Ask question

Ask question

All categories

3 years ago

9

A charge feels a 7.89 x 10-7 N force when it moves 2090 m/s at a 29.4° angle to a 4.23 x 10-3 T magnetic field. What is the amou

nt of the charge?

2 answers:

tamaranim1 [39]3 years ago

6 0

We have that the amount of the charge q is

$q=1.8*10^{-7}$

From the Question we are told that

Force $F=7.89 x*10^{-7}$

Velocity $V=2090m/s$

Angle $\theta=29.4$

Magnetic field $B=4.23 * 10^{-3} T$

Generally, the equation for Force F is mathematically given by

$F=qVBsin\theta\\\\q=\frac{F}{VBsin\theta}$

$q=\frac{7.89 x*10^{-7}}{4.23 * 10^{-3} T*sin29.4*2090m/s}$

$q=1.8*10^{-7}$

In conclusion

The amount of the charge q is

$q=1.8*10^{-7}$

For more information on this visit

brainly.com/question/1470439

sasho [114]3 years ago

3 0

From the question, the given charge moving within the magnetic field would have a value of 1.82 x $10^{-7}$ C.

A charge is either a positively or negatively charged particle. When a charge is moving through a magnetic field, it would experience a force which depends on the amount of the charge, its speed, the magnetic field strength and angle. The force on a charge in a magnetic field can be determined by:

F = qvBSin θ

where: q is the charge, v is its velocity/ speed in the field and θ is the angle of deflection of the charge.

Given that: F = 7.89 x $10^{-7}$ N, v = 2090 m/s, B = 4.23 x $10^{-3}$ T and θ = 29.4°.

Thus, the amount of the charge can be determined by;

q = $\frac{F}{vBsin O}$

= $\frac{7.89*10^{-7} }{2090*4.23*10^{-3}* Sin 29.4 }$

= $\frac{7.89*10^{-7} }{4.340}$

q = 1.818 x $10^{-7}$

q = 1.82 x $10^{-7}$ C

Therefore, the amount of the charge that moved within the magnetic field is 1.82 x $10^{-7}$ C.

For more understanding, visit: brainly.com/question/22403676

You might be interested in

Which trait is determined entirely by heredity ?

quester [9]

Shape of the nose is entirely hereditary, as genes have nothing to do with any of the other attributes.

7 0

3 years ago

Read 2 more answers

The manufacturer of a 6V dry-cell flashlight battery says that the battery will deliver 15 mA for 60 continuous hours. During th

Vaselesa [24]

Answer:

16200 J

Explanation:

t = Time the battery is on = 60 hours

I = Current = $15\times 10^{-3}\ A$

Average voltage

$V=\dfrac{6+4}{2}=5\ V$

Energy is given by

$E=V\times I\times t$

$\\\Rightarrow E=5\times 15\times 10^{-3}\times 60\times 3600$

$\\\Rightarrow E=16200\ J$

The energy delivered in the given time is 16200 J

6 0

3 years ago

The planet should move around the elliptical orbit, and two segments of the orbit should become shaded in green. What aspect(s)

strojnjashka [21]

Answer: not sure

Explanation:

4 0

3 years ago

many lichens are composed of fungi and algae. The fungi get sugars from the algae, and the algae get water, minerals, and protei

Tcecarenko [31]

Answer:

The answer is mutualism because they are both on the receiving and giving ends

6 0

3 years ago

Un the way to the moon, the Apollo astro-

kherson [118]

Answer:

Distance = 345719139.4[m]; acceleration = 3.33*10^{19} [m/s^2]

Explanation:

We can solve this problem by using Newton's universal gravitation law.

In the attached image we can find a schematic of the locations of the Earth and the moon and that the sum of the distances re plus rm will be equal to the distance given as initial data in the problem rt = 3.84 × 108 m

$r_{e} = distance earth to the astronaut [m].\\r_{m} = distance moon to the astronaut [m]\\r_{t} = total distance = 3.84*10^8[m]$

Now the key to solving this problem is to establish a point of equalisation of both forces, i.e. the point where the Earth pulls the astronaut with the same force as the moon pulls the astronaut.

Mathematically this equals:

$F_{e} = F_{m}\\F_{e} =G*\frac{m_{e} *m_{a}}{r_{e}^{2} } \\$

$F_{m} =G*\frac{m_{m}*m_{a} }{r_{m} ^{2} } \\where:\\G = gravity constant = 6.67*10^{-11}[\frac{N*m^{2} }{kg^{2} } ] \\m_{e}= earth's mass = 5.98*10^{24}[kg]\\ m_{a}= astronaut mass = 100[kg]\\m_{m}= moon's mass = 7.36*10^{22}[kg]$

When we match these equations the masses cancel out as the universal gravitational constant

$G*\frac{m_{e} *m_{a} }{r_{e}^{2} } = G*\frac{m_{m} *m_{a} }{r_{m}^{2} }\\\frac{m_{e} }{r_{e}^{2} } = \frac{m_{m} }{r_{m}^{2} }$

To solve this equation we have to replace the first equation of related with the distances.

$\frac{m_{e} }{r_{e}^{2} } = \frac{m_{m} }{r_{m}^{2} } \\\frac{5.98*10^{24} }{(3.84*10^{8}-r_{m} )^{2} } = \frac{7.36*10^{22} }{r_{m}^{2} }\\81.25*r_{m}^{2}=r_{m}^{2}-768*10^{6}* r_{m}+1.47*10^{17} \\80.25*r_{m}^{2}+768*10^{6}* r_{m}-1.47*10^{17} =0$

Now, we have a second-degree equation, the only way to solve it is by using the formula of the quadratic equation.

$r_{m1,2}=\frac{-b+- \sqrt{b^{2}-4*a*c } }{2*a}\\ where:\\a=80.25\\b=768*10^{6} \\c = -1.47*10^{17} \\replacing:\\r_{m1,2}=\frac{-768*10^{6}+- \sqrt{(768*10^{6})^{2}-4*80.25*(-1.47*10^{17}) } }{2*80.25}\\\\r_{m1}= 38280860.6[m] \\r_{m2}=-2.97*10^{17} [m]$

We work with positive value

rm = 38280860.6[m] = 38280.86[km]

<u>Second part</u>

<u />

The distance between the Earth and this point is calculated as follows:

re = 3.84 108 - 38280860.6 = 345719139.4[m]

Now the acceleration can be found as follows:

$a = G*\frac{m_{e} }{r_{e} ^{2} } \\a = 6.67*10^{11} *\frac{5.98*10^{24} }{(345.72*10^{6})^{2} } \\a=3.33*10^{19} [m/s^2]$

6 0

3 years ago