 = 700 ohm

Now R' and R'' are in parallel.

So, the equivalent resistance is

Req = R' x R'' / ( R' + R'')

Req = 700 / 2 = 350 ohm

Thus, the reading of ohmmeter is 350 ohm.

6 0

3 years ago