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Mademuasel [1]
3 years ago
6

A wire along the z axis carries a current of 4.9 A in the z direction Find the magnitude and direction of the force exerted on a

3.3 cm long length of this wire by a uniform magnetic field pointing in the x direction having a magnitude 0.43T
Physics
1 answer:
AURORKA [14]3 years ago
5 0

Answer:

<h2>0.069 N, in the X direction</h2>

Explanation:

According to Flemming's left hand rule, it sates that if the first three fingers of the left hand are held mutually at right angles to one another, the fore finger will point in the direction of magnetic field, the middle finger will point in direction of current, while the thumb will point to the direction of force.

Mathematically the law is stated as

F= BIL

given data

Magnetic field B=  0.43T

Current I= 4.9 A

length of conductor L= 3.3cm to meter , 3.3/100=  0.033 m

Applying the formula the force is calculated as

F= 0.43*4.9* 0.033= 0.069 N

According to Flemming's rule the direction of all parameters are mutually perpendicular to one another, then the Force is in the X direction

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In a television picture tube, electrons strike the screen after being accelerated from rest through a potential difference of 27
Vera_Pavlovna [14]

Answer:

9.74 x 10^7 m/s

Explanation:

V = 27000 V

energy of electrons = e x V

K = 1.6 x 10^-19 x 27000 = 43200 x 10^-19 J

Energy = 1/2 m v^2

43200 x 10^-19 = 0.5 x 9.1 x 10^-31 x v^2

v^2 = 9.495 x 10^15

v = 9.74 x 10^7 m/s

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3 years ago
The maximum mass of a white dwarf is ______ times the mass of the sun.
Savatey [412]

Answer:

(A)

Explanation:

about 1.4 times the mass of our sun.

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3 years ago
An object moving at a velocity of 32 m/s slows to a stop in 4 seconds. What was its acceleration
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Answer:

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Explanation:

Because you divide 32/4 which gives you 8.

6 0
3 years ago
A 75-g bullet is fired from a rifle having a barrel 0.540 m long. choose the origin to be at the location where the bullet begin
lyudmila [28]
Part a) The work done by the gas on the bullet is the integral of the force in dx, where x is the distance covered by the bullet inside the barrel with respect to the origin:
W= \int\limits^{0.540m}_{0} {F} \, dx =  \int\limits^{0.540m}_{0} {(16000+10000x-26000x^2)} \, dx =
=16000x+10000  \frac{x^2}{2} - 26000  \frac{x^3}{3}
By substituting the length of the barrel, L=0.540 m, we find the total work done by the gas on the bullet:
W=16000(0.540m)+10000  \frac{(0.540m)^2}{2} - 26000  \frac{(0.540m)^3}{3}  =
=8733 J=8.73 kJ

part b) The resolution of the problem is the same, we just have to use the new length of the barrel (L=0.95 m) inside the final formula, and we find the new value of the work:
W=16000(0.95m)+10000  \frac{(0.95m)^2}{2} - 26000  \frac{(0.95m)^3}{3}  =
=12280 J=12.28 kJ
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<span>The lighter object is in orbit around the heaviest, and the Sun is the heaviest object in the Solar System</span>
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3 years ago
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