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kotykmax [81]
4 years ago
14

An experimentalist fires a beam of electrons, creating a visible path in the air that can be measured. The beam is fired along a

direction parallel to a current-carrying wire, and the electrons travel in a circular path in response to the wire's magnetic field. Assuming the mass and charge of the electrons is known, what quantities would you need to measure in order to deduce the current in the wire and the magnetic field due to that current?
Physics
1 answer:
gtnhenbr [62]4 years ago
4 0

Answer:  

Velocity of the electron in the beam.

Radius of the circulating electrons due to the magnetic field.

Explanation:

We have a Mathematical expression for the force on a moving charge in a magnetic field as:

F=q.v.B.sin \theta ...........................(1)

where:

q= charge on the particle in coulomb

v= velocity of the charge projected into the magnetic field

B= intensity of the magnetic field in tesla

\theta= angle between the velocity and direction of magnetic field

For the forces on rotating mass we have the formula:

F=m.\frac{v^2}{r}..........................................(2)

where:

m= mass of the charged particle

v= velocity of projection of charge into the magnetic field

r= radius of the path traced  by the charge in the magnetic field

From eq. (1) and (2) we can calculate the magnetic field .

Now,

Using Ampere's Law we have:

B = \frac{\mu_0 .I}{2 \pi r}

where:

I= current in the wire

\mu_0= The permeability of free space.

r= radial distance from the current carrying wire( in this case it is same as the radius of the circular path)

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Answer:

β2= β1+10*f

Explanation:

comparing β2 and β1, it is said that β2 is increased by a factor of f.

for each factor of f, there is a 10*f dB increase.

therefore if the β1 is increases by an intensity of factor f

the new intensity would be β1+ 10*f

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3 years ago
Suppose a ball is dropped from shoulder height, falls, makes a perfectly elastic collision with the floor, and rebounds to shoul
Bezzdna [24]

Answer:Same magnitude

Explanation:

When ball is dropped from shoulder height h then velocity at the bottom is given by

v_1=\sqrt{2gh}

if it makes elastic collision then it will acquire the same velocity and riser up to the same height

If m is the mass of ball then impulse imparted is given by

J=m(v_2-v_1)

J=2m\sqrt{2gh}

Thus impulse imparted by gravity and Floor will have same magnitude of impulse but direction will be opposite to each other.

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4 years ago
What type of configuration is used in the HEVs available in the US today?
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Answer:

Currently in the united states using parallel system

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because you can walk with the twomodes with internal combustion engine or running on electric power.

6 0
2 years ago
_______ are different forms of a single element. A) Atoms B) Elements C) Ions D) Isotopes
ELEN [110]
The answer is D. Isotopes.
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7 0
4 years ago
Read 2 more answers
A 10-kg package drops from chute into a 25-kg cart with a velocity of 3 m/s. The cart is initially at rest and can roll freely w
amid [387]

Answer:

(a) the final velocity of the cart is 0.857 m/s

(b) the impulse experienced by the package is 21.43 kg.m/s

(c) the fraction of the initial energy lost is 0.71

Explanation:

Given;

mass of the package, m₁ = 10 kg

mass of the cart, m₂ = 25 kg

initial velocity of the package, u₁ = 3 m/s

initial velocity of the cart, u₂ = 0

let the final velocity of the cart = v

(a) Apply the principle of conservation of linear momentum to determine common final velocity for ineleastic collision;

m₁u₁  + m₂u₂ = v(m₁  +  m₂)

10 x 3   + 25 x 0   = v(10  +  25)

30  = 35v

v = 30 / 35

v = 0.857 m/s

(b) the impulse experienced by the package;

The impulse = change in momentum of the package

J = ΔP = m₁v - m₁u₁

J = m₁(v - u₁)

J = 10(0.857 - 3)

J = -21.43 kg.m/s

the magnitude of the impulse experienced by the package = 21.43 kg.m/s

(c)

the initial kinetic energy of the package is calculated as;

K.E_i = \frac{1}{2} mu_1^2\\\\K.E_i = \frac{1}{2} \times 10 \times (3)^2\\\\K.E_i = 45 \ J\\\\

the final kinetic energy of the package;

K.E_f = \frac{1}{2} (m_1 + m_2)v^2\\\\K.E_f = \frac{1}{2} \times (10 + 25) \times 0.857^2\\\\K.E_f = 12.85 \ J

the fraction of the initial energy lost;

= \frac{\Delta K.E}{K.E_i} = \frac{45 -12.85}{45} = 0.71

7 0
3 years ago
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