The current in the circuit is 5 A
Explanation:
The intensity of current is given by the equation:

where
I is the current
q is the amount of charge passing through a given point of the circuit in a time interval of t
For the cell in this problem, we have
q = 150 C is the charge
t = 30 s is the time interval
Substituting into the equation, we f ind

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Answer:
Intensity of the light (first polarizer) (I₁) = 425 W/m²
Intensity of the light (second polarizer) (I₂) = 75.905 W/m²
Explanation:
Given:
Unpolarized light of intensity (I₀) = 950 W/m²
θ = 65°
Find:
a. Intensity of the light (first polarizer)
b. Intensity of the light (second polarizer)
Computation:
a. Intensity of the light (first polarizer)
Intensity of the light (first polarizer) (I₁) = I₀ / 2
Intensity of the light (first polarizer) (I₁) = 950 / 2
Intensity of the light (first polarizer) (I₁) = 425 W/m²
b. Intensity of the light (second polarizer)
Intensity of the light (second polarizer) (I₂) = (I₁)cos²θ
Intensity of the light (second polarizer) (I₂) = (425)(0.1786)
Intensity of the light (second polarizer) (I₂) = 75.905 W/m²
From the given equation we can deduce what changes will occur if the frequency of the sound is doubled
V= f (λ)
Speed = frequency. Wavelength
When the frequency is doubled, speed will not change. Because speed depends on factors like temperature, air pressure, density of the gas. Since all these factors are unchanged thus speed will remain unchanged
Frequency is the number of waves produced per second. Frequency and wavelength are inversely proportional .Thus, if the frequency is doubled the wavelength would be halved.
Answer:0.58 m
Explanation:
The initial velocity of the ball is u = 2.0 m/s
The height of the table is, h = 1.0 m
The ball falls in vertical direction under acceleration due to gravity.
Time taken for ball to hit the floor:
h= ut + 0.5gt² ( from the equation of motion)
1.0 m=2.0 m/s × t+0.5 × 9.8 m/s²× t²
Solving this for t,
t = 0.29 s ( we have neglected the negative value of t)
In the same time, the ball would cover a horizontal distance of :
s = u t
⇒s = 2.0 m/s×0.29 s = 0.58 m
Thus, the landing spot is 0.58 m away from the table.
Gamma rays then x rays then UVA rays then visible light then IR then radio waves (from highest to lowest frequency).