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Masteriza [31]
3 years ago
5

A point charge of 6.8 C moves at 6.5 × 104 m/s at an angle of 15° to a magnetic field that has a field strength of 1.4 T.

Physics
2 answers:
Alexxx [7]3 years ago
5 0

The correct answer is A) 1.6 x 10-1 N

Gekata [30.6K]3 years ago
4 0

As per the question, the point charge is given as [q] = 6.8 C

The velocity of the charged particle [v] = 6.5*10^{4}\ m/s

The magnetic field [B] = 1.4 T

The angle made between magnetic field and velocity [\theta] = 15 degree.

We are asked to calculate the magnetic force experienced by the charged  particle.

The magnetic force experienced by the charged particle is calculated as -

Magnetic force \vec F=q(\ \vec V \times \vec B)

                        i.e F = =qVBsin\theta

                                   =6.8*6.5*10^{4}*1.4*sin15

                                   =61.88*10^{4}sin15

                                   =61.88*10^4*0.2588\ N

                                   =16.016*10^4\ N

                                   =1.6*10^5\ N

Hence, the force experienced by the charged particle is C i.e 1.6*10^5 N

                                   

             

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Determine the binding energy of an F-19 nucleus. The F-19 nucleus has a mass of 18.99840325 amu. A proton has a mass of 1.00728
Anvisha [2.4K]

Answer:

Energy = 1.38*10^13 J/mol

Explanation:

Total number of proton in F-19 = 9

Total number of neutron in F-19 = 10

Expected Mass of F-19  

= 9*1.007 + 10*1.008 = 19.152 u

Actual  mass of F-19 = 18.998 u

Energy of one particle of F-19 = 931.5*Δm = 931.5*(19.152-18.998)

= 143.234 MeV

Energy of one mole of F-19 = 143.234*10^6*1.6*10^-19*6.022*10^23  

= 1.38*10^13 J/mol

8 0
3 years ago
How do sinkholes occur? Overburden falls into the space left behind when bedrock is dissolved. Bedrock falls under the pressure
BabaBlast [244]

Answer: over burden is dissolved by water wind and acids

6 0
2 years ago
Which of the following modifications to a solenoid would be most likely to decrease the strength of its magnetic field?
Goryan [66]

By reading the fine details of the question, carefully and analytically, I have determined that there's no list of modifications to choose from.

The strength of the magnetic field of a solenoid depends on the electric current in its coil windings, the number of wire turns in its coil windings, and the material in its core.

In order to <em>DE</em>crease the strength of its magnetic field, any one or more of these steps could do the job:

-- DEcrease the electric current in its coil windings.  This can be accomplished by decreasing the voltage of the power source that energizes the coil, and/or increasing the resistance of the wire in the coil.

-- DEcrease the number of wire turns in the coil.

-- If the solenoid has anything in its core, change the core to something with a lower magnetic 'permeability'.  An Iron core will produce the greatest magnetic field strength.  Air, vacuum, or NO core will produce the lowest magnetic field strength.

8 0
3 years ago
If a machine is doing 500 J of work with 75 watts of power, how long did it take to complete its task?
kompoz [17]

Answer:

time = 6.67secs

Explanation:

power = workdone( energy) / time

therefore :

Time = workdone / power

time = 500 / 75

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3 0
3 years ago
A wheel is rotating about an axis that is in the
Schach [20]

Answer:

A) \alpha=+0.625rad/s^2 is Positive

B) t2 = 6.4s

C) t1 = 9.6s

D) \Delta \theta =-16rad

Explanation:

Let's first calculate the aangular acceleration:

\alpha=\frac{\omega f-\omega o}{\Delta t} = +0.625rad/s^2

So, acceleration is positive

Let Δt be divided into 2 time intervals t1 and t2 where t1 is the interval when the wheel goes from -6.00 rad/s to 0.00 rad/s

t1 = \frac{\omega 1 - \omega o}{\alpha}=\frac{0rad/s - (-6rad/s)}{0.625rad/s^2}

t1 = 9.6s   This is the interval where speed is decreasing

The other time interval t2 will be:

t2 = 16s - t1 = 6.4s  This is the interval where speed is increasing

The angular displacement is:

\Delta \theta=\omega o *\Delta t+\alpha*\Delta t^2/2

\Delta \theta =-16rad

4 0
3 years ago
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