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Minchanka [31]
3 years ago
11

Two interfering sound waves have frequencies of 223 Hz and 214 Hz. Determine the beat period (in s) of the composite wave. (Hint

: first determine the beat frequency.)
Physics
1 answer:
Nataliya [291]3 years ago
5 0

Answer:

The beat period of the composite wave is 0.11 s.

Explanation:

Given;

frequency of the first sound wave, F₁ = 223 Hz

frequency of the second sound wave, F₂ = 214 Hz

The beat frequency is given as;

Fb  = F₁ - F₂

Fb = 223 Hz - 214 Hz

Fb = 9 Hz

The beat period is given as;

T_b = \frac{1}{F_b} \\\\T_b = \frac{1}{9} \\\\T_b = 0.11 \ s

Therefore, the beat period of the composite wave is 0.11 s.

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What causes a decrease in an objects weight
Artist 52 [7]

The mass of an object can neither be change but the weight of the object can be changed.

<h3>What causes a decrease in an object weight?</h3>

The weight of the body or of an object can be changed if the body is placed farther away from the earth or placed in the planet which is far away from the earth's gravitational field, so  the force of gravity on the object will change. However the mass of the body or mass of the object will remains the same  regardless of whether the object is on Earth, in outer space, or on the Moon. By doing so the weight of the object or body will change but the mass remains the same.

So we can conclude that: The mass of an object can neither be change but the weight of the object can be changed.

Learn more about Weight here: brainly.com/question/86444

#SPJ1

4 0
1 year ago
If the center of mass passes outside the area of support of an object, what will happen to it?
defon

Answer:

If a vertical line extending down from an object's CG extends outside its area of support, the object will topple

Explanation:

We can understand better this situation using a diagram with the forces acting on it.

In the attached image we can see that when the gravity center is bouncing outside from the area of the pedestal, the object will be out of balance and will fall.

6 0
3 years ago
The common electrical wall receptacle voltage in North America is often referred to as 120 volts AC. One hundred twenty volts is
Vika [28.1K]

Answer:

120 volts is the root mean square (rms) average of the voltage as it varies with time.

Explanation:

A. The average voltage over many weeks of time (false)

Reason: Average AC voltage over one cycle is cycle (from one peak to other) is zero and so over many weeks of time  it is zero.

B. The peak voltage from an AC wall receptacle (false)

Reason: The peak voltage of an AC source in North America is zero.

C. The arithmetic mean of the voltage as it varies with time (false)

Reason: Arithmetic mean AC voltage over one cycle is cycle (from one peak to other) is zero and so over many weeks of time  it is zero.

D.  One-half the peak voltage (false)

Peak voltage =170 Volts

One-half the peak voltage = 85 volts

E. The root mean square (rms) average of the voltage as it varies with time (True)

Reason:

The peak voltage and root mean square voltage are related by:

V_{rms}=\frac{V_{p}}{\sqrt{2} }\\\\V_{rms}=\frac{170}\sqrt{2}V_{rms}\\\\V_{rms}=120 Volts

Average value of voltage over one cycle is zero, so instead of calculating average voltage for AC peak voltage is first squared and the mean is calculated.

5 0
3 years ago
If you were to mix 10 grams of salt into 100 grams of water, how much will the total mass of the solution be?
Ede4ka [16]

110 grams

mass of salt + mass of water

10 + 100

110g

4 0
2 years ago
A clock battery wears out after moving 10,400 C of charge through the clock at a rate of 0.530 mA. (a) How long (in s) did the c
Sati [7]

Answer:

Explanation:

Given that:

Charge Q = 10400 C

current I = 0.530 mA

From our previous knowledge, we know that charge Q can be expressed as:

Q = I×t

where

I = current and t = time

∴

replacing our values

10400 = (0.530×10⁻³) t

t = 10400/(0.530×10⁻³)

t = 19622641.51 seconds

However, the required flow of electrons per seconds can be calculated by using the formula:

number of electrons/second = current (I)/ charge on electron

= (0.530×10⁻³)/(1.6×10⁻¹⁹)

= 3.3125×10¹⁵ electrons

7 0
3 years ago
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