Answer:
1.2 amps :)
Explanation:
A heater has a resistance of 10.0 Ω. It operates on a 12.0 V. What is the current through the resistor?
Known:
Unknown:
I = V/R
= 12.0 V / 10.0 Ω
= 1.2 amps
Answer:
The current of the solenoid is 0.0129 A.
Explanation:
The movement of the electron within the solenoid in a circle is produced by equaling the magnetic force and the centripetal force, as follows:


Where:
I: is the current
m: is the electron's mass = 9.1x10⁺³¹ kg
v: is the electron's speed = 3.0x10⁵ m/s
μ₀: is the permeability magnetic = 4πx10⁻⁷ T.m/A
n: is the number of turns per unit length = 35/cm
r: is the radius of the circle = 3.0 cm
e: is the electron's charge = 1.6x10⁻¹⁹ C
Therefore, the current of the solenoid is 0.0129 A.
I hope it helps you!
She should use shorter focal length to fit the entire landscape which she is trying to photograph into her picture.
What is focal length?
The focal length is a measure of how strongly the system converges or diverges light.
A positive focal length indicates that a system converges light, while a negative focal length indicates that the system diverges light.
For a standard rectilinear lens,
FOV = 2 arctan (x/2f)
FOV ∝ 1 / f
where x is the diagonal of the film.
Focal length (f) and field of view (FOV) of a lens are inversely proportional.
From the equation we can say that,
A shorter focal length gives you a wide angle of view which allows more view to fit in the frame.
Hence,
She should use shorter focal length to fit the entire landscape which she is trying to photograph into her picture.
Learn more about focal length here
<u>brainly.com/question/13885819</u>
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Answer:
1.5 * 10^-2 Tm^2
Explanation:
Electric Flux = B.A cos(theta)
B = 0.055 T
A = 0.32 m^2
theta = 30
Electric Flux = (0.055 T).(0.32 m^2).Cos(30) = 0.0152 = 1.5 * 10^-2 Tm^2
Answer:

Explanation:
Given data
Force F=2 N
Length L=17 cm = 0.17 m
Spring Constant k=42 N/m
To find
Relaxed length of the spring
Solution
From Hooke's Law we know that
