The electric force on the electron is opposite in direction to the electric field E. E points in the -y direction, so the electric force will point in the +y direction. The magnitude of the electric force is given by:
F = Eq
F = electric force, E = electric field strength, q = electron charge
We need to set up a magnetic field such that the magnetic force on the electron balances out the electric force. Since the electric force points in the +y direction, we need the magnetic force to point in the -y direction. Using the reversed right hand rule, the magnetic field must point in the -z direction for this to happen. Since the direction is perpendicular to the +x direction of the electron's velocity, the magnetic force is given by:
F = qvB
F = magnetic force, q = charge, v = velocity, B = magnetic field strength
The electric force must equal the magnetic force.
Eq = qvB
Do some algebra to isolate B:
E = vB
B = E/v
Let's solve for the electron's velocity. Its kinetic energy is given by:
KE = 0.5mv²
KE = kinetic energy, m = mass, v = velocity
Given values:
KE = 2.9keV = 4.6×10⁻¹⁶J
m = 9.1×10⁻³¹kg
Plug in and solve for v:
4.6×10⁻¹⁶ = 0.5(9.1×10⁻³¹)v²
v = 3.2×10⁷m/s
B = E/v
Given values:
E = 7500V/m
v = 3.2×10⁷m/s
Plug in and solve for B:
B = 7500/3.2×10⁷
B = 0.00023T
B = 0.23mT
Answer:
The magnitude of force is 1593.4N
Explanation:
The sum of the horizontal components of the friction and the normal force will be equal to the centripetal force on the car. This can be represented as
fcostheta + Nsintheta = mv^2/r
Where F = force of friction
Theta = angle of banking
N = normal force
m = mass of car
v = velocity of car
r = radius of curve
The car has no motion in the vertical direction so the sum of forces = 0
The vertical component of the normal force acts upwards whereas the weight of the car and the vertical component friction acts downwards.
Taking the upward direction to be positive,rewrite the equation above to get:
Ncos thetha = mg - fsintheta =0
Ncistheta = mg + fain theta
N = mg/cos theta + sintheta/ costheta
fcostheta +[mg/costheta + ftan theta] sin theta = mv^2/r
Substituting gives:
f = (1/(costheta + tanthetasintheta) + mgtantheta = mv^2/r - mgtantheta)
Substituting given values into the above equation
f = 1/(cos25 + tan 25 )(sin25)[ 600×30/120 - (600×9.81)tan
f = 1593.4N25
Answer:
What is the best description for the volume of air volume of air provided in a high quality rescue breath?
Explanation:
A) only Enough air to create a visible rise of the chest.
B) Until you can no longer force air in.
C) plenty of air make sure it is adequate to sustain life
D) clear and obvious rise of the chest, sustained over a few seconds
<h2>
Answer: B. Gravitational potential energy </h2>
Explanation:
<em>The gravitational potential energy is the energy that a body or object possesses, due to its position in a gravitational field.
</em>
That is why this energy depends on the relative height of an object with respect to some point of reference and associated with the gravitational force.
In the case of the <u>Earth</u>, in which <u>the gravitational field is considered constant</u>, the value of the gravitational potential energy 
will be:
Where 
is the mass of the object, 
the acceleration due gravity and 
the height of the object.
As we can see, the value of is directly proportional to the height.
Answer:
The S.I unit of heat is Joule .
Hope it helps you :--)
Explanation:
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