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natulia [17]
3 years ago
8

If two oscillators connected to speakers create frequencies of 594 hz and 600 hz, what is the beat frequency when you hear both

at the same time? hz
Physics
1 answer:
Annette [7]3 years ago
3 0
When we have two sound waves with frequencies f1 and f2 that interfere, the beat frequency is equal to the absolute value of the difference of the two frequencies, so in our problem:
f_{beat}=f_2-f_1 = 600 Hz-594 Hz = 6Hz
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A boy whirls a stone in a horizontal circle of radius 1.1 m and at height 2.1 m above ground level. The string breaks, and the s
DedPeter [7]

Answer:

212.8 m/s^{2}

Explanation:

Time taken by stone to cover horizontal distance

t=\sqrt{\frac {2h}{g}} where t is time, h is height of whirling the stone in horizontal circle, g is gravitational constant, Substituting h for 2.1 m and g for 9.81

t=\sqrt{\frac {2*2.1}{9.81}}= 0.654654 seconds

t=0.65 s

Velocity, v= distance/time

v=10/0.65= 15.27525 m/s

v=15.3 m/s

a=\frac {v^{2}}{r} where r is radius of circle, substituting r with 1.1m

a=\frac {15.3^{2}}{1.1}

a=212.8 m/s^{2}

Therefore, centripetal acceleration is 212.8 m/s^{2}

8 0
4 years ago
Amanda spent 2/5 of her time after school doing homework and ¼ of her remaining time riding her bike. If she rode her bike for 4
Doss [256]

Answer:120 min

Explanation:

Given

Amanda  spent \frac{2}{5} of her time after school doing Home work

And \frac{1}{4} of her remaining  time riding her bike

It is given that she rode her bike for 45 minutes in a week

Let t be the time after school

therefore Amanda spend \frac{2t}{5} in home work and  \frac{3t}{5} time is left

From remaining \frac{3t}{5} time she spends \frac{1}{4} time riding her bike

therefore \frac{3t}{5}\times \frac{1}{4}=45

thus t=300 min

therefore time  spent on home work is \frac{2}{5}\times 300=120 min

6 0
3 years ago
Name the principle in which rocket work.
REY [17]
The rocket engine works on the basic principle proposed by Newton which is Newton’s Third Law.
8 0
3 years ago
Is the force of friction the sum of all the microscopic contact forces between the bottom of the block and the surface?
kotegsom [21]

The force of friction is that force that tends to oppose the motion of the body with the surface in contact... are you clear with it?

6 0
3 years ago
Question: Self-test 3.12 Calculate the change in G for ice at -10°C, with density 917 kg mº, when the pressure is increased from
Akimi4 [234]

The change in the Gibb's free energy per mole (G) is 1.96 J.

The given parameters:

  • Density of the ice, ρ = 917 kg/m³
  • Initial pressure, P₁ = 1.0 bar
  • Final pressure, P₂ = 2.0 bar
  • Temperature, T = - 10 C
  • Mass of water = 18 g

The change in the Gibb's free energy per mole (G) is calculated as follows;

\Delta G = V(P_2-P_1) \\\\

where;

V is the volume of the ice

Density = \frac{Mass}{Volume} \\\\Volume = \frac{Mass}{Density} \\\\Volume = \frac{18 \times 10^{-3} \ kg}{917 \ m^3} \\\\Volume = 1.96 \times 10^{-5} \ m^3\\\\Volume = 1.96 \times 10^{-5} \ m^3 \times \frac{1000 \ L}{m^3} \\\\Volume = 0.0196 \ L

Change in pressure;

P_2 - P_1 = 2.0 \ bar \ - \ 1.0 \ bar = 1.0 \ bar = 0.987 \ atm

The change in the Gibb's free energy per mole (G);

\Delta G= V(P_2-P_1)\\\\\Delta G = 0.0196\ L \times 0.987\ atm \\\\\Delta G = 0.0193 \ L.atm\\\\1 \ L.atm = 101.325 \ J\\\\\Delta G =  0.0193 \ L.atm \times \frac{101.325 \ J}{1 \ L.atm} \\\\\Delta G = 1.96 \ J

Thus, the change in the Gibb's free energy per mole (G) is 1.96 J.

Learn more about Gibb's free energy here: brainly.com/question/10012881

3 0
3 years ago
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