All categories

2 years ago

What is the rate of vibration measured in

Physics

2 answers:

jek_recluse [69]2 years ago

7 0

Answer:

I'm not 100% sure tbh but the only thing I think makes sense to represent vibration would be frequency which is measure in Hertz (Hz)

Explanation:

Reptile [31]2 years ago

4 0

The rate of vibration is measured in hertz (Hz)

You might be interested in

n alpha particle (q = +2e, m = 4.00 u) travels in a circular path of radius 5.94 cm in a uniform magnetic field with B = 1.10 T.

9966 [12]

Answer:

a). V = 3.13*10⁶ m/s

b). T = 1.19*10^-7s

c). K.E = 2.04*10⁵

d). V = 1.02*10⁵V

Explanation:

q = +2e

M = 4.0u

r = 5.94cm = 0.0594m

B = 1.10T

1u = 1.67 * 10^-27kg

M = 4.0 * 1.67*10^-27 = 6.68*10^-27kg

a). Centripetal force = magnetic force

Mv / r = qB

V = qBr / m

V = [(2 * 1.60*10^-19) * 1.10 * 0.0594] / 6.68*10^-27

V = 2.09088 * 10^-20 / 6.68 * 10^-27

V = 3.13*10⁶ m/s

b). Period of revolution.

T = 2Πr / v

T = (2*π*0.0594) / 3.13*10⁶

T = 1.19*10⁻⁷s

c). kinetic energy = ½mv²

K.E = ½ * 6.68*10^-27 * (3.13*10⁶)²

K.E = 3.27*10^-14J

1ev = 1.60*10^-19J

xeV = 3.27*10^-14J

X = 2.04*10⁵eV

K.E = 2.04*10⁵eV

d). K.E = qV

V = K / q

V = 2.04*10⁵ / (2eV).....2e-

V = 1.02*10⁵V

7 0

2 years ago

If a calorie is equivalent to 4.184 joules how many joules are contained in 250 kg calories slice of pizza

Andreyy89

In order to answer this, we will set up a simple ratio as such:

1 calorie = 4.184 joules

1 kilocalorie = 1000 calories

1 kilocalorie = 4,184 joules

250 kilocalories = x joules

Cross multiplying the second and third equations, we get:

x joules = 4,184 * 250

250 kilocalories are equivalent to 1,046 kJ

6 0

2 years ago

Because the earth's orbit is slightly elliptical, the earth actually gets closer to the sun during part of the year. When the ea

BabaBlast [244]

Closer to the sun . . . orbital speed is faster.

Farther from the sun . . . orbital speed is slower.

Flag answer: Answer 13 Answer 13

5 0

2 years ago

Two 10-cm-diameter charged rings face each other, 21.0 cm apart. Both rings are charged to +40.0 nC. What is the electric field

Katyanochek1 [597]

Complete question:

Two 10-cm-diameter charged rings face each other, 21.0 cm apart. Both rings are charged to +40.0 nC. What is the electric field strength at the midpoint between the two rings ?

Answer:

The electric field strength at the mid-point between the two rings is zero.

Explanation:

Given;

diameter of each ring, d = 10 cm = 0.1 m

distance between the rings, r = 21.0 cm = 0.21 m

charge of each ring, q = 40 nC = 40 x 10⁻⁹ C

let the midpoint between the two rings = x

The electric field strength at the midpoint between the two rings is given as;

$E_{mid} = E_{right} +E_{left}\\\\E_{right} = \frac{KQ}{(x^2 + r^2)^\frac{2}{3} } \\\\E_{leftt} = -\ \frac{KQ}{(x^2 + r^2)^\frac{2}{3} }\\\\E_{mid} = \frac{KQ}{(x^2 + r^2)^\frac{2}{3} } - \frac{KQ}{(x^2 + r^2)^\frac{2}{3} } = 0$

Therefore, the electric field strength at the mid-point between the two rings is zero.

7 0

2 years ago

What is the displacement current in the capacitor if the potential difference across the capacitor is increasing at 500,000V/s?

konstantin123 [22]

Answer:

$I=2.71\times 10^{-5}\ A$

Explanation:

A 6.0-cm-diameter parallel-plate capacitor has a 0.46 mm gap.

What is the displacement current in the capacitor if the potential difference across the capacitor is increasing at 500,000V/s?

Let given is,

The diameter of a parallel plate capacitor is 6 cm or 0.06 m

Separation between plates, d = 0.046 mm

The potential difference across the capacitor is increasing at 500,000 V/s

We need to find the displacement current in the capacitor. Capacitance for parallel plate capacitor is given by :

$C=\dfrac{A\epsilon_o}{d}\\\\C=\dfrac{\pi r^2\epsilon_o}{d}$ , r is radius

Let I is the displacement current. It is given by :

$I=C\dfrac{dV}{dt}$

Here, $\dfrac{dV}{dt}$ is rate of increasing potential difference

$I=\dfrac{\pi r^2\epsilon_o}{d}\times \dfrac{dV}{dt}\\\\I=\dfrac{\pi (0.03)^2\times 8.85\times 10^{-12}}{0.46\times 10^{-3}}\times 500000\\\\I=2.71\times 10^{-5}\ A$

So, the value of displacement current is $2.71\times 10^{-5}\ A$ .

4 0

2 years ago