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Korolek [52]
3 years ago
8

What is the relationship between the current passes through the

Physics
1 answer:
yaroslaw [1]3 years ago
8 0
Increasing the number of bulbs in a series circuit decreases the brightness of the bulbs. In a series circuit, the voltage is equally distributed among all of the bulbs. Bulbs in parallel are brighter than bulbs in series. In a parallel circuit the voltage for each bulb is the same as the voltage in the circuit.
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the average time required for radioactive elements, whereby they decay to half their original mass is called the "average life"
VikaD [51]
I'm not sure what your question is. But, the half life is the amount of time required for half the material to decay. For U238 this is 4.5 billion years, whilst for Fr-223 (Francium) its about 22 minutes. To calculate the time for something to decay you need to use the equation:
Mass (after time t) = Mass (initial) * (0.5)^(time/half life)
Hope this helps
4 0
3 years ago
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Why does air blows from one place to another​
Maslowich

Air blows from one place to another because gases move from high-pressure areas to low-pressure areas

In simple words

it happens because of pressure differences.

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2 years ago
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Teams red and blue are having a tug-of-war. According to Newton's third law, the force with which the red team pulls on the blue
Zielflug [23.3K]
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4 years ago
I NEED a LOT of help PLEASE!
Minchanka [31]
Infrared, visible light, then ultraviolet. Infrared is light that the human eye can not see and visible light is clearly light we can see then ultraviolet is has such a high frequency we can't see it either.
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3 years ago
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A square coil ℓ = 2cm on a side with 30 turns rotates in a uniform magnetic field, B~ = B0zˆ = 0.1Tˆz, such that the normal of t
kow [346]

Answer:

a) 1.2*10^{-3}cos(1.25t)

b) 0.49mV

Explanation:

a) The coil rotates periodically with period T. Hence, we can write the variation of the magnetic flux with a sinusoidal function, and with max flux NAB. Thus, we have that:

\Phi_B(t)=NABcos(\omega t)\\\\\omega=\frac{2\pi}{T}=1.25\frac{rad}{s}\\\\A=l^2=(0.02m)^2=4*10^{-4}m^2\\\\B=0.1T\\\\\Phi_B(t)=1.2*10^{-3}cos(1.25 t) W

where we have used the values given by the information of the problem for N B and A.

b)

the emf is given by:

emf=-\frac{d\Phi_B}{dt}=-NBA\omega sin(\omega t)\\\\emf(t=12.5s)=-(30)(0.1T)(4*10^{-4})(1.25\frac{rad}{s})sin(1.25*12.5)=1.49*10^{-4}V=0.49mV

hope this helps!!

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3 years ago
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