All categories

3 years ago

What universal standard is the basis of the atomic mass unit

Physics

2 answers:

ki77a [65]3 years ago

4 0

<span>The mass of Avogadro's number of Carbon-12 atoms, which exactly equals 12.000</span>

MakcuM [25]3 years ago

3 0

Answer:

Explanation:

AMU stands for atomic mass unit. It is used to define the mass of an atom.

1 amu is defined as the mas of one twelfth of the mass of one carbon - 12 atom.

1 amu = mass of 1/12 of one carbon 12 atom

It is said to be that 1 amu is the average mass of one proton and one neutron.

You might be interested in

Find the energy in Joules required to lift a 55.0 Megagram object a distance of 500 cm.

fredd [130]

Energy to lift something =

(mass of the object) x (gravity) x (height of the lift).

BUT ...

This simple formula only works if you use the right units.

Mass . . . kilograms
Gravity . . . meters/second²
Height . . . meters

For this question . . .

Mass = 55 megagram = 5.5 x 10⁷ grams = 5.5 x 10⁴ kilograms

Gravity (on Earth) = 9.8 m/second²

Height = 500 cm = 5.0 meters

So we have ...

Energy = (5.5 x 10⁴ kilogram) x (9.8 m/s²) x (5 m)

= 2,696,925 joules .

That's quite a large amount of energy ... equivalent to
straining at the rate of 1 horsepower for almost exactly an
hour, or burning a 100 watt light bulb for about 7-1/2 hours.

The reason is the large mass that's being lifted.
On Earth, that much mass weighs about 61 tons.

7 0

2 years ago

Draw a schematic diagram of a circuit consisting of a battery of five 2 V cells, a 5 ohm resistor, a 10 ohm resistor, a 15 ohm r

MakcuM [25]

Answer:

Current = 0.33 A

Explanation:

For diagram refer the attachment.

It is given that five cells of 2V are connected in series, so total voltage of the battery:

$\dashrightarrow \: \: \sf V = 2 \times 5 = 10 V$

Three resistor of 5 $\Omega$ , 10 $\Omega$ , 15 $\Omega$ are connected in Series, so the net resistance:

$\dashrightarrow \: \: \sf R_{n} = R_{1} + R_{2} + R_{3}$

$\dashrightarrow \: \: \sf R = 5 + 10 + 15$

${ \pink{\dashrightarrow \sf \: \: { \underbrace{R = 30 \: \Omega}}}}$

According to ohm's law:

$\dashrightarrow \sf\: \: V = IR$

$\dashrightarrow \sf \: \: I = \dfrac{V}{R}$

On substituting resultant voltage (V) as 10 V and resultant resistant, as 30 ${\pmb{\sf{\Omega}}}$ we get:

$\dashrightarrow \sf \: \: I = \dfrac{10V}{30\Omega}$

${ \pink{\dashrightarrow \sf \: \: { \underbrace{I = 0.33 A}}}}$

$\therefore$ The electric current passing through the above circuit when the key is closed will be <u>0.33 A</u>

4 0

2 years ago

Read 2 more answers

A parallel-plate capacitor stores charge Q. The capacitor is then disconnected from its voltage source, and the space between th

Stells [14]

Answer:

The relationship between the initial stored energy $PE_{i}$ and the stored energy after the dielectric is inserted $PE_{f}$ is:

c) $PE_{f} =0.5\ PE_{i}$

Explanation:

A parallel plate capacitor with $C_{o}$ that is connected to a voltage source $V_{o}$ holds a charge of $Q_{o} =C_{o} V_{o}$ . Then we disconnect the voltage source and keep the charge $Q_{o}$ constant . If we insert a dielectric of $\kappa=2$ between the plates while we keep the charge constant, we found that the potential decreases as:

$V=\frac{V_{o}}{\kappa}$

The capacitance is modified as:

$C=\frac{Q}{V} =\kappa\frac{Q_{o}}{V_{o}}=\kappa\ C_{o}$

The stored energy without the dielectric is

$PE_{i}=\frac{1}{2}\frac{Q_{o}^{2}}{C_{o}}=\frac{1}{2}C_{o}V_{o}^{2}$

The stored energy after the dielectric is inserted is:

$PE_{f}=\frac{1}{2}\frac{Q^{2}}{C}=\frac{1}{2}CV^{2}$

If we replace in the above equation the values of V and C we get that

$PE_{f}=\frac{1}{2}\kappa\ C_{o}(\frac{V_{o}}{\kappa})^{2}=\frac{1}{\kappa}(\frac{1}{2}C_{o}V_{o}^{2})$

$PE_{f} =\frac{PE_{i}}{\kappa}$

Finally

$PE_{f} =0.5\ PE_{i}$

5 0

3 years ago

At an amusement park there is a ride in which cylindrically shaped chambers spin around a central axis. People sit in seats faci

Klio2033 [76]

Answer:

The radius of a chamber is 2.36 meters.

Explanation:

Given that,

The outer wall moves at a speed of 2.72 m/s.

Mass of the person, m = 75.1 kg

The person feels a force of 235 N force pressing against his back. The force acting on the person is centripetal force. It is given by the below formula :

$F=\dfrac{mv^2}{r}$