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butalik [34]
3 years ago
7

The focused, penetrating sound of the koto, which is similar to the tone of Japanese vocal music, comes from the_______________

a. Tension of the stringsc. Relaxation of strings d. Sound wave overlapping d. None of the above
Physics
1 answer:
Arte-miy333 [17]3 years ago
5 0

Answer:

a. Tension of the strings

Explanation:

The koto is a pluckedstring instrument. In these instruments, the sound is obtained causing the vibration in several strings. Therefore, its sound depends on the thickness, extension and tension of the strings. The lower the tension the deeper the sound will be.

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A point P is placed exactly between two charges, Q1 and Q2. If the electric field experienced by point P due to charge Q1 is 1.5
iren2701 [21]

<em>Answer:  D</em>

<em>Given data:</em>

The electric field due to charge q₁  is (E₁) = 1.5 × 10⁵ N/C

The electric field due to charge q₂ is (E₂) = 7.2 × 10⁵ N/C

Determine the net electric field at point P (Enet) = ?

electric field measured in Newtons/coulomb

<em>We know that,</em>

        The net electric field (Enet) is vector sum of E₁ and E₂ caused by the charge q₁ and q₂ respectively

             Enet = E₁ + E₂

                       = (1.5 × 10⁵)+ (7.2 × 10⁵)

                       = 8.7 × 10⁵ Newtons/coulomb

<em>The net electric field at point P is  8.7 × 10⁵ N/C</em>

4 0
4 years ago
A meter stick balances horizontally on a knife-edge at the 50.0 cm mark. With two 5.12 g coins stacked over the 24.2 cm mark, th
pashok25 [27]

Answer:

The mass of the meter stick is  1.66054054054g

Explanation:

Here since the meter stick is in equilibrium position its net torque and net force should be equal to zero

      Since at the begging the meter stick is balanced at center of the meter stick that means its center of mass should be present at 50.0cm

Now lets consider the later case where stick is balanced by two 5.12 g coins .

Here torque due to two coins = t_{c} = 5.12\times(27.8-24.2)\times2\times9.8

  Torque due to weight of meter stick =  t_{m} =  m\times(50-27.8)\times9.8

  where m = mass of the meter stick

    Here t_{c} = t_{m}.

Upon equating we will be getting mass of the meter stick =1.66054054054g

4 0
3 years ago
A form letter can be customized by using different fields in a __________.
Juli2301 [7.4K]
The answer is C. Email
8 0
3 years ago
Read 2 more answers
A 1.2kg stone is tied to a string and swung in a vertical circle with a radius of 0.75m. The string can withstand a tension of 4
san4es73 [151]

Hello!

We can begin by summing the forces acting on the stone when it is at the bottom of its trajectory.

Refer to the free-body diagram in the image below for clarification.

We have the force of tension (produced by the string) and the force of gravity acting in opposite directions, so:
\Sigma F = T - F_g

The net force is equivalent to the centripetal force experienced by the stone. Recall the equation for centripetal force for uniform circular motion:
F_c = \frac{mv^2}{r}

m = mass of object (1.2 kg)
v = velocity of object (? m/s)
r = radius of circle (0.75 m)

The centripetal force is the resultant of the forces of tension and gravity, and points upward (same direction as the tension force) since the tension force is greater.

Therefore:
\frac{mv^2}{r} = T - Mg

We can solve the equation for 'v':

mv^2 = r(T - Mg) \\\\v^2 = \frac{r(T - Mg)}{m}\\\\v = \sqrt{\frac{r(T - Mg)}{m}}

Plug in values and solve.

v = \sqrt{\frac{(0.75)(40 - 1.2(9.8))}{1.2}} = \boxed{4.201 \frac{m}{s}}

3 0
2 years ago
To what potential should you charge a 2.0 μF capacitor to store 1.0 J of energy?
Bess [88]
E = (1/2)CV²
1 = (1/2)*(2*10⁻⁶)V²
10⁶ = V²
1000 = V

You should charge it to 1000 volts to store 1.0 J of energy.
6 0
3 years ago
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