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3 years ago

Sound waves are a- Transverse b- Longitudinal c- Transverse and longitudinal

Physics

2 answers:

kari74 [83]3 years ago

6 0

Answer:

hey sally!

Explanation:

Sound waves are C) Transverse and longitudinal.

We can say that a wave is produced by the vibrations of the particles of the medium through which it passes.

There are two types of waves: Longitudinal waves and Transverse waves.

Longitudinal Waves: A wave in which the particles of the medium vibrate back and forth in the ‘same direction’ in which the wave is moving. Medium can be solid, liquid or gases. Therefore, sound waves are longitudinal waves.

Transverse Waves: A wave in which the particles of the medium vibrate up and down ‘at right angles’ to the direction in which the wave is moving. These waves are produced only in a solids and liquids but not in gases.

Katen [24]3 years ago

3 0

Answer:

Sound waves are longitudinal in nature.

Explanation:

There are many types of waves like transverse, longitudinal, electromagnetic wave etc.

Sound waves are longitudinal in nature. In longitudinal type of wave, the medium particles moves parallel to the propagation of the wave. This type of waves move in the form of compression and rarefaction.

In compression, the particle density at a point is very less while in rarefaction, the particle density at a point is very high.

So, the correct option is (b) "longitudinal wave".

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a sprinter with a mass of 80kg accelerates uniformly from 0 m/s to 9 m/s in 3 s. a.)what is the runners acceleration? b.) what i

Katyanochek1 [597]

Part A:
Acceleration can be calculated by dividing the difference of the initial and final velocities by the given time. That is,
a = (Vf - Vi) / t
where a is acceleration,
Vf is final velocity,
Vi is initial velocity, and
t is time

Substituting,
a = (9 m/s - 0 m/s) / 3 s = 3 m/s²
ANSWER: 3 m/s²

Part B:
From Newton's second law of motion, the net force is equal to the product of the mass and acceleration,
F = m x a
where F is force,
m is mass, and
a is acceleration

Substituting,
F = (80 kg) x (3 m/s²) = 240 kg m/s² = 240 N

ANSWER: 240 N 

Part C:
The distance that the sprinter travel is calculated through the equation,
d = V₀t + 0.5at²

Substituting,
d = (0 m/s)(3 s) + 0.5(3 m/s²)(3 s)²
d = 13.5 m

ANSWER: d = 13.5 m

3 0

3 years ago

An object of known mass M with speed v0 travels toward a wall. The object collides with it and bounces away from the wall in the

Bingel [31]

Neither side of the equation may be used because there are too many unknown quantities before, during, and after the collision

Explanation:

The impulse theorem states that the change in momentum of an object is equal to the impulse, which is the product between the average force applied and the duration of the collision:

$\Delta p = F \Delta t$

where

$\Delta p$ is the change in momentum

F is the average force

$\Delta t$ is the duration of the collision

In this problem, neither side of the equation can be used to measure the change in momentum. In fact:

- The change in momentum (left side) is given by

$\Delta p = m(v-u)$

where

m is the mass of the object

u is the initial velocity

v is the final velocity

Here the final velocity is not known, so it's not possible to use this side of the equation

- The impulse (right side) is given by

$F\Delta t$

here the average force is known, however the duration of the collision is not known, so it's not possible to use this side of the equation.

Learn more about momentum:

brainly.com/question/9484203

#LearnwithBrainly

3 0

3 years ago

A camera lens with focal length f = 50 mm and maximum aperture f>2

Brut [27]

Answer:

The minimum distance between two points on the object that are barely resolved is 0.26 mm

The corresponding distance between the image points = 0.0015 m

Explanation:

Given

focal length f = 50 mm and maximum aperture f>2

s = 9.0 m

aperture = 25 mm = 25 *10^-3 m

Sin a = 1.22 *wavelength /D

Substituting the given values, we get –

Sin a = 1.22 *600 *10^-9 m /25 *10^-3 m

Sin a = 2.93 * 10 ^-5 rad

Now

Y/9.0 m = 2.93 * 10 ^-5

Y = 2.64 *10^-4 m = 0.26 mm

Y’/50 *10^-3 = 2.93 * 10 ^-5

Y’ = 0.0015 m

8 0

3 years ago

When a certain element is excited with electricity, we see three main lines in its emission spectrum: two red lines and one oran

Eddi Din [679]

The absorption spectrum would have all the wavelengths of the light source but would have black lines where the two red and one orange lines were in the emission spectrum

4 0

3 years ago

Each plate of a parallel‑plate capacitor is a square of side 4.19 cm, 4.19 cm, and the plates are separated by 0.407 mm. 0.407 m

alexandr1967 [171]

Answer:

The electric field strength inside the capacitor is 49880.77 N/C.

Explanation:

Given:

Side length of the capacitor plate (a) = 4.19 cm = 0.0419 m

Separation between the plates (d) = 0.407 mm = $0.407\times 10^{-3}\ m$

Energy stored in the capacitor (U) = $7.87\ nJ=7.87\times 10^{-9}\ J$

Assuming the medium to be air.

So, permittivity of space (ε) = $8.854\times 10^{-12}\ F/m$

Area of the square plates is given as:

$A=a^2=(0.0419\ m)^2=1.75561\times 10^{-3}\ m^2$

Capacitance of the capacitor is given as:

$C=\dfrac{\epsilon A}{d}\\\\C=\frac{8.854\times 10^{-12}\ F/m\times 1.75561\times 10^{-3}\ m^2 }{0.407\times 10^{-3}\ m}\\\\C=3.819\times 10^{-11}\ F$

Now, we know that, the energy stored in a parallel plate capacitor is given as:

$U=\frac{CE^2d^2}{2}$

Rewriting in terms of 'E', we get:

$E=\sqrt{\frac{2U}{Cd^2}}$

Now, plug in the given values and solve for 'E'. This gives,

$E=\sqrt{\frac{2\times 7.87\times 10^{-9}\ J}{3.819\times 10^{-11}\ F\times (0.407\times 10^{-3})^2\ m^2}}\\\\E=49880.77\ N/C$

Therefore, the electric field strength inside the capacitor is 49880.77 N/C

8 0

3 years ago