All categories

3 years ago

A wave has a frequency of 270 Hz and a wavelength of 60 m. What is the speed of the wave?

Physics

2 answers:

mr_godi [17]3 years ago

4 0

Answer:

(none of the options is correct... see below)

Explanation:

Recall that frequency, wavelength and speed are related by

Velocity (i.e speed) = frequency x wavelength

in our case

Speed = 270 Hz x 60m = 16,200 m/s

Zigmanuir [339]3 years ago

4 0

Answer:

If the question is actually A wave has a frequency of 270 Hz and a wavelength of 6.0 m. What is the speed of the wave?

Explanation:

then the answer would be 1600 m/s

You might be interested in

A 782-kg satellite is in a circular orbit about Earth at a height above Earth equal to Earth's mean radius. (a) Find the satelli

Vadim26 [7]

Answer:

a) v = 5.59x10³ m/s

b) T = 4 h

c) F = 1.92x10³ N

Explanation:

a) We can find the satellite's orbital speed by equating the centripetal force and the gravitation force as follows:

$F_{c} = F_{G}$

$\frac{mv^{2}}{r + h} = \frac{GMm}{(r + h)^{2}}$

$v = \sqrt{\frac{gr^{2}}{r+h}$

Where:

g is the gravity = 9.81 m/s²

r: is the Earth's radius = 6371 km

h: is the satellite's height = r = 6371 km

$v = \sqrt{\frac{gr^{2}}{2r}} = \sqrt{\frac{gr}{2}} = \sqrt{\frac{9.81 m/s^{2}*6.371 \cdot 10^{6} m}{2}} = 5.59 \cdot 10^{3} m/s$

b) The period of its revolution is:

$T = \frac{2\pi}{\omega} = \frac{2\pi (r + h)}{v} = \frac{2\pi (2*6.371 \cdot 10^{6} m)}{5.59 \cdot 10^{3} m/s} = 14322.07 s = 4 h$

c) The gravitational force acting on it is given by:

$F = \frac{GMm}{(r + h)^{2}}$

Where:

M is the Earth's mass = 5.97x10²⁴ kg

m is the satellite's mass = 782 kg

G is the gravitational constant = 6.67x10⁻¹¹ Nm²kg⁻²

$F = \frac{GMm}{(r + h)^{2}} = \frac{6.67 \cdot 10^{-11} Nm^{2}kg^{-2}*5.97 \cdot 10^{24} kg*782 kg}{(2*6.371 \cdot 10^{6} m)^{2}} = 1.92 \cdot 10^{3} N$

I hope it helps you!

3 0

3 years ago

When the frequency of an electromagnetic wave increases, its energy

OverLord2011 [107]

The energy stays the same

8 0

3 years ago

Read 2 more answers

Two resistors of resistances R1 and R2, with R2>R1, are connected to a voltage source with voltage V0. When the resistors are

Contact [7]

Answer:

r=0.127

Explanation:

When connected in series

Current = I

When connected in parallel

Current = 10 I

We know that equivalent resistance

In series R = R₁+R₂

in parallel R= R₁R₂/(R₂+ R₁)

Given that voltage is constant (Vo)

V = I R

Vo = I (R₁+R₂) ------------1

Vo = 10 I (R₁R₂/(R₂+ R₁)) -------2

From above equations

10 I (R₁R₂/(R₂+ R₁)) = I (R₁+R₂)

10 R₁R₂ = (R₁+R₂) (R₂+ R₁)

10 R₁R₂ = 2 R₁R₂ + R₁² + R₂²

8 R₁R₂ = R₁² + R₂²

Given that

r = R₁/R₂

Divides by R₂²

8R₁/R₂ = ( R₁/R₂)²+ 1

8 r = r ² + 1

r ² - 8 r+ 1 =0

r= 0.127 and r= 7.87

But given that R₂>R₁ It means that r<1 only.

So the answer is r=0.127

8 0

4 years ago

A balloon of hydrogen is put into to pressure chamber. The initial pressure and volume of hydrogen is 1 atm and .5 cm3. The pres

Elis [28]

Answer:

0.25 cm³.

Explanation:

We shall apply Boyle's law to find the solution . According to it

PV = constant where P is pressure and V is volume of the gas.

P₁ V₁ = P₂V₂

1 x .5 = 2 x V₂

V₂ = 0.25 cm³.

4 0

3 years ago

The main illustration in the video shows the life track of a one-solar mass star. Each point along this track represents _______

REY [17]

Each point along the track of one solar mass star represents the star's surface temperature and luminosity at one time.

<h3>What is the one-solar mass star?</h3>

A star having a mass equal to the mass of the Sun is called a one-solar mass star.

Its life track shows the luminous intensity as well as the surface temperature.

Learn more about one-solar mass star.

brainly.com/question/14984575

#SPJ1

7 0

2 years ago