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Xelga [282]
3 years ago
12

Can you find the magnetic field based on force? a straight segment of wire 35.0 cm long carrying a current of 1.40 a is in a uni

form magnetic field. the segment makes an angle of 53° with the direction of the magnetic field. if the force on the segment is 0.20 n, what is the magnitude of the magnetic field?
Physics
1 answer:
Lostsunrise [7]3 years ago
3 0
The force exerted by a magnetic field on a wire carrying current is:
F=ILB \sin \theta
where I is the current, L the length of the wire, B the magnetic field intensity, and \theta the angle between the wire and the direction of B.

In our problem, the force is F=0.20 N. The current is I=1.40 A, while the length of the wire is L=35.0 cm=0.35 m. The angle between the wire and the magnetic field is 53 ^{\circ}, so we can re-arrange the formula and substitute the numbers to find B:
B= \frac{F}{IL \sin \theta}= \frac{0.20 N}{(1.40 A)(0.35 m)(\sin 53^{\circ})}=0.51 T
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The force that contributed to the formation of planets, determines the motion of bodies in the solar system, and pulls objects t
kykrilka [37]

Answer:

The force of gravity

Explanation:

Gravity was studied, by early scientists such as Copernicus and others, Galileo was the first to ensure that planets moved according to a physical equation that depended on a force that caused celestial bodies to move and interact with each other. But years later Newton based on studies conducted deciphering what Galileo assumed, he was able to find the equation of the force of gravity in any body in the universe. This equation depends on the masses of the two interacting bodies, the distance between them and a constant, which I call universal gravitation constant.

F_{g}=G*\frac{m_{1}*m_{2}}{r^2}

Fg = gravity force [N]

G =  universal gravitation constant = 6.67*10^(-11) [N*m^2/kg^2]

m1 = mass of the 1st body [kg]

m2 = mass of the 2nd body [kg]

r = distance between the bodies [meters]

6 0
3 years ago
Q4: Two fixed charges, 1 c and -3 C are
White raven [17]

Answer:

B

Explanation:

Is the equilibrium stable or unstable for the third charge

5 0
2 years ago
One electron collides elastically with a second electron initially at rest. After the collision, the radii of their trajectories
morpeh [17]

Answer:

63.750KeV

Explanation:

We are given that

Initial velocity of second electron,u_2=0

Radius,r_1=0

r_2=2.3 cm=\frac{2.3}{100}=0.023 m

1 m=100 cm

Magnetic field,B=0.0370 T

We have to determine the energy of the incident electron.

Mass of electron,m=9.1\times 10^{-31} kg

Charge on an electron,q=-1.6\times 10^{-19} C

Velocity,v=\frac{Bqr}{m}

Using the formula

Speed of electron,v_1=\frac{Bqr_1}{m}=\frac{0.0370\times 1.6\times 10^{-19}\times 0}{9.1\times 10^{-31}}=0

Speed of second electron,v_2=\frac{0.0370\times 1.6\times 10^{-19}\times 0.023}{9.1\times 10^{-31}}

v_2=1.5\times 10^8 m/s

Kinetic energy of incident electron=\frac{1}{2}mv^2_1+\frac{1}{2}mv^2_2

Kinetic energy of incident electron=0+\frac{1}{2}(9.1\times 10^{-31})(1.5\times 10^8)^2=1.02\times 10^{-14} J

Kinetic energy of incident electron=\frac{1.02\times 10^{-14}}{1.6\times 10^{-19}}=63750eV=\frac{63750}{1000}=63.750KeV

1KeV=1000eV

3 0
3 years ago
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il63 [147K]
Hyaline, elastic, & fibrocartilage!
3 0
3 years ago
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5. What is the density of 4.5 mL of a liquid that has a mass of 1.3 grams?
antoniya [11.8K]

Answer:

A. 0.289g/mL

Explanation:

Using the equation for density which is d = m/v  or density = mass/volume, we input 1.3g/4.5mL and get 0.289g/mL.

5 0
3 years ago
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