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3 years ago

Which statement describes a way electromagnetic waves are different from mechanical waves?

Physics

1 answer:

Goryan [66]3 years ago

7 0

a. Wrong. EM waves can travel through matter OR vacuum.

b. True, but so can mechanical waves.

c. Wrong. Waves transfer energy, not matter.

d. This is it. EM waves can travel through empty space, and mechanical waves can't.

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A current is flowing in a wire in direction 3i + 4j where the direction of the magnetic field is 5j + 12k. The force on the wire

nasty-shy [4]

Answer:

F = 0.768 i ^ - 0.576 j ^ + 0.24 k ^

the correct answer is "b"

Explanation:

The magnetic force is

F = i l x B

The bold are vectors, in this case they give us the direction of the current and the magnetic field, for which we can solve as a determinant

$F = i \left[\begin{array}{ccc}x&y&z\\3&4&0\\0&5&12\end{array}\right]$

resolver

F = i ^ (4 12 - 0) + j ^ (0- 3 12) + k ^ (3 5 - 0)

F = i (48 i ^ - 36 j ^ + 15 k⁾

in this case i is the value of the current flowing through the cable

i = 16 mA = 0.016 A

F = 0.768 i ^ - 0.576 j ^ + 0.24 k ^

When reviewing the different answers, the correct answer is "b"

6 0

3 years ago

Machines

Ainat [17]

Explanation:

Hey there!!

Here, Given is,

Efficiency = 75%

VR = no. of pulleys = 5

Now,

$effeciency = (\frac{ma}{vr} ) \times 100\%$

$75\% = \frac{ma}{5} \times 100\%$

100% ma = 75%×5

$ma = \frac{75\% \times5}{100\%}$

$ma = \frac{375\%}{100\%}$

Therefore, MA is 3.75.

Hope it helps...

3 0

3 years ago

Read 2 more answers

A child in danger of drowning in a river is being carried down-stream by a current that flows uniformly with a speed of 2.20 m/s

zmey [24]

Answer:

The angle is 65.6°.

Explanation:

Given that,

Speed = 2.20 m/s

Distance from the shore= 500 m

Distance from the bottom= 1100 m

Speed of boat = 7.30 m/s

According to figure,

We need to calculate the angle with shore

Using formula of angle

$\tan\theta=\dfrac{y}{x}$

Put the value into the formula

$\tan\theta=\dfrac{500}{1100}$

$\theta=\tan^{-1}(\dfrac{500}{1100})$

$\theta=24.4^{\circ}$

We need to calculate the angle

$\alpha=90-\theta$

Put the value into the formula

$\alpha=90-24.4$

$\alpha=65.6^{\circ}$

Hence, The angle is 65.6°.

5 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)

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

$\frac{I_{2} }{I_{1} }= 0.159$

b) When the sunbather lies on his side, the vertical component vanishes, i.e $E_{v} = 0$

$\frac{I_{2} }{I_{1} } = \frac{E_{h} ^{2} }{E_{h} ^{2} + E_{v} ^{2}}$

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

$\frac{I_{2} }{I_{1} } = \frac{5.29E_{v}^{2} }{E_{v} ^{2} +5.29E_{v}^{2} }$

$\frac{I_{2} }{I_{1} } = \frac{5.29E_{v}^{2} }{6.29E_{v}^{2} }\\\frac{I_{2} }{I_{1} } = \frac{5.29}{6.29} \\\frac{I_{2} }{I_{1} } = 0.84$

8 0

3 years ago

Suppose you dissolve 4.75 g of another solid (not urea) in 50.0 mL of water. You note that the temperature changes from 25.0 °C

torisob [31]

Answer:

Mass of the solution = 54.75 g

Explanation:

Mass of solid dissolved = 4.75 g

Mass of the solution = Mass of solid dissolved + Mass of water.

Mass of water = Volume x Density

Volume = 50 mL = 50 cm³

Density = 1 g/cm³

Mass of water = 50 x 1 = 50 g

Mass of the solution = 4.75 + 50 = 54.75 g

8 0

4 years ago