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vladimir2022 [97]
2 years ago
13

What is common between transverse waves and longitudinal waves?

Physics
2 answers:
rjkz [21]2 years ago
7 0

Answer:

b

Explanation:

i took the quiz

balandron [24]2 years ago
3 0

Answer:

b

Explanation:

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a car travel uniformly speed of 30 km/h for 30 minutes and then at uniform speed of 60km/h for next 30min calculate the average
algol13

Explanation:

i hope this will help you

3 0
3 years ago
Which statement correctly describes the relationship between frequency and wavelength?
Len [333]
The relationship between the frequency and wavelength of a wave is given by the equation:

v=λf, where v is the velocity of the wave, λ is the wavelength and f is the frequency. 

If we divide the equation by f we get:

λ=v/f

From here we see that the wavelength and frequency are inversely proportional. So as the frequency increases the wavelength decreases. 

So the second statement is true: As the frequency of a wave increases, the shorter the wavelength is.  
3 0
3 years ago
Read 2 more answers
A disk 7.90 cm in radius rotates at a constant rate of 1 190 rev/min about its central axis. (a) Determine its angular speed. 12
Tanya [424]

Answer:

124.62\ \text{rad/s}

3.71\ \text{m/s}

1.23\ \text{km/s}^2

20.28\ \text{m}

Explanation:

r = Radius of disk = 7.9 cm

N = Number of revolution per minute = 1190 rev/minute

Angular speed is given by

\omega=N\dfrac{2\pi}{60}\\\Rightarrow \omega=1190\times \dfrac{2\pi}{60}\\\Rightarrow \omega=124.62\ \text{rad/s}

The angular speed is 124.62\ \text{rad/s}

r = 2.98 cm

Tangential speed is given by

v=r\omega\\\Rightarrow v=2.98\times 10^{-2}\times 124.62\\\Rightarrow v=3.71\ \text{m/s}

Tangential speed at the required point is 3.71\ \text{m/s}

Radial acceleration is given by

a=\omega^2r\\\Rightarrow a=124.62^2\times 7.9\times 10^{-2}\\\Rightarrow a=1226.88\approx 1.23\ \text{km/s}^2

The radial acceleration is 1.23\ \text{km/s}^2.

t = Time = 2.06 s

Distance traveled is given by

d=vt\\\Rightarrow d=\omega rt\\\Rightarrow d=124.62\times 7.9\times 10^{-2}\times 2.06\\\Rightarrow d=20.28\ \text{m}

The total distance a point on the rim moves in the required time is 20.28\ \text{m}.

8 0
3 years ago
How do you calculate energy lost due to friction in an experiment?
Snowcat [4.5K]

Answer:

A treadmill get it? but its   Ff * d cos theta

Explanation:

6 0
3 years ago
A block has a volume of 0.09 m3 and a density of 4,000 kg/m3. What's the force of gravity acting on the block in water?
12345 [234]

                                       Density = (mass) / (volume)

                                4,000 kg/m³ = (mass) / (0.09 m³)

Multiply each side
by  0.09 m³ :           (4,000 kg/m³) x (0.09 m³) = mass

                                 mass = 360 kg .

Force of gravity = (mass) x (acceleration of gravity)

                           = (360 kg) x (9.8 m/s²)

                           = (360 x 9.8)  kg-m/s²

                           =   3,528 newtons . 

That's the force of gravity on this block, and it doesn't matter
what else is around it.  It could be in a box on the shelf or at
the bottom of a swimming pool . . . it's weight is 3,528 newtons
(about 793.7 pounds).

Now, it won't seem that heavy when it's in the water, because
there's another force acting on it in the upward direction, against
gravity.  That's the buoyant force due to the displaced water.

The block is displacing 0.09 m³ of water.  Water has 1,000 kg of
mass in a m³, so the block displaces 90 kg of water.  The weight
of that water is  (90) x (9.8) = 882 newtons (about 198.4 pounds),
and that force tries to hold the block up, against gravity.

So while it's in the water, the block seems to weigh

       (3,528  -  882) = 2,646 newtons  (about 595.2 pounds) .

But again ... it's not correct to call that the "force of gravity acting
on the block in water".  The force of gravity doesn't change, but
there's another force, working against gravity, in the water.
5 0
3 years ago
Read 2 more answers
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