Answer:
The force when θ = 33° is 1.7625 times of the force when θ = 18°
Explanation:
The force on a moving charge through a magnetic field is given by
F = qvB sin θ
q = charge of the moving particle
v = Velocity of the moving charge
B = Magnetic field strength
θ = angle between the magnetic field and the velocity (direction of the motion) of the moving charge
Because qvB are all constant, we can call the expression K.
F = K sinθ
when θ = 18°,
F = K sin 18° = 0.309K
when θ = 33°, let the force be F₁
F₁ = K sin 33° = 0.5446K
(F₁/F) = (0.5446K/0.309K) = 1.7625
F₁ = 1.7625 F
Hope this Helps!!!
An animal is a lion, it may compete for food and it’s territory. The animal may have to fight other animals to get these things
The focal length would likely decrease as the refractive index and the increase when the curvature radius of the lens increases. The decrease in focal length happens since a higher index of refraction would signify that the rays of the sun striking to an object would tend to bend more.
Answer: The shortest possible wavelength of sound the organ can produce is 0.224 m
Explanation:
To calculate the wavelength of light, we use the equation:
where,
= wavelength of the light
c = speed of light =
As wavelength and frequency follows inverse relation, shortest wavelength is produced by highest frequency.
= highest frequency of light =
Thus the shortest possible wavelength of sound the organ can produce is 0.224 m
Answer:
The highest electric field is experienced by a 2 C charge acted on by a 6 N electric force. Its magnitude is 3 N.
Explanation:
The formula for electric field is given as:
E = F/q
where,
E = Electric field
F = Electric Force
q = Charge Experiencing Force
Now, we apply this formula to all the cases given in question.
A) <u>A 2C charge acted on by a 4 N electric force</u>
F = 4 N
q = 2 C
Therefore,
E = 4 N/2 C = 2 N/C
B) <u>A 3 C charge acted on by a 5 N electric force</u>
F = 5 N
q = 3 C
Therefore,
E = 5 N/3 C = 1.67 N/C
C) <u>A 4 C charge acted on by a 6 N electric force</u>
F = 6 N
q = 4 C
Therefore,
E = 6 N/4 C = 1.5 N/C
D) <u>A 2 C charge acted on by a 6 N electric force</u>
F = 6 N
q = 2 C
Therefore,
E = 6 N/2 C = 3 N/C
E) <u>A 3 C charge acted on by a 3 N electric force</u>
F = 3 N
q = 3 C
Therefore,
E = 3 N/3 C = 1 N/C
F) <u>A 4 C charge acted on by a 2 N electric force</u>
F = 2 N
q = 4 C
Therefore,
E = 2 N/4 C = 0.5 N/C
The highest field is 3 N, which is found in part D.
<u>A 2 C charge acted on by a 6 N electric force</u>