Answer:
Conditions that result in the emission of electrons from a conductor:
Heating the conductor to a suitable temperature
Exposing the conductor to a strong light
Subjecting the conductor to a very high applied voltage
Subjecting the conductor to high-speed electrons from another source
Both diodes and triodes influence current flow and contain electrodes. Diodes involve only two active pieces; triodes contain three electrodes within a tube. Current can flow through a diode in only one direction. Diodes tend to act as rectifiers. In a triode, the third electrode is a grid located between the cathode and the anode. A small difference of potential between the grid and the cathode controls the number of electrons that reach the anode. Adjusting the charge on the grid affects the number of electrons that can reach the anode. Triodes tend to act as amplifiers.
Transmitting antenna
Modulator
Oscillator
Microphone
RF amplifier
Receiving antenna
Loudspeaker
Demodulator
Tuner
hue, saturation, intensity/brightness
red, blue, green
Explanation:
Penn Foster
Ways to increase friction
- increase the area of contact
- increase the roughness of the contact materials
- dry up or take away any lubricant
- increase the pressure on the contact
Answer:
v = 2.974
Explanation:
Perhaps the formula should be
v = √(2*g*d (sin(θ) - uk*cos(θ) ) This is a bit easier to read.
v = √(2* 9.80*0.725(0.707 - 0.12*0.707) ) Substitute values. Find 2*g*d
v = √14.21 * (0.707 - 0.0849) Figure out Sin(θ) - uk cos(θ)
v = √14.21 * (0.6222)
v = √8.8422 Take the square root of the value
v = 2.974
If an icy surface means no friction, then Newton's second law tells us the net forces on either block are
• <em>m</em> = 1 kg:
∑ <em>F</em> (parallel) = <em>mg</em> sin(45°) - <em>T</em> = <em>ma</em> … … … [1]
∑ <em>F</em> (perpendicular) = <em>n</em> - <em>mg</em> cos(45°) = 0
Notice that we're taking down-the-slope to be positive direction parallel to the surface.
• <em>m</em> = 0.4 kg:
∑ <em>F</em> (vertical) = <em>T</em> - <em>mg</em> = <em>ma</em> … … … [2]
<em />
Adding equations [1] and [2] eliminates <em>T</em>, so that
((1 kg) <em>g</em> sin(45°) - <em>T </em>) + (<em>T</em> - (0.4 kg) <em>g</em>) = (1 kg + 0.4 kg) <em>a</em>
(1 kg) <em>g</em> sin(45°) - (0.4 kg) <em>g</em> = (1.4 kg) <em>a</em>
==> <em>a</em> ≈ 2.15 m/s²
The fact that <em>a</em> is positive indicates that the 1-kg block is moving down the slope. We already found the acceleration is <em>a</em> ≈ 2.15 m/s², which means the net force on the block would be ∑ <em>F</em> = <em>ma</em> ≈ (1 kg) (2.15 m/s²) = 2.15 N directed down the slope.
-- It accelerates.
-- Its speed increases.
-- It gains momentum.
-- It loses altitude.
-- It loses potential energy.
-- It gains kinetic energy.
-- Its wheels make a lot of noise.
-- Everybody screams.
