Transition

What would a drummer do to make the sound of a drum give a note of lower pitch?

stiks02 [169]

A drummer would loosen the drum skin in order to make the sound of a drum give a note of lower pitch.
However, I am not entirely sure.

5 0

3 years ago

Read 2 more answers

The circuit shown in the figure (Figure 1) uses a neon-filled tube. This neon lamp has a threshold voltage V0 for conduction, be

Svetllana [295]

Part 1)

Answer:

Explanation:

As we know by equation of charging of the capacitor we will have

$V = E(1 - e^{-t/RC})$

so we will have

$87 = 105(1 - e^{-t/RC})$

here we know that

$R = 3.00 \times 10^6 ohm$

$C = 0.250 \mu F$

so we have

$t = 1.32 s$

Part b)

Answer:

The time will increase.

Explanation:

As we know that on increasing the value of the resistance the the product of the resistance and capacitance will increase so the time will increase to get the above voltage.

Part c)

Answer:

The capacitor discharges through a very low resistance (the lamp filled with ionized gas), and so the discharge time constant is very short. Thus the flash is very brief.

Explanation:

Since the lamp resistance is very small so the energy across the lamp will totally lost in very short interval of time

Part d)

Answer:

Once the lamp has flashed, the stored energy in the capacitor is gone, and there is no source of charge to maintain the lamp current. The lamp "goes out", the lamp resistance increases, and the capacitor starts to recharge. It charges again and the process will repeat.

Explanation:

Since we know that the battery is connected to the given system so after whole energy of capacitor is flashed out it is again charged by the battery and the process will continue

3 0

3 years ago

Free charges do not remain stationary when close together. To illustrate this, calculate the magnitude of the instantaneous acce

ASHA 777 [7]

Answer:

a=2.304×10¹⁶m/s²

Explanation:

Given data

Distance d=2.5 nm=2,5×10⁻⁹m

Mass of proton m=1.6×10⁻²⁷kg

charge of proton q=1.6×10⁻¹⁹C

To find

acceleration a

Solution

Apply the Coulombs Law

$F=k\frac{q_{1}q_{2} }{r^{2} }$

Where k is coulombs constant (k=9×10⁹Nm²/C²)

q=q₁=q₂

r=d

So

$F=k\frac{|q^{2} |}{d^{2} }\\ as \\F=ma\\ma=k\frac{|q^{2} |}{d^{2} }\\a=\frac{k}{m} \frac{|q^{2} |}{d^{2} }\\a=\frac{(9*10^{9} )*(1.6*10^{-19} )^{2} }{(1.6*10^{-27} )*(2.5*10^{-9} )^{2} }\\ a=2.304*10^{16}m/s^{2}$

4 0

3 years ago

Which observational tool helped astronomers Arno Penzias and Robert Wilson discover the existence of the cosmic microwave backgr

Softa [21]

<span>In 1964, while experimenting with the Holmdel Horn Antenna, which was used as a radio telescope, Penzias and Wilson accidentally discovered the microwave background radiation that exists universally. The Holmdel Horn Antenna was used to support the "Big Bang Theory" as opposed to the "Steady State Theory".</span>

4 0

2 years ago

Hi im a little stuck on this question that came from my textbook in class it got a little confusing for me because i really dont

koban [17]

Consult the attached free body diagram.

By Newton's second law, the net force on the crate acting parallel to the surface is

∑ F[para] = (370 N) cos(-20°) - f = 0

(this is the x-component of the resultant force)

where

• (370 N) cos(-20°) = magnitude of the horizontal component of the pushing force

• f = magnitude of kinetic friction

The crate is moving at a constant speed and thus not accelerating, so the crate is in equilibrium.

Solve for f :

f = (370 N) cos(-20°) ≈ 347.686 N

The net force acting perpendicular to the surface is

∑ F[perp] = n - 1480 N - (370 N) sin(-20°) = 0

(this is the y-component of the resultant force)

where

• n = magnitude of normal force

• 1480 N = weight of the crate

• (370 N) sin(-20°) = magnitude of the vertical component of push

The crate doesn't move up or down, so it's also in equilibrium in this direction.

Solve for n :

n = 1480 N + (370 N) sin(-20°) ≈ 1606.55 N ≈ 1610 N

Then the coefficient of kinetic friction is µ such that

f = µn ⇒ µ = f/n ≈ 0.216

7 0

2 years ago