Answer
given,
frequency from Police car= 1240 Hz
frequency of sound after return = 1275 Hz
Calculating the speed of the car = ?
Using Doppler's effect formula
Frequency received by the other car
..........(1)
u is the speed of sound = 340 m/s
v is the speed of the car
Frequency of the police car received
now, inserting the value of equation (1)
1.02822(340 - v) = 340 + v
2.02822 v = 340 x 0.028822
2.02822 v = 9.799
v = 4.83 m/s
hence, the speed of the car is equal to v = 4.83 m/s
Miniature circuit breakers is called the developed form of fuse because MCBs are more sensitive to current than fuses. They immediately detect any abnormality and switch off the electrical circuit automatically. This prevents any permanent damage to electrical appliances and human beings
The electric field at the surface of the cylinder is 51428V/m
Given data:
• The length of the charge is l= 7m.
• The charge is q = 2μC..
• The radius the cylinder is r = 10 cm
Since the filament length is so large as compared to the cylinder length that the infinite line of charge can be assumed.
The expression to calculate the electric field is given as,
E=2kλ/r
Here, λ is the linear charge density.
Substitute the values in the above equation,
E = (2×9×109N⋅m^2/C^2×2×10^−6C)/0.1m×7m
E = 51428N/C×(V/m)/(N/C)
=51428V/m
An electric charge is the property of matter where it has more or fewer electrons than protons in its atoms. Electrons carry a negative charge and protons carry a positive charge. Matter is positively charged if it contains more protons than electrons, and negatively charged if it contains more electrons than protons.
Learn more about charge here:
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Answer:
Q = 1.35*10⁻¹¹ C.
Explanation:
By definition, the capacitance of a capacitor, is the charge on one of the plates, divided by the potential difference between them, as follows:
At the same time, we can show (applying Gauss' Law to the surface of one of the plates), that the capacitance of a parallel-plate capacitor (with a dielectric of air), can be written as follows:
C = ε₀*A / d
Replacing by the values of A, and d, and taking into account that
ε₀ = 8.85*10⁻¹² F/m,
we get the value of the capacitance as follows:
C = 8.97*10⁻¹² F
As the voltage of an AA battery is 1.5 V, and is all applied to the capacitor, we can conclude that the charge on one of the plates is as follows:
Q = C* V = 8.97*10⁻¹² F* 1.5 V = 1.35*10⁻¹¹ C
