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2 years ago

Which of the following is an inorganic compound?

Physics

2 answers:

DiKsa [7]2 years ago

8 0

Answer:

b. aluminium, hydrogen and oxygen

Explanation:

Organic compounds are generally formed by the combination of these elements: Carbon, Hydrogen, Oxygen, Nitrogen and Sulphur. This discards options a and d.

As special case, organobromides are organic compounds that contain carbon bonded to bromine. This discards option c.

Then the correct option is b. An example of an inorganic compound formed with aluminium, hydrogen and oxygen is aluminium hydroxide, Al(OH)3, which is found in nature as the mineral gibbsite.

Lyrx [107]2 years ago

7 0

Answer:

It is b

Explanation: aluminum is a man made object or metal

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The magnitude of the gravitational field strength near Earth's surface is represented by

Zanzabum

Answer:

The magnitude of the gravitational field strength near Earth's surface is represented by approximately $9.82\,\frac{m}{s^{2}}$ .

Explanation:

Let be M and m the masses of the planet and a person standing on the surface of the planet, so that M >> m. The attraction force between the planet and the person is represented by the Newton's Law of Gravitation:

$F = G\cdot \frac{M\cdot m}{r^{2}}$

Where:

$M$ - Mass of the planet Earth, measured in kilograms.

$m$ - Mass of the person, measured in kilograms.

$r$ - Radius of the Earth, measured in meters.

$G$ - Gravitational constant, measured in $\frac{m^{3}}{kg\cdot s^{2}}$ .

But also, the magnitude of the gravitational field is given by the definition of weight, that is:

$F = m \cdot g$

Where:

$m$ - Mass of the person, measured in kilograms.

$g$ - Gravity constant, measured in meters per square second.

After comparing this expression with the first one, the following equivalence is found:

$g = \frac{G\cdot M}{r^{2}}$

Given that $G = 6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}}$ , $M = 5.972 \times 10^{24}\,kg$ and $r = 6.371 \times 10^{6}\,m$ , the magnitude of the gravitational field near Earth's surface is:

$g = \frac{\left(6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}} \right)\cdot (5.972\times 10^{24}\,kg)}{(6.371\times 10^{6}\,m)^{2}}$

$g \approx 9.82\,\frac{m}{s^{2}}$

The magnitude of the gravitational field strength near Earth's surface is represented by approximately $9.82\,\frac{m}{s^{2}}$ .

4 0

2 years ago

The voltage across the secondary winding is 10,000 V, and the voltage across the primary winding is 25,000 V. If the primary win

Allisa [31]

Answer

80 coils

Explanation

The turn rule of a transformer says that, the amount of voltage induced in the secondary coil is proportional to the number of coil. It is simplified by the ratio:

Np/Ns = Vp/Vs

200/Ns = 25,000/10,000

200 × 10,000 = 25,000Ns

2,000,000 = 25000Ns

Dividing both sides by 25,000,

Ns = 2,000,000/25,000

= 80 coils

7 0

3 years ago

Read 2 more answers

You want the current amplitude through a inductor with an inductance of 4.70 mH (part of the circuitry for a radio receiver) to

goldenfox [79]

Answer:

f = 1.69*10^5 Hz

Explanation:

In order to calculate the frequency of the sinusoidal voltage, you use the following formula:

$V_L=\omega iL=2\pi f i L$ (1)

V_L: voltage = 12.0V

i: current = 2.40mA = 2.40*10^-3 A

L: inductance = 4.70mH = 4.70*10^-3 H

f: frequency = ?

you solve the equation (1) for f and replace the values of the other parameters:

$f=\frac{V_L}{2\pi iL}=\frac{12.0V}{2\pi (2.4*10^{-3}A)(4.70*10^{-3}H)}=1.69*10^5Hz$

The frequency of the sinusoidal voltage is f

3 0

3 years ago

Help me please i dont understand

solniwko [45]

Answer:

It is b

Explanation:

The question tells you that the distance was increased 3 times the original distance, which means it was moved 3 times

5 0

2 years ago

An ultrasound unit is being used to measure a patient's heartbeat by combining the emitted 2.0 MHz signal with the sound waves r

alexandr402 [8]

Hi there,
for this question we have:
Signal 2.0 MHz = Emitted so we can call it $f_e$
and we need the Reflected = $f_{r}$
In this question, we have a source which goes to the heart and a reflected which comes back from the heart and we need the speed of the reflected.
So you should know that the speed of reflected is lower than the source(Emitted).
we also know: ΔBeat frequency(max) = 560 Hz = $f_{b}$
so we have:
$f_{e}$ - $f_{r}$ = $f_{b}$
so frequency of Reflected is:
2.0 × 10^6 Hz - 560 Hz = 1.99 × 10^6 Hz = $f_{r}$
now you know that Lambda = v/f
so if we find the lambda with our Emitted then we can find v with the Reflected:
Lambda = 1540(m/s) / 2.0 × 10^6 Hz = 7.7 × 10^-4 m
=> $v_{max}$ = (lambda)( $f_{r}$
=> 7.7 × 10^-4m (1.99 × 10^6Hz) = 1532 m/s
so the $v_{max}$ is equal to 1532 m/s :)))
This question is solved by two top teachers as fast as they could :))
I hope this is helpful
have a nice day

8 0

3 years ago