The one that does that is the photoelectric effect. This allows the sun to hit the surface and produces enough energy for electrons to be knocked off the atom and allows a current to move.
The receiver of a signal must understand the code or language being used to avoid confusion and losses.
<h3>What is a Signal?</h3>
This is usually in the form of a sound or body movement and is involved in conveying messages to people.
The receiver must understand the code or language in order to prevent confusion or loss of lives and properties.
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Nitrogen oxides play a critical role in photochemical smog. They give the smog its yellowish-brown hue. Indoor residential appliances like gas stoves and gas or wood heaters can be significant emitters of nitrogen oxides in poorly ventilated environments.
- Nitrogen dioxide (NO₂), ozone (O₃), peroxyacetyl nitrate (PAN), and chemical compounds with the -CHO group are the main harmful elements of photochemical smog (aldehydes). If present in high enough amounts, PAN and aldehydes can harm plants and irritate the eyes.
- The greatest sources of emissions are power plants, heavy construction equipment driven by diesel, other moveable engines, and industrial boilers. Cars, trucks, and buses are next in line.
Therefore , on conclusion i.e. two gases with molecules consisting of nitrogen and oxygen atoms are nitric oxide (NO) and nitrogen dioxide (NO₂). These nitrogen oxides play a part in the development of smog and acid rain, adding to the issue of air pollution.
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Given Information:
Length of wire = 132 cm = 1.32 m
Magnetic field = B = 1 T
Current = 2.2 A
Required Information:
(a) Torque = τ = ?
(b) Number of turns = N = ?
Answer:
(a) Torque = 0.305 N.m
(b) Number of turns = 1
Explanation:
(a) The current carrying circular loop of wire will experience a torque given by
τ = NIABsin(θ) eq. 1
Where N is the number of turns, I is the current in circular loop, A is the area of circular loop, B is the magnetic field and θ is angle between B and circular loop.
We know that area of circular loop is given by
A = πr²
where radius can be written as
r = L/2πN
So the area becomes
A = π(L/2πN)²
A = πL²/4π²N²
A = L²/4πN²
Substitute A into eq. 1
τ = NI(L²/4πN²)Bsin(θ)
τ = IL²Bsin(θ)/4πN
The maximum toque occurs when θ is 90°
τ = IL²Bsin(90)/4πN
τ = IL²B/4πN
torque will be maximum for N = 1
τ = (2.2*1.32²*1)/4π*1
τ = 0.305 N.m
(b) The required number of turns for maximum torque is
N = IL²B/4πτ
N = 2.2*1.32²*1)/4π*0.305
N = 1 turn
The tangential velocity of the car's tire is the product of the angular velocity and radius of the car's tire which is 11(r) m/s.
<h3>
Angular velocity of the tire</h3>
The angular velocity of the tire is the rate of change of angular displacement of the tire with time.
The magnitude of the angular velocity of the tire is calculated as follows;
ω = 2πN
where;
- N is the number of revolutions per second
ω = 2π x (5.25 / 3)
ω = 11 rad/s
<h3>Tangential velocity of the tire</h3>
The tangential velocity of the car's tire is the product of the angular velocity and radius of the car's tire.
The magnitude of the tangential velocity is caculated as follows;
v = ωr
where;
- r is the radius of the car's tire
v = 11r m/s
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