Imagine you are standing on a bathroom scale inside an elevator. What would happen to your weight reading when the elevator acce
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Answer:
Magnetic field = 0.534 T
Explanation:
The solving is on the attach document.
Explanation:
Draw a free body diagram of the bar.
There are 3 reaction forces at O in the x, y, and z direction (Ox, Oy, and Oz).
There is a tension force Tac at A in the direction of the rope. There are also tension forces Tbd and Tbe at B in the direction of the ropes.
Finally, there is a weight force mg pulling down halfway between A and B, where m = 400 kg.
There are 6 unknown variables, so we'll need 6 equations to solve. Summing the forces in the x, y, and z direction will give us 3 equations. Summing the torques about the x, y, and z axes will give us 3 more equations.
First, let's find the components of the tension forces.
Tbe is purely in the z direction.
Tbd has components in the y and z directions. The length of Tbd is √8.
(Tbd)y = 2/√8 Tbd
(Tbd)z = 2/√8 Tbd
Tac has components in the x, y, and z directions. The length of Tac is √6.
(Tac)x = 1/√6 Tac
(Tac)y = 1/√6 Tac
(Tac)z = 2/√6 Tac
Sum of the forces in the +x direction:
∑F = ma
Ox − (Tac)x = 0
Ox − 1/√6 Tac = 0
Sum of the forces in the +y direction:
∑F = ma
Oy + (Tac)y + (Tbd)y − mg = 0
Oy + 1/√6 Tac + 2/√8 Tbd − mg = 0
Sum of the forces in the +z direction:
∑F = ma
Oz − (Tac)z − (Tbd)z − Tbe = 0
Oz − 2/√6 Tac − 2/√8 Tbd − Tbe = 0
Sum of the torques counterclockwise about the x-axis:
∑τ = Iα
mg (2 m) − (Tac)y (2 m) − (Tbd)y (2 m) = 0
mg − (Tac)y − (Tbd)y = 0
mg − 1/√6 Tac − 2/√8 Tbd = 0
Sum of the torques counterclockwise about the y-axis:
∑τ = Iα
-(Tac)x (2 m) + (Tbd)z (1.5 m) + Tbe (1.5 m) = 0
-4 (Tac)x + 3 (Tbd)z + 3 Tbe = 0
-4/√6 Tac + 6/√8 Tbd + 3 Tbe = 0
Sum of the torques counterclockwise about the z-axis:
∑τ = Iα
-mg (0.75 m) + (Tbd)y (1.5 m) = 0
-mg + 2 (Tbd)y = 0
-mg + 4/√8 Tbd = 0
As you can see, by summing the torques about axes passing through O, we were able to write 3 equations independent of those reaction forces. We can solve these equations for the tension forces, then go back and find the reaction forces.
-mg + 4/√8 Tbd = 0
4/√8 Tbd = mg
Tbd = √8 mg / 4
Tbd = √8 (400 kg) (9.8 m/s²) / 4
Tbd = 2772 N
mg − 1/√6 Tac − 2/√8 Tbd = 0
1/√6 Tac = mg − 2/√8 Tbd
Tac = √6 (mg − 2/√8 Tbd)
Tac = √6 ((400 kg) (9.8 m/s²) − 2/√8 (2772 N))
Tac = 4801 N
-4/√6 Tac + 6/√8 Tbd + 3 Tbe = 0
3 Tbe = 4/√6 Tac − 6/√8 Tbd
Tbe = (4/√6 Tac − 6/√8 Tbd) / 3
Tbe = (4/√6 (4801 N) − 6/√8 (2772 N)) / 3
Tbe = 653 N
Now, using our sum of forces equations to find the reactions:
Ox − 1/√6 Tac = 0
Ox = 1/√6 Tac
Ox = 1/√6 (4801 N)
Ox = 1960 N
Oy + 1/√6 Tac + 2/√8 Tbd − mg = 0
Oy = mg − 1/√6 Tac − 2/√8 Tbd
Oy = (400 kg) (9.8 m/s²) − 1/√6 (4801 N) − 2/√8 (2772 N)
Oy = 0 N
Oz − 2/√6 Tac − 2/√8 Tbd − Tbe = 0
Oz = 2/√6 Tac + 2/√8 Tbd + Tbe
Oz = 2/√6 (4801 N) + 2/√8 (2772 N) + 653 N
Oz = 6533 N
Answer:
1) A time-varying magnetic field will produce an electric field.
4) A time-varying electric field will produce a magnetic field.
Explanation:
1) A time-varying magnetic field will produce an electric field.
TRUE
time varying magnetic field will produce electric field which is given as
2) Time-varying electric and magnetic fields can propagate through space only if there is no matter in their path.
FALSE
Time varying electric field and magnetic field will induce each other and it can travel through any medium as well as it can travel without any medium also
3) Electric and magnetic fields can be treated independently only if they vary in time.
FALSE
electric field can be due to stationary charge and magnetic field due to current carrying elements so it is not compulsory to have time varying
4) A time-varying electric field will produce a magnetic field.
TRUE
Time varying electric field will produce magnetic field given as