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3 years ago

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A rectangular coil of wire (a = 22.0 cm, b = 46.0 cm) containing a single turn is placed in a uniform 4.60 T magnetic field, as

the drawing shows. The current in the loop is 10.0 A. Determine the magnitude of the magnetic force on the bottom side of the loop.

1 answer:

frez [133]3 years ago

4 0

Explanation:

Below is an attachment containing the solution.

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Explain two ways you could reduce the friction between two surfaces

abruzzese [7]

<span>A lubricant such as oil, grease, graphite powder can reduce the friction between two surfaces. Or using metal balls to space them and reduce the contact surface area as used in ball bearings.</span>

3 0

2 years ago

Adjacent antinodes of a standing wave on a string are 15.0 cm apart. A particle at an antinode oscillates in simple harmonic mot

timama [110]

Answer:

Explanation:

A ) Distance between two adjacent anti-node will be equal to distance between two adjacent nodes . So the required distance is 15 cm .

B ) wave-length, amplitude, and speed of the two traveling waves that form this pattern are as follows

wave length = same as wave length of wave pattern formed. so it is 30 cm

amplitude = 1/2 the amplitude of wave pattern formed so it is .850 / 2 = .425 cm

Speed = frequency x wavelength ( frequency = 1 / time period )

= 1 / .075) x 30 cm

400 cm / m

C ) maximum speed

= ω A

= (2π / T) x A

= 2 X 3.14 x .85 / .075 cm / s

= 71.17 cm / s

minimum speed is zero.

D ) The shortest distance along the string between a node and an antinode

= Wavelength / 4

= 30 / 4

= 7.5 cm

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2 years ago

A person 1.8m tall stands 0.75m from a reflecting globe in a garden.

Maru [420]

Answer:

1. The image of the person is 1.41 m, virtual and formed at the back of the surface of the globe.

2. The person's image is 3.38 m tall.

Explanation:

From the given question, object distance, u = 0.75 m, object height = 1.8 m, radius of curvature of the reflecting globe, r = 8 cm = 0.08 m.

f = $\frac{r}{2}$ = $\frac{0.08}{2}$ = 0.04 m

1. The image distance, v, can be determined by applying mirror formula:

$\frac{1}{f}$ = $\frac{1}{u}$ + $\frac{1}{v}$

$\frac{1}{0.04}$ = $\frac{1}{0.75}$ + $\frac{1}{v}$

$\frac{4}{100}$ - $\frac{75}{100}$ = $\frac{1}{v}$

$\frac{1}{v}$ = $\frac{4 - 75}{100}$

= - $\frac{71}{100}$

⇒ v = - $\frac{100}{71}$

= - 1.41 m

The image of the person is 1.41 m, virtual and formed at the back of the surface of the globe.

2. $\frac{image distance}{object distance}$ = $\frac{image height}{object height}$

$\frac{1.41}{0.75}$ = $\frac{v}{1.8}$

v = $\frac{2.538}{0.75}$

= 3.384

v = 3.38 m

The person's image is 3.38 m tall.

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3 years ago

If the mass of the book is 50 sliding with acceleration 1.2 m/s ^ 2 then the friction force is

Tatiana [17]

Answer:

73N

Explanation:Just multiply 1.2^2 by 50

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2 years ago

Show that rigid body rotation near the Galactic center is consistent with a spherically symmetric mass distribution of constant

irakobra [83]

To solve this problem we will use the concepts related to gravitational acceleration and centripetal acceleration. The equality between these two forces that maintains the balance will allow to determine how the rigid body is consistent with a spherically symmetric mass distribution of constant density. Let's start with the gravitational acceleration of the Star, which is

$a_g = \frac{GM}{R^2}$

Here

$M = \text{Mass inside the Orbit of the star}$

$R = \text{Orbital radius}$

$G = \text{Universal Gravitational Constant}$

Mass inside the orbit in terms of Volume and Density is

$M =V \rho$

Where,

V = Volume

$\rho =$ Density

Now considering the volume of the star as a Sphere we have

$V = \frac{4}{3} \pi R^3$

Replacing at the previous equation we have,

$M = (\frac{4}{3}\pi R^3)\rho$

Now replacing the mass at the gravitational acceleration formula we have that

$a_g = \frac{G}{R^2}(\frac{4}{3}\pi R^3)\rho$

$a_g = \frac{4}{3} G\pi R\rho$

For a rotating star, the centripetal acceleration is caused by this gravitational acceleration. So centripetal acceleration of the star is

$a_c = \frac{4}{3} G\pi R\rho$

At the same time the general expression for the centripetal acceleration is

$a_c = \frac{\Theta^2}{R}$

Where $\Theta$ is the orbital velocity

Using this expression in the left hand side of the equation we have that

$\frac{\Theta^2}{R} = \frac{4}{3}G\pi \rho R^2$

$\Theta = (\frac{4}{3}G\pi \rho R^2)^{1/2}$

$\Theta = (\frac{4}{3}G\pi \rho)^{1/2}R$

Considering the constant values we have that

$\Theta = \text{Constant} \times R$

$\Theta \propto R$

As the orbital velocity is proportional to the orbital radius, it shows the rigid body rotation of stars near the galactic center.

So the rigid-body rotation near the galactic center is consistent with a spherically symmetric mass distribution of constant density

6 0

3 years ago