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lina2011 [118]
3 years ago
13

A wire perpendicular to the screen carries a current in the direction shown.

Physics
2 answers:
mrs_skeptik [129]3 years ago
8 0

Answer:

A) Up

Explanation:

The direction of the current in the wire is going out of the screen. We can use the right-hand rule to find the direction of the magnetic field at point Z. In fact, we have to:

- Put the thumb of the right hand in the same direction as the current (out of the screen)

- Wrap the other fingers around the thumb. The direction of the fingers will be the same as the direction of the magnetic field.

By doing this procedure with out right hand, we see that the other fingers at point Z are directed up, so the magnetic field is also directed up.

____ [38]3 years ago
4 0

A) Up is the direction of the magnetic field at point Z.

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Force X has a magnitude of 1260 ​pounds, and Force Y has a magnitude of 1530 pounds. They act on a single point at an angle of 4
weeeeeb [17]

Answer:

Fe= 2579.68 P

α= 24.8°

Explanation:

Look at the attached graphic

we take the forces acting on the x-y plane and applied at the origin of coordinates

FX = 1260 P , horizontal (-x)

FY = 1530  P , forming 45° with positive x axis

x-y components FY

FYx= - 1530*cos(45)° = - 1081.87 P

FYy= -  1530*sin(45)° = - 1081.87 P

Calculation of the components of net force (Fn)

Fnx= FX + FYx

Fnx= -1260 P -1081.87 P

Fnx= -2341.87 P

Fny=FYy

Fny= -1081.87 P

Calculation of the components of equilibrant force (Fe)

the x-y components of the  equilibrant force are equal in magnitude but in the opposite direction to the net force components:

Fnx= -2341.87 P, then, Fex= +2341.87 P

Fny=  -1081.87 P P, then, Fex= +1081.87 P

Magnitude of the equilibrant (Fe)

F_{n} = \sqrt{(F_{nx})^{2} +(F_{ny})^{2}  }

F_{e} =\sqrt{(2341.87)^{2}+(1081.87)^{2}  }

Fe= 2579.68 P

Calculation of the direction of  equilibrant force (α)

\alpha =tan^{-1} (\frac{F_{ny} }{F_{nx} } )

\alpha =tan^{-1} (\frac{1081.87 }{2341.87} )

α= 24.8°

Look at the attached graphic

6 0
3 years ago
How does altitude, distance from the ocean, amount of sunlight, distance from the equator, and ocean currents affect polar clima
drek231 [11]
The altitude is usually low. Tropical places are mainly on the ocean, that's usually why it's so hot. They are usually close to the equator, but not right on it. The tropics get a lot of direct sunlight, so wear that sunscreen! The ocean currents are warm, so they bring along warm water. All of those help make the tropics the way they are.
 
Hope this helps 
7 0
3 years ago
If the horse lifts a 50 kilogram mass a vertical distance of 3 meters in 2 seconds, what is the horse’s minimum power output?
yaroslaw [1]

Answer:

735watts

Explanation:

Power = Force*distance/time

Given

Force = mg = 50 * 9.8'

Force = 490N

distance = 3m

time = 2s

Substitute

Power = 490*3/2

Power = 1470/2

Power = 735Watts

Hence the minimum power output is 735watts

7 0
3 years ago
A form of energy is stored in the bonds between atoms. What is the name for this stored energy?
tester [92]

Answer:

chemical energy

Explanation:

A form of energy is stored in the bonds between atoms is known as chemical energy.

3 0
3 years ago
Read 2 more answers
An ideal spring hangs from the ceiling. A 2.15 kg mass is hung from the spring, stretching the spring a distance d = 0.0895 m fr
Igoryamba

Answer:

The kinetic energy of the mass at the instant it passes back through the equilibrium position is 0.06500 J.

Explanation:

Given that,

Mass = 2.15 kg

Distance = 0.0895 m

Amplitude = 0.0235 m

We need to calculate the spring constant

Using newton's second law

F= mg

Where, f = restoring force

kx=mg

k=\dfrac{mg}{x}

Put the value into the formula

k=\dfrac{2.15\times9.8}{0.0895}

k=235.41\ N/m

We need to calculate the kinetic energy of the mass

Using formula of kinetic energy

K.E=\dfrac{1}{2}mv^2

Here, v = A\omega

K.E=\dfrac{1}{2}m\times(A\omega)^2

Here, \omega=\sqrt{\dfrac{k}{m}}^2

K.E=\dfrac{1}{2}m\times A^2\sqrt{\dfrac{k}{m}}^2

K.E=\dfrac{1}{2}kA^2

Put the value into the formula

K.E=\dfrac{1}{2}\times235.41\times(0.0235)^2

K.E=0.06500\ J

Hence, The kinetic energy of the mass at the instant it passes back through the equilibrium position is 0.06500 J.

8 0
3 years ago
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