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3 years ago

"what is the period of one vibration of this tone?"

1 answer:

7 0

The full question is:
On a keyboard, you strike middle C, whose frequency is 256 Hz. What is the period of one vibration of this tone?
The period of a vibration is the time it takes for the particle to make one full oscillation. Frequency is by definition number of full oscillations per unit of time.
When the frequency is expressed in Hz that unit of time is one second.
So there is the following relation between frequency and period:
$T=\frac{1}{f}$
When we plug in the numbers we get:
$T=\frac{1}{256}=0.0039$s$

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Batman chased the joker on his batbike for 20 minutes, traveling at the speed of 30 kilometers per minute. How far did Batman go

AURORKA [14]

Answer:

123 kilometer

Explanation:

none

8 0

3 years ago

HELP PLEASE 20 POINTS SHOW WORK, ALL EQUATIONS

nataly862011 [7]

Answer:

s = 3 m

Explanation:

Let t be the time the accelerating car starts.

Let's assume the vehicles are point masses so that "passing" takes no time.

the position of the constant velocity and accelerating vehicles are

s = vt = 40(t + 2) cm

s = ½at² = ½(20)(t)² cm

they pass when their distance is the same

½(20)(t)² = 40(t + 2)

10t² = 40t + 80

0 = 10t² - 40t - 80

0 = t² - 4t - 8

t = (4±√(4² - 4(1)(-8))) / 2(1)

t = (4± 6.928) / 2 ignore the negative time as it has not occurred yet.

t = 5.464 s

s = 40(5.464 + 2) = 298.564 cm

300 cm when rounded to the single significant digit of the question numerals.

7 0

3 years ago

Where do you think the government should put the greatest support: solar energy, wind energy, clean coal, oil exploration, gas e

True [87]

The government should put it support in a combination of sources, as no source in the present can fully provide all energy requirements.

7 0

3 years ago

Read 2 more answers

A 594 Ω resistor, an uncharged 1.3 μF capacitor, and a 6.53 V emf are connected in series. What is the current in milliamps afte

ivanzaharov [21]

Answer:

6.88 mA

Explanation:

Given:

Resistance, R = 594 Ω

Capacitance = 1.3 μF

emf, V = 6.53 V

Time, t = 1 time constant

Now,

The initial current, I₀ = $\frac{\textup{V}}{\textup{R}}$

I₀ = $\frac{\textup{6.53}}{\textup{594}}$

I₀ = 0.0109 A

also,

I = $I_0[1-e^{-\frac{t}{\tau}}]$

here,

τ = time constant

e = 2.717

on substituting the respective values, we get

I = $0.0109[1-e^{-\frac{\tau}{\tau}}]$

I = $0.0109[1-2.717^{-1}]$

I = 0.00688 A

I = 6.88 mA

5 0

3 years ago

If a 1000-pound capsule weighs only 165 pounds on the moon, how much work is done in propelling this capsule out of the moon's g

Bas_tet [7]

Answer:

W = 1,307 10⁶ J

Explanation:

Work is the product of force by distance, in this case it is the force of gravitational attraction between the moon (M) and the capsule (m₁)

F = G m₁ M / r²

W = ∫ F. dr

W = G m₁ M ∫ dr / r²

we integrate

W = G m₁ M (-1 / r)

We evaluate between the limits, lower r = R_ Moon and r = ∞

W = -G m₁ M (1 /∞ - 1 / R_moon)

W = G m1 M / r_moon

Body weight is

W = mg

m = W / g

The mass is constant, so we can find it with the initial data

For the capsule

m = 1000/32 = 165 / g_moon

g_moom = 165 32/1000

.g_moon = 5.28 ft / s²

I think it is easier to follow the exercise in SI system

W_capsule = 1000 pound (1 kg / 2.20 pounds)

W_capsule = 454 N

W = m_capsule g

m_capsule = W / g

m = 454 /9.8

m_capsule = 46,327 kg

Let's calculate

W = 6.67 10⁻¹¹ 46,327 7.36 10²² / 1.74 10⁶

W = 1,307 10⁶ J

7 0

3 years ago