Are S waves transvers? (Need answer fast, hurry)

A ball is rolling along at speed v without slipping on a horizontal surface when it comes to a hill that rises at a constant ang

MissTica

Answer:

The ball has an initial linear kinetic energy and initial rotational kinetic energy which can both be converted into gravitational potential energy. Therefore the hill with friction will let the ball reach higher.

Explanation:

The ball has an initial linear kinetic energy and initial rotational kinetic energy which can both be converted into gravitational potential energy. Therefore the hill with friction will let the ball reach higher.

This is because:

If we consider the ball initially at rest on a frictionless surface and a force is exerted through the centre of mass of the ball, it will slide across the surface with no rotation, and thus, there will only be translational motion.

Now, if there is friction and force is again applied to the stationary ball, the frictional force will act in the opposite direction to the force but at the edge of the ball that rests on the ground. This friction generates a torque on the ball which starts the rotation.

Therefore, static friction is infact necessary for a ball to begin rolling.

Now, from the top of the ball, it will move at a speed 2v, while the centre of mass of the ball will move at a speed v and lastly, the bottom edge of the ball will instantaneously be at rest. So as the edge touching the ground is stationary, it experiences no friction.

So friction is necessary for a ball to start rolling but once the rolling condition has been met the ball experiences no friction.

6 0

3 years ago

) Music. When a person sings, his or her vocal cords vibrate in a repetitive pattern that has the same frequency as the note tha

vaieri [72.5K]

(a) 0.0021 s, 2926.5 rad/s

The frequency of the B note is

$f= 466 Hz$

The time taken to make one complete cycle is equal to the period of the wave, which is the reciprocal of the frequency:

$T=\frac{1}{f}=\frac{1}{466 Hz}=0.0021 s$

The angular frequency instead is given by

$\omega = 2\pi f$

And substituting

f = 466 Hz

We find

$\omega = 2\pi (466 Hz)=2926.5 rad/s$

(b) 20 Hz, 125.6 rad/s

In this case, the period of the sound wave is

T = 50.0 ms = 0.050 s

So the frequency is equal to the reciprocal of the period:

$f=\frac{1}{T}=\frac{1}{0.050 s}=20 Hz$

While the angular frequency is given by:

$\omega = 2\pi f = 2 \pi (20 Hz)=125.6 rad/s$

(c) $4.30\cdot 10^{14} Hz, 7.48\cdot 1^{14} Hz, 2.33\cdot 10^{-15} s, 1.34\cdot 10^{-15}s$

The minimum angular frequency of the light wave is

$\omega_1 = 2.7\cdot 10^{15}rad/s$

so the corresponding frequency is

$f=\frac{\omega}{2 \pi}=\frac{2.7\cdot 10^{15}rad/s}{2\pi}=4.30\cdot 10^{14} Hz$

and the period is the reciprocal of the frequency:

$T=\frac{1}{f}=\frac{1}{4.30\cdot 10^{14}Hz}=2.33\cdot 10^{-15}s$

The maximum angular frequency of the light wave is

$\omega_2 = 4.7\cdot 10^{15}rad/s$

so the corresponding frequency is

$f=\frac{\omega}{2 \pi}=\frac{4.7\cdot 10^{15}rad/s}{2\pi}=7.48\cdot 10^{14} Hz$

and the period is the reciprocal of the frequency:

$T=\frac{1}{f}=\frac{1}{7.48\cdot 10^{14}Hz}=1.34\cdot 10^{-15}s$

(d) $2.0\cdot 10^{-7}s, 3.14\cdot 10^{7} rad/s$

In this case, the frequency is

$f=5.0 MHz = 5.0 \cdot 10^6 Hz$

So the period in this case is

$T=\frac{1}{f}=\frac{1}{5.0\cdot 10^6 Hz}=2.0 \cdot 10^{-7} s$

While the angular frequency is given by

$\omega = 2\pi f=2 \pi (5.0\cdot 10^{6}Hz)=3.14\cdot 10^{7} rad/s$

7 0

3 years ago

A bicycle tire has a pressure of 7.00×105 N/m2 at a temperature of 18.0ºC and contains 2.00 L of gas. What will its pressure be

stepan [7]

Answer:

$p_2 = 664081 N/m^{2}$

Explanation:

from the ideal gas law we have

PV = mRT

$P = \rho RT$

$\rho = \frac{P}{RT}$

HERE R is gas constant for dry air = 287 J K^{-1} kg^{-1}

$\rho = \frac{7.00 10^{5}}{287(18+273)}$

$\rho = 8.38 kg/m^{3}$

We know by ideal gas law

$\rho = \frac{m_1}{V_1}$

$m_1 = \rho V_1 = 8.38 *2*10^{-3}$

$m_1 = 0.0167 kg$

for m_2

$m_2 = \rho V_i - V_removed$

$m_2 = 8.38*(.002 - 10^{-4})$

$m_2 = 0.0159 kg$

WE KNOW

PV = mRT

V, R and T are constant therefore we have

P is directly proportional to mass

$\frac{p_2}{p_1}=\frac{m_2}{m_1}$

$p_2 = p_1 * \frac{m_2}{m_1}$

$p_2 =7*10^{5} * \frac {.0159}{0.0167}$

$p_2 = 664081 N/m^{2}$

8 0

2 years ago

What type of energy conversion results when friction slows down a ball rolling across the floor

ale4655 [162]

Friction between the ball and the floor is stealing some of the kinetic energy of the ball, and turning it into heat.

8 0

2 years ago

A laser used to dazzle the audience at a rock concert emits blue light with a wavelength of 463 nm . calculate the frequency of

ZanzabumX [31]

The wavelength of light is given as 463 nm or can also be written as 463 x 10^-9 m. [wavelength = ʎ]

We know that the speed of light is 299 792 458 m / s or approximately 3 x 10^8 m / s. [speed of light = c]

Given the two values, we can calculate for the frequence (f) using the formula:

f = c / ʎ

Substituting the given values:

f = (3 x 10^8 m / s) / 463 x 10^-9 m

f = 6.48 x 10^14 / s = 6.48 x 10^14 s^-1

5 0

3 years ago