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3 years ago

Which two quantities can be expressed using the same units?

Physics

1 answer:

lara [203]3 years ago

4 0

I think it’s 3. Work and energy
I’m so sorry if I’m wrong

Hope you found this helpful !

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The pressure of a gas changes from 120 kPa. The volume changes from 45 L to 40 L. If the initial temperature is 81 c, what is th

irina [24]

Explanation:

we are not given the pressure change, check yhe question please

7 0

3 years ago

Two sound waves have equal displacement amplitudes, but wave 1 has two-thirds the frequency of wave 2. What is the ratio of the

zlopas [31]

Answer:

$\dfrac{I_1}{I_2}=\dfrac{4}{9}$

Explanation:

c = Speed of wave

$\rho$ = Density of medium

A = Area

$\nu$ = Frequency

$\nu_1=\dfrac{2}{3}\nu_2$

Intensity of sound is given by

$I=\dfrac{1}{2}\rho c(A\omega)^2\\\Rightarrow I=\dfrac{1}{2}\rho c(A2\pi \nu)^2$

So,

$I\propto \nu^2$

We get

$\dfrac{I_1}{I_2}=\dfrac{\nu_1^2}{\nu_2^2}\\\Rightarrow \dfrac{I_1}{I_2}=\dfrac{\dfrac{2}{3}^2\nu_2^2}{\nu_2^2}\\\Rightarrow \dfrac{I_1}{I_2}=\dfrac{4}{9}$

The ratio is $\dfrac{I_1}{I_2}=\dfrac{4}{9}$

8 0

3 years ago

Why is Pluto no longer a planet?

Step2247 [10]

The International Astronomical Union (IAU) downgraded the status of Pluto to that of a dwarf planet because it did not meet the three criteria the IAU uses to define a full-sized planet. Essentially Pluto meets all the criteria except one—it “has not cleared its neighboring region of other objects.

5 0

3 years ago

Read 2 more answers

If the speed of sound in air at 20 °C is 342m/s , what will be the increase in speed at 30 °C? with steps

andrew-mc [135]

Answer:

Hmm

Explanation:

I don't know sorry forgive me.

6 0

3 years ago

An electron moving parallel to a uniform electric field increases its speed from 2.0 × 10^7 m/s to 4.0 × 10^7 m/s over a distanc

julia-pushkina [17]

Answer:

$E = 2.84 * 10^5 N/C$

Explanation:

The speed increased from 2.0 * 10^7 m/s to 4.0 * 10^7 m/s over a 1.2 cm distance.

Let us find the acceleration:

$v^2 = u^2 + 2as$

$(4.0 * 10^7)^2 = (2.0 * 10^7)^2 + 2 * a * 0.012\\\\(4.0 * 10^7)^2 - (2.0 * 10^7)^2 = 0.024a\\\\1.2 * 10^{15}= 0.024a\\\\a = 1.2 * 10^{15} / 0.024\\\\a = 5 * 10^{16} m/s^2$

Electric force is given as the product of charge and electric field strength:

F = qE

where q = electric charge

E = Electric field strength

Force is generally given as:

F = ma

where m = mass

a = acceleration

Equating both:

ma = qE

E = ma / q

For an electron:

m = 9.11 × 10^{-31} kg

q = 1.602 × 10^{-19} C

Therefore, the electric field strength of the electron is:

$E = \frac{9.11 * 10^{-31} * 5 * 10^{16}}{1.602 * 10^{-19}} \\\\E = 2.84 * 10^5 N/C$

8 0

3 years ago