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3 years ago

Would someone just answer this now?

Physics

1 answer:

Vitek1552 [10]3 years ago

6 0

Answer:

Explanation:

<u>Rectangular Components of a Vector</u>

A vector can be expressed as an ordered pair (x,y) where x is the x-coordinate, and y is the y-coordinate of a rectangular system.

It can also be given as a magnitude-angle pair (r,θ) where θ is measured from the positive x-axis. It's also known as polar coordinates.

The conversion from polar to rectangular coordinates is:

$x=r\cos \theta$

$y=r\sin\theta$

We are given the vector as a magnitude r=47.3 m and at an angle θ=39.4°. The x-component of the vector is:

$x=47.3\cos (39.4^\circ)$

Using a scientific calculator to find the trigonometric function:

$x=47.3\cdot 0.7716$

x=36.5 m

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A potential difference of 53 mV is developed across the ends of a 12.0-cm-longwire as it moves through a 0.27 T uniform magnetic

Klio2033 [76]

Answer:

The angle between the magnetic field and the wire’s velocity is 19.08 degrees.

Explanation:

Given that,

Potential difference, V = 53 mV

Length of the wire, l = 12 cm = 0.12 m

Magnetic field, B = 0.27 T

Speed of the wire, v = 5 m/s

Due to its motion, an emf is induced in the wire. It is given by :

$\epsilon=Blv\sin\theta$

Here,

$\theta$ is the angle between magnetic field and the wire’s velocity

$\sin\theta=\dfrac{\epsilon}{Blv}\\\\\sin\theta=\dfrac{53\times 10^{-3}}{0.27\times 0.12\times 5}\\\\\sin\theta=0.327\\\\\theta=19.08^{\circ}$

So, the angle between the magnetic field and the wire’s velocity is 19.08 degrees.

8 0

3 years ago

Find the direction of this vector.

Lyrx [107]

The direction of this vector is 170° from the positive x-axis.

Vectors are physical quantities with both magnitude and direction
To identify the direction, we usually use angles and take a reference axis
The vector in the diagram is 80° to the left of the positive y-axis, so the direction of the vector is 80° + 90° = 170° anti-clockwise from the positive x-axis.
Scalars are physical quantities which have only magnitude
Examples of vector quantities are displacement, velocity, acceleration, force etc
Examples of scalar quantities are distance, speed, volume, density, mass, time etc
The direction of the given vector is 170° from the positive x-axis.

Learn more about vectors here:

brainly.com/question/26700114

#SPJ10

3 0

2 years ago

The Nichrome wire is replaced by a wire of the same length and diameter, and same mobile electron density but with electron mobi

Marina CMI [18]

Answer:

The electric field inside the wire will remain the same or constant, while the drift velocity will by a factor of four.

Explanation:

Electron mobility, μ = $\frac{v_d}{E}$

where

$v_d$ = Drift velocity

E = Electric field

Given that the electric field strength = 1.48 V/m,

Therefore since the electric potential depends on the length of the wire and the attached potential difference, then when the electron mobility is increased 4 times the Electric field E will be the same but the drift velocity will increase four times. That is

4·μ = $\frac{4*v_d}{E}$

6 0

3 years ago

Read 2 more answers

A plumber uses a spanner on a tap. She puts a force of 200N on the spanner 30cm from the tap. What is the size of the moment?

zaharov [31]

Answer:

60Nm

Explanation:

Given data

Applied force= 200N

length of spanner= 30cm to meter

= 30/100= 0.3m

We know that the formula for the moment is

P=Fl

that is

P= force * length

P= 200*0.3

P= 60Nm

Hence the moment is 60Nm

7 0

3 years ago

At a beach the light is generallypartially polarized owing to reflections off sand and water. At a particular beach on a particu

Ne4ueva [31]

Answer:

a) 0.159

b) 0.84

Explanation:

The Horizontal component is 2.3 times the vertical component

Let the horizontal electric field component = $E_{h}$

Let the vertical electric field component = $E_{v}$

The formula for light intensity is given by:

$I = \frac{E_{m} ^{2} }{2c \mu}$ ..............................(1)

$E_{m}$ is the resolution of the vertical and horizontal components, $E_{h} and E_{v}$

$E_{m} ^{2} = E_{h} ^{2} + E_{v} ^{2}$ ..................(2)

Light intensity before the glasses were put on:

$I_{1} = \frac{E_{m} ^{2} }{2c \mu_{1} }$ .............................(3)

Put equation (2) into equation (3)

$I_{1} = \frac{E_{h} ^{2} + E_{v} ^{2}}{2c \mu_{1} }$ .............................(4)

After the glasses were put on the horizontal component vanishes, i.e. $E_{h} = 0$

$I_{2} = \frac{ E_{v} ^{2}}{2c \mu_{2} }$ ...................................(5)

Divide equation (5) by equation (4)

$\frac{I_{2} }{I_{1} } = \frac{E_{v} ^{2} }{E_{h} ^{2} + E_{v} ^{2}}$ ...............................(6)

But $E_{h} = 2.3E_{v}$ ......................(7)

Insert equation (7) into (6)

$\frac{I_{2} }{I_{1} } = \frac{E_{v}^{2} }{(2.3E_{v})^{2} + E_{v} ^{2} } \\\frac{I_{2} }{I_{1} } = \frac{E_{v}^{2} }{5.29E_{v}^{2} + E_{v} ^{2} }\\\frac{I_{2} }{I_{1} } = \frac{E_{v}^{2} }{6.29E_{v}^{2} }\\\frac{I_{2} }{I_{1} } =\frac{1}{6.29} \\$