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3 years ago

Two tuning forks produce sounds of wavelengths of 3.4 meters and 3.3 meters.

Physics

1 answer:

KengaRu [80]3 years ago

5 0

Answer:

Explanation:

wavelength, λ = 3.4 m

wavelength, λ' = 3.3 m

Speed, v = 340 m/s

f = v / λ = 340 / 3.4 = 100 Hz

f' = v / λ' = 340 / 3.3 = 103.03 Hz

Frequency of beat, n = f' - f = 103.03 - 100 = 3.03 Hz

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How far will a runner travel if she had an average speed of 10 km/hour and runs for 2.1 hours?

Rasek [7]

10 Km.
S= Speed
D= distance
T= time

S= d/t
but since you are solving for "d" the equation is d=st so you plug in 10 km/h for speed and 2.1 hours for time and just multiply them. The hours cancel out so you are left with 10km.

8 0

2 years ago

Read 2 more answers

The potential difference between the plates of a capacitor is 145 V. Midway between the plates, a proton and an electron are rel

aniked [119]

Answer:

= 2.52 x 10^ 6 m/s

Explanation:

The force that acts on charged particles between capacitor plates =

F = (q) (Δv) ÷ d

Here, d = distance between the two plates

q = charge of the charged particle

Δv = voltage

Normally, the force that makes both proton and electron released from rest, giving the charge acceleration is F=m X a. where m= mass and a = acceleration

Poting this equation with the first one, we have:

m X a = (q) (Δv) ÷ d

So, the acceleration of a proton when moving towards a negatively charged plate is

a = (q) (Δv) ÷ (d) (m) {proton}

Likewise, the acceleration of an electron when moving towards a positively charged plate is

a = (q) (Δv) ÷ (d) (m) {electron}

Dividing the proton acceleration formula by the electron acceleration formula we have:

a (proton) / a (electron) = m (proton) / m(electron)

inserting equation of motion to get distance, s

s = ut + 1/2 at^2

recall that electron travel distance, d/2

d/2 = 1/2 at^2

making t the subject of the formula

we have, t =√(d ÷ a(electron))

The distance of proton:

d/2 = ut + 1/2 at^2 [proton}

put d/2 = ut + 1/2 at^2 [proton} into t =√(d ÷ a(electron))

Initial speed, ui = √(d ÷ a(electron)) = (d/2) - (1/2) x (d) (a(proton) + a(electron))

since acceleration wasn't given in the question, lets use mass(elect

ron) ÷ mass(proton) rather than use (a(proton) + a(electron))

Therefore, intial speed= 1/2√((e X Δv) ÷ m(electron)) (1- m(electron)/ m(proton))

Note, e = 1.60 x 10^-19

m(electron) = 9.11 X 10^-31

m(proton) = 1.67 X 10^-27

Input these values into the formula above, initial speed, UI =

= 2.52 x 10^ 6 m/s

7 0

3 years ago

what percentage of North American adults may be functioning below their potential due to prolonged exposure to stress

Yuliya22 [10]

I believe the percentage is between 15-20%. Stress is a well known factor that affects the performance of people.

6 0

3 years ago

You observe that a mass suspended by a spring takes 0.25 s to make a full oscillation. What is the frequency of this oscillation

Katarina [22]

Answer:

Frequency of oscillation, f = 4 Hz

time period, T = 0.25 s

Angular frequency, $\omega = 25.13 rad/s$

Given:

Time taken to make one oscillation, T = 0.25 s

Solution:

Frequency, f of oscillation is given as the reciprocal of time taken for one oscillation and is given by:

f = $\frac{1}{T}$

f = $\frac{1}{0.25}$

Frequency of oscillation, f = 4 Hz

The period of oscillation can be defined as the time taken by the suspended mass for completion of one oscillation.

Therefore, time period, T = 0.25 s

Angular frequency of oscillation is given by:

$\omega = 2\pi \times f$

$\omega = 2\pi \times 4$

$\omega = 25.13 rad/s$

5 0

3 years ago

A large asteroid of mass 98700 kg is at rest far away from any planets or stars. A much smaller asteroid, of mass 780 kg, is in

Citrus2011 [14]

Answer:

1.81 x 10^-4 m/s

Explanation:

M = 98700 kg

m = 780 kg

d = 201 m

Let the speed of second asteroid is v.

The gravitational force between the two asteroids is balanced by the centripetal force on the second asteroid.

$\frac{GMm}{d^{2}}=\frac{mv^2}{d}$

$v=\sqrt{\frac{GM}{d}}$

Where, G be the universal gravitational constant.

G = 6.67 x 10^-11 Nm^2/kg^2

$v=\sqrt{\frac{6.67 \times 10^{-11}\times 98700}{201}}$

v = 1.81 x 10^-4 m/s

7 0

3 years ago