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suter [353]
1 year ago
13

A vertical wire carries current in the upward direction. An electron is traveling parallel to the wire. What is the angle ααalph

a between the velocity of the electron and the magnetic field of the wire? Express your answer in degrees. View Available Hint(s) ααalpha = nothing degreesdegrees
Physics
1 answer:
Katen [24]1 year ago
5 0

Answer:

First of all note that The magnetic field produced by the vertical wire will be into on the right hand side and it will be out of the page on the left hand side

Assuming that the electron beam is coming from the right hand side of the page parallel to the wire, The direction of the velocity vector(V) is left and the direction of magnetic field due the wire(B) is into the page. If u use right hand rule, you will get the direction downwards but as the formula also depends on q , the charge on electron is negative .Therefore the direction will be inverted i.e Upwards.

If you assume the electron beam coming from left hand side.Then also u will get the same answer.

So, the angle α between the velocity of the electron and the magnetic field of the wire is 90°.

Explanation:

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7 0
2 years ago
A 8.00g sample of substance (substance, molar mass = 152.0 g/mol) was combusted in a bomb calorimeter with a heat capacity of 6.
aleksandrvk [35]

Answer:

ΔH°comb=-5899.5 kJ/mol

Explanation:

First, consider the energy balance:

m_{c} *Cp*(T_{2}-T_{1})=-n_{s} *H_{c} Where m_{c} is the calorimeter mass and n_{s} is the number of moles of the samples; H_{c} is the combustion enthalpy. The energy balance says that the energy that the reaction release is employed in rise the temperature of the calorimeter, which is designed to be adiabatic, so it is suppose that the total energy is employed rising the calorimeter temperature.

The product m_{c} *Cp is the heat capacity, so the balance equation is:

6.21\frac{kJ}{K}*(75-25)=-8.00g*\frac{mol}{152.0g}*H_{c}

So, the enthalpy of combustion can be calculated:

H_{c}=-5899.5\frac{kJ}{mol}

I will be happy to solve any doubt you have.

4 0
2 years ago
Object A of mass M is moving east at speed v. It collides with object B of mass 2M that was initially at rest. The motion of the
Firlakuza [10]

Answer:

v_B=\frac{v}{3}

Explanation:

Given that:

mass of object A, m_A=M

mass of object B, m_B=2M

speed of object A, v_A=v

So, according to the conservation of momentum, the momentum before collision is equal to the momentum after conservation.

m_A.v+m_B\times 0=m_A\times v_A +m_B\times v_B

M\times v+0 = M\times \frac{v}{3}+2M\times v_B

2M\times v_B= \frac{2M\times v}{3}

v_B=\frac{v}{3}

3 0
2 years ago
Assume that the length of the magnet is much smaller than the separation between it and the charge. As a result of magnetic inte
faltersainse [42]

Answer:

Assuming that the length of the magnet is much smaller than the separation between it and the charge. As a result of magnetic interaction (i.e., ignore pure Coulomb forces) between the charge and the bar magnet, the magnet will not experience any torque at all - option A

Explanation:

Assuming that the length of the magnet is much smaller than the separation between it and the charge. As a result of magnetic interaction (i.e., ignore pure Coulomb forces) between the charge and the bar magnet, the magnet will not experience any torque at all; the reason being that: no magnetic field is being produced by a charge that is static. Only a moving charge can produce a magnetic effect. And the magnet can not have any torque due to its own magnetic lines of force.

5 0
1 year ago
An elevator held by a single cable is ascending but slowing down. Is the work done by tension positive, negative, or zero? What
inessss [21]

Explanation:

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2.) The direction of motion lies in the same direction as the tension in the cable. So the work done by tension would be positive.

3.) The direction of the weight (due to gravity) would act opposite to the direction of motion of the elevator, so work done by gravity becomes negative.

8 0
2 years ago
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