Answer:
(a) 1.3 x 10^6 Hz
(b) 76.73 cm
Explanation:
(a)
the formula for the frequency is given by
f = B q / 2 π m
where, B be the strength of magnetic field, q be the charge on one electron, m is the mass of one electron.
B = 46.7 micro tesla = 46.7 x 10^-6 T
q = 1.6 x 10^-19 C
m = 9.1 x 10^-31 kg
f = (46.7 x 10^-6 x 1.6 x 10^-19) / (2 x 3.14 x 9.1 x 10^-31) = 1.3 x 10^6 Hz
(b) K = 114 eV = 114 x 1.6 x 10^-19 J = 182.4 x 10^-19 J
K = 1/2 mv^2
182.4 x 10^-19 = 0.5 x 9.1 x 10^-31 x v^2
v = 6.3 x 10^6 m/s
r = m v / B q
Where, r be the radius of circular path
r = (9.1 x 10^-31 x 6.3 x 10^6) / (46.7 x 10^-6 x 1.6 x 10^-19)
r = 0.7673 m = 76.73 cm
Have them give up there phone for 15 minutes
talk to them when there in a good mood
watch a movie then have a disscution about it
Answer:
Explanation:
Given that,
Mass of the object, m = 100 grams
Volume of the object, V = 20 cm³
We need to find the density of the object. We know that, density is equal to mass per unit volume. So,
So, the density of the object is equal to 
.
Explanation:
Take F=ma
a = F/m
For a higher, F higher or m lower
Means higher horse power for engine or lower mass for the car
Answer:
The average induced emf in the coil is 0.0286 V
Explanation:
Given;
diameter of the wire, d = 11.2 cm = 0.112 m
initial magnetic field, B₁ = 0.53 T
final magnetic field, B₂ = 0.24 T
time of change in magnetic field, t = 0.1 s
The induced emf in the coil is calculated as;
E = A(dB)/dt
where;
A is area of the coil = πr²
r is the radius of the wire coil = 0.112m / 2 = 0.056 m
A = π(0.056)²
A = 0.00985 m²
E = -0.00985(B₂-B₁)/t
E = 0.00985(B₁-B₂)/t
E = 0.00985(0.53 - 0.24)/0.1
E = 0.00985 (0.29)/ 0.1
E = 0.0286 V
Therefore, the average induced emf in the coil is 0.0286 V
