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3 years ago

What is true of transverse waves?

Physics

2 answers:

topjm [15]3 years ago

7 0

<h2>Answer:D</h2>

Explanation:

Option A:

Surface waves are neither transverse nor longitudinal.They traverse perpendicularly or parallel to the wave's motion along the interface between different media.

Option B:

Transverse waves vibrate perpendicularly to the direction of the propagation of the wave.

Option C:

Sound is a longitudinal wave.Not a transverse wave.

Option D:

Transverse waves don't require a medium for propagation.But they propagate in medium too.

wariber [46]3 years ago

4 0

“Transverse waves travel through a medium of the wave's motion” is true about transverse waves.

Answer: Option D

<u>Explanation: </u>

A transverse wave is nothing but a wave, which moves vertical to its direction of propagation. Most waves are transverse, and have a crest and trough. Surface waves are not longitudinal, but they are not transverse too. Sound waves are longitudinal, because they move in same direction of propagation.

Only transverse waves travel through a medium of the wave's motion. The propagation of a transverse wave is quite peculiar. The wave, while moving, sets the particles of medium, into the perpendicular direction, and hence, the motion takes place.

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A speeding car is pulling away from a police car. The police car is moving at 30 m/s. The radar gun in the police car emits an e

Gekata [30.6K]

Answer:

65.625 m/s

Explanation:

$F=20\times 10^9 Hz$

$\Delta F=F_r-F_t=4750 Hz$

c = Speed of light = 3×10⁸ m/s

Relative speed

$v=\frac{\Delta F}{2F}\times c\\\Rightarrow v=\frac{4750}{2\times 20\times 10^9}\times 3\times 10^8\\\Rightarrow v=35.625\ m/s$

Relative speed = 35.625 m/s

Velocity of speeder = Velocity of police car + Relative speed

⇒Velocity of speeder = 30 + 35.625 = 65.625 m/s

∴ Velocity of speeder is 65.625 m/s

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3 years ago

Write any three applicataion pressure in our daily life ?

Nina [5.8K]

Answer:

because the gravity of the earth

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2 years ago

Where is the near point of an eye for which a contact lens with a power of +2.55 diopters is prescribed?Where is the far point o

tankabanditka [31]

Answer:

(a). The eye's near point is 68.98 cm from the eye.

(b). The eye's far point is 33.33 cm from the eye.

Explanation:

Given that,

Power = 2.55 D

Object distance = 25 cm for near point

Object distance = ∞ for far point

Suppose where is the far point of an eye for which a contact lens with a power of -3.00 D is prescribed for distant vision?

(a) We need to calculate the focal length

Using formula of power

$f =\dfrac{1}{P}$

Put the value into the formula

$f=\dfrac{100}{2.55}$

$f=39.21\ cm$

We need to calculate the image distance

Using formula of lens

$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$

Put the value into the formula

$\dfrac{1}{39.21}=\dfrac{1}{v}-\dfrac{1}{-25}$

$\dfrac{1}{39.21}-\dfrac{1}{25}=\dfrac{1}{v}$

$\dfrac{1}{v}=-\dfrac{1421}{98025}$

$v=-68.98\ cm$

The eye's near point is 68.98 cm from the eye.

(b). We need to calculate the focal length

Using formula of power

$f =\dfrac{1}{P}$

Put the value into the formula

$f=-\dfrac{100}{3.00}$

$f=-33.33\ cm$

We need to calculate the image distance

Using formula of lens

$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$

Put the value into the formula

$\dfrac{1}{-33.33}=\dfrac{1}{v}-\dfrac{1}{\infty}$

$-\dfrac{1}{33.33}+\dfrac{1}{\infty}=\dfrac{1}{v}$

$\dfrac{1}{v}=-\dfrac{1}{33.33}$

$v=-33.33\ cm$

The eye's far point is 33.33 cm from the eye.

Hence, (a). The eye's near point is 68.98 cm from the eye.

(b). The eye's far point is 33.33 cm from the eye.

5 0

3 years ago

A car of mass 1800 kg can be just be lifted. What is the least force that the electromagnet must use to lift the car? (1 g = 10

Doss [256]

Answer:

m=1800kg

1800000g×10N/kg

18000000N force is required to life the car

8 0

3 years ago

A uniform rod is hung at one end and is partially submerged in water. If the density of the rod is 5/9 that of water, find the f

VashaNatasha [74]

Answer:

$\frac{y}{L}$ = 0.66

Hence, the fraction of the length of the rod above water = $\frac{y}{L}$ = 0.66

and fraction of the length of the rod submerged in water = 1 - $\frac{y}{L}$ = 1 - 0.66 = 0.34

Explanation:

Data given:

Density of the rod = 5/9 of the density of the water.

Let's denote density of Water with w

And density of rod with r

So,

r = 5/9 x w

Required:

Fraction of the length of the rod above water.

Let's denote total length of the rod with L

and length of the rod above with = y

Let's denote the density of rod = r

And density of water = w

So, the required is:

Fraction of the length of the rod above water = y/L

y/L = ?

In order to find this, we first need to find out the all type of forces acting upon the rod.

We know that, a body will come to equilibrium if the net torque acting upon a body is zero.

As, we know

F = ma

Density = m/v

m = Density x volume

Volume = Area x length = X ( L-y)

So, let's say X is the area of the cross section of the rod, so the forces acting upon it are:

F = mg

F = (Density x volume) x g

g = gravitational acceleration

F1 = X(L-y) x w x g (Force on the length of the rod submerged in water)

where,

X (L-y) = volume

w = density of water.

Another force acting upon it is:

F = mg

F2 = X x L x r x g

Now, the torques acting upon the body:

T1 + T2 = 0

F1 ( y + $(\frac{L-y}{2})$ ) g sinФ - F2 x ( $\frac{L}{2}$ ) x gsinФ = 0

plug in the equations of F1 and F2 into the above equation and after simplification, we get:

$(L^{2} - y^{2} ) . w$ = $L^{2}$ . r

where, w is the density of water and r is the density of rod.

As we know that,

r = 5/9 x w

So,

$(L^{2} - y^{2} ) . w$ = $L^{2}$ . 5/9 x w

Hence,

$(L^{2} - y^{2} )$ = $\frac{5L^{2} }{9}$

$\frac{L^{2} - y^{2} }{L^{2} }$ = $\frac{5}{9}$

Taking $L^{2}$ common and solving for $\frac{y}{L}$ , we will get

$\frac{y}{L}$ = 0.66

Hence, the fraction of the length of the rod above water = $\frac{y}{L}$ = 0.66

and fraction of the length of the rod submerged in water = 1 - $\frac{y}{L}$ = 1 - 0.66 = 0.34

8 0

3 years ago