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3 years ago

explain how a maglev train provides evidence that magnetic fields can apply forces between objects that do not touch.

Physics

1 answer:

Furkat [3]3 years ago

4 0

Answer:

Powerful electromagnets are fitted on top of guideways either ttract or repel the magents fitted on the bottom of train. The froce of attraction/repulsion rasies the train in hovering position.

Explanation:

Magentic fields exist between two opposite/like poles. The poles do not have to be touching each other. They can be separated by a distance in a medium that doesn' block magnetic field.

In magelv train the magnetic forces between two poles are used to operate the train. The two poles are separated by train body and air.

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Vector has a magnitude of 6.0 m and points 30° north of east. Vector has a magnitude of 4.0 m and points 30° east of north. The

Brut [27]

Answer:

The resultant vector is $\vec R = \vec A + \vec B = 7.196\,i + 6.464\,j$ .

Explanation:

First, each vector is determined in terms of absolute coordinates:

6-meter vector with direction: 30º north of east.

$\vec A = (6\,m)\cdot (\cos30^{\circ} \,i + \sin 30^{\circ}\,j)$

$\vec A = 5.196\,i + 3\,j$

4-meter vector with direction: 30º east of north.

$\vec B = (4\,m)\cdot (\cos 60^{\circ}\,i + \sin 60^{\circ}\,j)$

$\vec B = 2\,i + 3.464\,j$

The resultant vector is obtaining by sum of components:

$\vec R = \vec A + \vec B = 7.196\,i + 6.464\,j$

The resultant vector is $\vec R = \vec A + \vec B = 7.196\,i + 6.464\,j$ .

5 0

4 years ago

Add a vector whose magnitude is 13 with angle 27 degrees to one whose magnitude is 11 with angle 45 degrees? Put the length firs

mafiozo [28]

Answer:

Magnitude of the vector is $23.75\ \text{units}$ and the direction is $35.23^{\circ}$

Explanation:

Magnitude of first vector = $|A| = 13\ \text{units}$

Angle = $\theta_1=27^{\circ}$

Magnitude of second vector = $|B| = 11\ \text{units}$

Angle = $\theta_2=45^{\circ}$

x component of first vector

$A_{x}=|A|\cos\theta_1\\\Rightarrow A_x=13\cos27^{\circ}\\\Rightarrow A_x=11.6\ \text{units}$

y component of first vector

$A_{y}=|A|\sin\theta_1\\\Rightarrow A_y=13\sin27^{\circ}\\\Rightarrow A_y=5.9\ \text{units}$

x component of second vector

$B_{x}=|B|\cos\theta_2\\\Rightarrow B_x=11\cos45^{\circ}\\\Rightarrow B_x=7.8\ \text{units}$

y component of first vector

$B_{y}=|B|\sin\theta_2\\\Rightarrow B_y=11\sin45^{\circ}\\\Rightarrow A_y=7.8\ \text{units}$

Adding the magnitudes

$C_x=A_x+B_x=11.6+7.8\\\Rightarrow C_x=19.4\ \text{units}$

$C_y=A_y+B_y=5.9+7.8\\\Rightarrow C_y=13.7\ \text{units}$

Magnitude of the sum of the vectors would be

$|C|=\sqrt{C_x^2+C_y^2}\\\Rightarrow |C|=\sqrt{19.4^2+13.7^2}=23.75\ \text{units}$

The direction would be

$\theta=\tan^{-1}\dfrac{C_y}{C_x}\\\Rightarrow \theta=\tan^{-1}\dfrac{13.7}{19.4}\\\Rightarrow \theta=35.23^{\circ}$

The magnitude of the vector is $23.75\ \text{units}$ and the direction is $35.23^{\circ}$

4 0

3 years ago

What is the mass of a 14.6 cm3 cube of mercury (density 13.6 g/cm3)?

otez555 [7]

Density = mass/ volume
13.6g/cm^3 = m /14.6cm^3
13.6g/cm^3 *14.6 cm^3 = m
198.56g = m

8 0

3 years ago

Consider a conical pendulum with a bob of mass m = 80.0 kg on a string of length L = 10.0 m that makes an angle of 5.0 degrees w

SCORPION-xisa [38]

Answer:

0.8582 m/s^2

Explanation:

mass of the bob m = 80 kg

length of the string L= 10.0 m

angle made with the vertical θ= 5.0

let the force exerted by the string = T

from the FBD

T cosθ= mg

T cos 6°= 80×9.81

⇒T= $\frac{70\times9.81}{cos5}$

= 787.79 N

how horizontal component of tension $T_h$

$T_h= Tsin\theta$

=787.79 sin 5°= 68.66 N

Now, radial acceleration, $a_c= \frac{T_h}{m}$

= 68.66/80= 0.8582 m/s^2

8 0

4 years ago

Fiona and her twin sister April are enjoying the bumper cars at an amusement park. Fiona drives her car toward her sisters and t

vlabodo [156]

The mass of April and her bumper car is 80 kg

Explanation:

This problem can be solved by using the law of conservation of momentum.

In fact, the total momentum of the system must be conserved before and after the collision, so we have:

$p_i = p_f\\m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$

where:

$m_1 = 80 kg$ is the mass of Fiona+her bumper

$u_1 = 3 m/s$ is the initial velocity of Fiona (we take her direction as positive direction)

$v_1 = -1 m/s$ is the final velocity of Fiona (she rebounds back)

$m_2$ is the mass of April+her bumper

$u_2 = - 1 m/s$ is the initial velocity of April (in the opposite direction)

$v_2 = 3 m/s$ is the final velocity of April

Re-arranging the equation, we can find the mass of April+bumper:

$m_2 = \frac{m_1 u_1 - m_1 v_1}{v_2-u_2}=\frac{(80)(3)-(80)(-1)}{3-(-1)}=80 kg$

Learn more about momentum here:

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5 0

3 years ago