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VikaD [51]
3 years ago
15

The observation location A is located at < 0.112, 0.247, 0 > m. The conventional current running through the wire is 7.5 a

mperes. In this exercise you will calculate the magnetic field at the observation location due only to segment 4 of the wire.
What is the direction of conventional current in this wire?
Physics
1 answer:
xenn [34]3 years ago
5 0

Answer:

Positive direction

Explanation:

We know that the direction of flow of current is always opposite the direction of the flow of electrons. Assuming that the electrons move in negative direction then the current will flow in the positive direction.

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40V because it will provide the same amount of power.
4 0
2 years ago
A toy cart is pulled a distance of 6.00 m in a straight line across the floor. The force pulling the cart has a magnitude of 20.
aivan3 [116]

Answer:

W = 95.8J

Explanation:

Given

S = 6.00m

F =20.0N

Theta = 37°

W = FSCos(theta)

W = 20×6.0×Cos(37)

W = 95.8J

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Organ system that is responsible for breaking down food
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3 years ago
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Challenge! A marshmallow is dropped from a 5-meter high pedestrian bridge and 0.83 seconds later, it lands right on the head of
WINSTONCH [101]

a₀).  You know ...
         -- the object is dropped from 5 meters
             above the pavement;
         --  it falls for 0.83 second.

a₁).  Without being told, you assume ...
         -- there is no air anyplace where the marshmallow travels,
             so it free-falls, with no air resistance;
         -- the event is happening on Earth,
            where the acceleration of gravity is  9.81 m/s² .

b).  You need to find how much LESS than 5 meters
       the marshmallow falls in 0.83 second.
    
c).  You can use whatever equations you like.
       I'm going to use the equation for the distance an object falls in
       ' T ' seconds, in a place where the acceleration of gravity is ' G '.

d).  To see how this all goes together for the solution, keep reading:


The distance that an object falls in ' T ' seconds
when it's dropped from rest is

                                 (1/2 G) x (T²) .

On Earth, ' G ' is roughly  9.81 m/s², so in 0.83 seconds,
such an object would fall

                               (9.81 / 2) x (0.83)² = 3.38 meters .

It dropped from 5 meters above the pavement, but it
only fell 3.38 meters before something stopped it.
So it must have hit something that was

                         (5.00 - 3.38)  =  1.62 meters

above the pavement.  That's where the head of the unsuspecting
person was as he innocently walked by and got clobbered.

7 0
3 years ago
Indiana Jones is in a temple searching for artifacts. He finds a gold sphere with a radius of 2 cm sitting on a pressure sensiti
Vlad1618 [11]

Answer:

Volume of Sand = 0.4 m³

Radius of Sand Sphere = 0.46 m

Explanation:

First we need to find the volume of gold sphere:

Vg = (4/3)πr³

where,

Vg = Volume of gold sphere = ?

r = radius of gold sphere = 2 cm = 0.02 m

Therefore,

Vg = (4/3)π(0.2 m)³

Vg = 0.0335 m³

Now, we find mass of the gold:

ρg = mg/Vg

where,

ρg = density of gold = 19300 kg/m³

mg = mass of gold = ?

Vg = Volume of gold sphere = 0.0335 m³

Therefore,

mg = (19300 kg/m³)(0.0335 m³)

mg = 646.75 kg

Now, the volume of sand required for equivalent mass of gold, will be given by:

ρs = mg/Vs

where,

ρs = density of sand = 1602 kg/m³

mg = mass of gold = 646.75 kg

Vs = Volume of sand = ?

Therefore,

1602 kg/m³ = 646.75 kg/Vs

Vs = (646.75 kg)/(1602 kg/m³)

<u>Vs = 0.4 m³</u>

Now, for the radius of sand sphere to give a volume of 0.4 m³, can be determined from the formula:

Vs = (4/3)πr³

0.4 m³ = (4/3)πr³

r³ = 3(0.4 m³)/4π

r³ = 0.095 m³

r = ∛(0.095 m³)

<u>r = 0.46 m</u>

4 0
2 years ago
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