Answer:
The voltage drop through which the proton moves is 39.1 V.
Explanation:
Given that,
Distance = 4.76 cm
Time 
We need to calculate the acceleration
Using equation of motion
Where, s = distance
a = acceleration
t = time
Put the value in the equation
We need to calculate the voltage drop
Using formula of electric field
....(I)
Using newton's second law
....(II)
Put the value of F in equation (I) from equation (II)
Where, q = charge
a = acceleration
d = distance
m= mass of proton
Put the value into the formula
Hence, The voltage drop through which the proton moves is 39.1 V.
Velocity = fλ
where f is frequency in Hz, and λ is wavelength in meters.
<span>2.04 * 10⁸ m/s = 5.09 * 10¹⁴ Hz * λ </span>
<span>(2.04 * 10⁸ m/s) / (5.09 * 10¹⁴ Hz ) = λ </span>
<span>4.007*10⁻⁷ m = λ </span>
<span>The wavelength of the yellow light = 4.007*10⁻⁷ m<span> </span></span>
Answer:
2100 m
2804 g
4 '2 mm squares'
20,000 photos
Explanation:
1.5 km = 1500 m
1500 m + 600 m = 2100 m
2.8 kg = 2800 g
2800 g + 4 g = 2804 g
8mm tube / 2 mm squares
4 can fit
1 Gb = 1000 mb
40 Gb = 40,000 mb
40,000 mb / 2 mb = 20,000 photos
Answer:
1. E x 4πr² = ( Q x r³) / ( R³ x ε₀ )
Explanation:
According to the problem, Q is the charge on the non conducting sphere of radius R. Let ρ be the volume charge density of the non conducting sphere.
As shown in the figure, let r be the radius of the sphere inside the bigger non conducting sphere. Hence, the charge on the sphere of radius r is :
Q₁ = ∫ ρ dV
Here dV is the volume element of sphere of radius r.
Q₁ = ρ x 4π x ∫ r² dr
The limit of integration is from 0 to r as r is less than R.
Q₁ = (4π x ρ x r³ )/3
But volume charge density, ρ = 
So, 
Applying Gauss law of electrostatics ;
∫ E ds = Q₁/ε₀
Here E is electric field inside the sphere and ds is surface element of sphere of radius r.
Substitute the value of Q₁ in the above equation. Hence,
E x 4πr² = ( Q x r³) / ( R³ x ε₀ )
The metal’s feely moving sea of electrons
