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fomenos
3 years ago
6

Magnetic field is measured in SI units of a tesla (T), and a current through a wire generates a field around the wire. The large

st fields that we can make with most normal conducting materials carrying current are limited by the resistance of the wire and are around 1 tesla. For comparison, Earth's magnetic field is roughly 50 micro tesla You may have heard of a field unit called a gauss. It takes 10,000 G to make 1 T. What would be the maximum force on a wire 0.1 m long carrying a current of 80 A in a uniform magnetic field of 1 T? 8 N when the wire is perpendicular to the field
80 N when the wire is perpendicular to the field
8 N when the wire is parallel to the field
80 N when the wire is parallel to the field
Physics
1 answer:
topjm [15]3 years ago
4 0

Answer:

The answer is 8 N

Explanation:

The Lorentz force for a current carrying wire is

f = I * L x B

So, for magnetic forces to manifest the current must not be parallel to the magnetic field. So the cases where the wire is parallel to the field would result in a force of zero applied on the wires  by the magnetic field because the cross product becomes zero.

For the perpendicular cases:

f = I * L * B

f = 80 * 0.1 * 1 = 8 N

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Suppose Tom Harmon17 is standing at the exact center of the Ohio State football field18 on the 50 yard line. The field is 300 fe
nignag [31]

Answer:

a) x = 40 t , y = 39 t ,  z = 6 + 32 t - 16 t ²,   b)     x = 80 feet ,  y = 78 feet , the ball came into the field  

Explanation:

a) This is a projectile launch exercise, where in the x and y axes there is no acceleration and in the z axis the acceleration of the acceleration of gravity, let's write the equations of motion in each axis

Since the cast is in the center of the field, let's place the coordinate system

          x₀ = 0

          y₀ = 0

          z₀ = 6 feet

x-axis (towards end zone,  GOAL zone)

         x = xo + v₀ₓ t

        x = 40 t

y-axis (field width)

        y = y₀ + v_{oy} t

        y = 39 t

z axis (vertical)

        z = z₀ + v_{oz} t - ½ g t²

        z = 6 + 32 t - ½ 32 t²

        z = 6 + 32 t - 16 t ²

b) The player catches the ball at the same height as it came out, so we can find the time it takes to arrive

          z = 6

          6 = 6 + 32 t - 16 t²

          (t - 2)t  = 0

           t=0 s

           t= 2 s

The ball position

          x = 40 2

          x = 80 feet

           

          y = 39 2

          y = 78 feet

the dimensions of the field from the coordinate system (center of the field) are

             x_total = 150 feet

             y _total = 80 feet

so we can see that the ball came into the field

6 0
3 years ago
A 4.87-kg ball of clay is thrown downward from a height of 3.21 m with a speed of 5.21 m/s onto a spring with k = 1570 N/m. The
Yuki888 [10]

Answer:

Approximately 0.560\; {\rm m}, assuming that:

  • the height of 3.21\; {\rm m} refers to the distance between the clay and the top of the uncompressed spring.
  • air resistance on the clay sphere is negligible,
  • the gravitational field strength is g = 9.81\; {\rm m\cdot s^{-2}}, and
  • the clay sphere did not deform.

Explanation:

Notations:

  • Let k denote the spring constant of the spring.
  • Let m denote the mass of the clay sphere.
  • Let v denote the initial speed of the spring.
  • Let g denote the gravitational field strength.
  • Let h denote the initial vertical distance between the clay and the top of the uncompressed spring.

Let x denote the maximum compression of the spring- the only unknown quantity in this question.

After being compressed by a displacement of x, the elastic potential energy \text{PE}_{\text{spring}} in this spring would be:

\displaystyle \text{PE}_{\text{spring}} = \frac{1}{2}\, k\, x^{2}.

The initial kinetic energy \text{KE} of the clay sphere was:

\displaystyle \text{KE} = \frac{1}{2}\, m \, v^{2}.

When the spring is at the maximum compression:

  • The clay sphere would be right on top of the spring.
  • The top of the spring would be below the original position (when the spring was uncompressed) by x.
  • The initial position of the clay sphere, however, is above the original position of the top of the spring by h = 3.21\; {\rm m}.

Thus, the initial position of the clay sphere (h = 3.21\; {\rm m} above the top of the uncompressed spring) would be above the max-compression position of the clay sphere by (h + x).

The gravitational potential energy involved would be:

\text{GPE} = m\, g\, (h + x).

No mechanical energy would be lost under the assumptions listed above. Thus:

\text{PE}_\text{spring} = \text{KE} + \text{GPE}.

\displaystyle \frac{1}{2}\, k\, x^{2} = \frac{1}{2}\, m\, v^{2} + m\, g\, (h + x).

Rearrange this equation to obtain a quadratic equation about the only unknown, x:

\displaystyle \frac{1}{2}\, k\, x^{2} - m\, g\, x - \left[\left(\frac{1}{2}\, m\, v^{2}\right)+ (m\, g\, h)\right] = 0.

Substitute in k = 1570\; {\rm N \cdot m^{-1}}, m = 4.87\; {\rm kg}, v = 5.21\; {\rm m\cdot s^{-1}}, g = 9.81\; {\rm m \cdot s^{-2}}, and h = 3.21\; {\rm m}. Let the unit of x be meters.

785\, x^{2} - 47.775\, x - 219.453 \approx 0 (Rounded. The unit of both sides of this equation is joules.)

Solve using the quadratic formula given that x \ge 0:

\begin{aligned}x &\approx \frac{-(-47.775) + \sqrt{(-47.775)^{2} - 4 \times 785 \times (-219.453)}}{2 \times 785} \\ &\approx 0.560\; {\rm m}\end{aligned}.

(The other root is negative and is thus invalid.)

Hence, the maximum compression of this spring would be approximately 0.560\; {\rm m}.

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