A proton travels through a constant magnetic field in the negative y-direction while moving in the negative x-direction. The proton will be subject to a magnetic pull that is directed into the page. Option B is correct.
<h3>What is the right-hand thumb rule?</h3>
Hold a current-carrying conductor in your right hand with your thumb pointing in the direction of the current then wrap your fingers around the conductor and orient them in the direction of the magnetic field lines.
A proton travels through a constant magnetic field in the negative y-direction while moving in the negative x-direction.
The proton will be subject to a magnetic pull that is directed into the page.
Hence, option B is correct.
To learn more about the right-hand thumb rule refer to the link;
brainly.com/question/11521829
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<span>Ohm's law deals with the relation between
voltage and current in an ideal conductor. It states that: Potential difference
across a conductor is proportional to the current that pass through it. It is
expressed as V=IR.
V = IR
200 = 20R
R = 10 ohms</span>
Answer:
a) 145.6kgm^2
b) 158.4kg-m^2/s
c) 0.76rads/s
Explanation:
Complete qestion: a) the rotational inertia of the merry-go-round about its axis of rotation
(b) the magnitude of the angular momentum of the child, while running, about the axis of rotation of the merry-go-round and
(c) the angular speed of the merry-go-round and child after the child has jumped on.
a) From I = MK^2
I = (160Kg)(0.91m)^2
I = 145.6kgm^2
b) The magnitude of the angular momentum is given by:
L= r × p The raduis and momentum are perpendicular.
L = r × mc
L = (1.20m)(44.0kg)(3.0m/s)
L = 158.4kg-m^2/s
c) The total moment of inertia comprises of the merry- go - round and the child. the angular speed is given by:
L = Iw
158.4kgm^2/s = [145kgm^2 + ( 44.0kg)(1.20)^2]
w = 158.6/208.96
w = 0.76rad/s
Answer:
Opposition of passing a electric circuit
Answer:
A -Added when in the same direction
Subtracted when in opposite directions.
Explanation:
