Directly proportional to pressure
Answer:
R = 0.93 ohms
Explanation:
Let's start by calculating the resistance of the coil,
V = I ( R + r )
Make R the subject of the above equation.
Hence,
V = IR + Ir
V - Ir = IR
( V - Ir )/I = R
Given that V = 10V
r = 0.07ohms
I = 10A ( this is an assumption )
( 10 - 10×0.07 ) ÷ 10
R = 0.93 ohms
Note: r is the internal resistance, R is the resistance, I is the current and the emf is represented as V. Since current was supposed to be included in the question but it was not, hence it was assumed to be 10A . Meaning whatsoever value is given for current will be substituted into the equation. Thanks
In a certain region of space, a uniform electric field is in the x direction. A particle with negative charge is carried from x=20.0 cm to x=60.0cm.
<h3>Where is the
electric potential, when the particle moved?</h3>
The charge field system's electric potential energy rose. The particle experiences an electric force that is directed against the x-axis. It is pushed uphill by an outside force, which raises the potential energy.
When a charge to be moved against an applied electric field, electric potential energy is needed. A charge must be moved through a stronger electric field with more energy than it would require to carry it via a weaker electric field.
In a certain region of space, a uniform electric field is in the x direction. A particle with negative charge is carried from x=20.0 cm to x=60.0cm.
The electric potential energy of the charge field system:
- (a) increase
- (b) remain constant
- (c) decrease
- (d) change unpredictably
The correct option is a).
To learn more about electric potential, refer to:
brainly.com/question/21808222
#SPJ4
Answer:
The moment of inertia I is
I = 2.205x10^-4 kg/m^2
Explanation:
Given mass m = 0.5 kg
And side lenght = 0.03 m
Moment of inertia I = mass x radius of rotation squared
I = mr^2
In this case, the radius of rotation is about an axis which is both normal (perpendicular) to and through the center of a face of the cube.
Calculating from the dimensions of the the box as shown in the image below, the radius of rotation r = 0.021 m
Therefore,
I = 0.5 x 0.021^2 = 2.205x10^-4 kg/m^2
