Batteries can supply a steady flow of electrons.<br> True<br> False

Having established that a sound wave corresponds to pressure fluctuations in the medium, what can you conclude about the directi

DaniilM [7]

Answer:

the pressure fluctuation is LONGITUDINAL

Explanation:

Sound waves are an oscillating movement of air particles, this can be analyzed in two different, but equivalent ways, as an air oscillation and with a pressure wave due to these oscillations.

The expression for the wave is

ΔP = Δo sin (kx - wt)

Therefore, the pressure variation is in the same direction as the displacement variation, consequently the pressure fluctuation is LONGITUDINAL

4 0

4 years ago

Use the impulse-momentum theorem to find how long a stone falling straight down takes to increase its speed from 5.6 m/s to 10.6

Nezavi [6.7K]

As stated, the impulse and momentum definitions will be used to later find the value of time through the force of gravity. According to the theory, the impulse formula is given as,

$I = F\Delta t$

Here,

F = Force

$\Delta t =$ Change in time

Now using the impulse theorem we have that,

Change in Impulse = Change in momentum

$F\Delta t = \Delta p$ (1)

The change in momentum is given as

$\Delta p = p_f -p_i$

$\Delta p = (-10.6)-(-5.6)$

$\Delta p = -5m/s$

The force due to gravity is through the Newton's second law

$F_g = -mg$

Here,

m = mass

g = Acceleration due to gravity

Substitute the value in (1)

$-mgt = -5m/s$

$9.8t = 4.9$

$t = \frac{5}{9.8}$

$t = 0.51s$

Therefore it will take 0.51s.

7 0

3 years ago

The fastest possible rate of rotation of a planet is that for which the gravitational force on material at the equator just bare

Brums [2.3K]

Answer:

6862.96871 seconds

Explanation:

M = Mass of Planet

G = Gravitational constant

r = Radius

$\rho$ = Density

T = Rotation period

In this system the gravitational force will balance the centripetal force

$G\frac{Mm}{r^2}=mr\omega^2$

$\omega=\frac{2\pi}{T}$ .

$M=\rho v\\\Rightarrow M=\rho \frac{4}{3}\pi r^3$

$\\\Rightarrow G\frac{Mm}{r^2}=mr\left(\frac{2\pi}{T}\right)^2\\\Rightarrow \frac{G\rho \frac{4}{3}\pi r^3}{r^3}=\frac{4\pi^2}{T^2}\\\Rightarrow T=\sqrt{\frac{3\pi}{G\rho}}$

Hence, proved

$T=\sqrt{\frac{3\pi}{6.67\times 10^{-11}\times 3000}}\\\Rightarrow T=6862.96871\ s$

The rotation period of the astronomical object is 6862.96871 seconds

7 0

3 years ago

Charge is placed on the surface of a 2.7-cm radius isolated conducting sphere. the surface charge density is uniform and has the

soldi70 [24.7K]

First, convert the radius of the sphere from centimeter to meter.
r = (2.7 cm)(1 m/ 100 cm)
r = 0.027 m

Then, calculate for the surface area of the given by the equation,
A = 4πr²

Substituting the known values,
A = 4(π)(0.027 m)²
A = 9.16 x 10⁻³ m²

Then, multiply the calculated area to the charge density given to determine the total charge.
C = (6.9 x 10⁻⁶ C/m²)(9.16 x 10⁻³ m²)
C = 6.32 x10⁻⁸ C

<em>Answer: C = 6.32 x 10⁻⁸ C</em>

5 0

3 years ago

What parameters affects the inductance of a coil

Kazeer [188]

Answer:

Number of Wire Turns in the Coil.

Explanation:

The greater the number of turns of wire in the coil, the greater the inductance. Fewer turns of wire in the coil results in lesser inductance. More coils of wires indicate a greater amount of magnetic field force for a given amount of coil current.

4 0

2 years ago

Batteries can supply a steady flow of electrons. True False