All categories

3 years ago

How does the speed of a wave relate to its wavelength and frequency?*

Physics

1 answer:

grigory [225]3 years ago

7 0

Answer: Wavespeed (V) = Frequency F x wavelength λ (V = F λ)

Explanation:

The wavespeed is the distance covered by a wave in one second. It is measured in metre per second, and represented by the symbol V. It is directly proportional to the wavelength and frequency

i.e Velocity (V) = Frequency F x wavelength λ

V = F λ

For instance:

Assume wavelength (λ)= 20 m

Frequency = 10 Hz.

To get the wavespeed, use the formula

V = F λ

V = 20 metres x 10 hertz

V = 200 metres per second

Thus, the wave travels at a speed of 200 metres per second

You might be interested in

Please help need answer asp

liraira [26]

The 3rd one Challenging
hope it helps!!

6 0

3 years ago

A wheel turns through 5.5 revolutions while being accelerated from rest at 20rpm/s.(a) What is the final angular speed ? (b) How

alina1380 [7]

Answer:

(a) The final angular speed is 12.05 rad/s

(b) The time taken to turn 5.5 revolutions is 5.74 s

Explanation:

Given;

number of revolutions, θ = 5.5 revolutions

acceleration of the wheel, α = 20 rpm/s

number of revolutions in radian is given as;

θ = 5.5 x 2π = 34.562 rad

angular acceleration in rad/s² is given as;

$\alpha = \frac{20 \ rev}{min} *\frac{1}{s} *(\frac{2\pi \ rad}{1 \ rev } *\frac{1 \ min}{60 \ s}) \\\\\alpha = 2.1 \ rad/s^2$

(a)

The final angular speed is given as;

$\omega _f^2 = \omega_i ^2 + 2\alpha \theta\\\\\omega _f^2 = 0 + 2\alpha \theta\\\\\omega _f^2 = 2\alpha \theta\\\\\omega _f = \sqrt{2\alpha \theta}\\\\ \omega _f = \sqrt{2(2.1) (34.562)}\\\\ \omega _f = 12.05 \ rad/s$

(b) the time taken to turn 5.5 revolutions is given as

$\omega _f = \omega _i + \alpha t\\\\12.05 = 0 + 2.1t\\\\t = \frac{12.05}{2.1} \\\\t = 5.74 \ s$

3 0

2 years ago

A string under a tension of 36 N is used to whirl a rock in a horizontal circle of radius 3.6 m at a speed of 16.12 m/s. The str

ExtremeBDS [4]

Answer:

6010.457N

Explanation:

Centripetal acceleration = a= V²/R

At a radius of 3.6m and velocity of 16.12m/s,

Acceleration is

a = 16.12²/ 3.6 = 72.182 m/s²

Force = Mass (m) * Acceleration (a)

36 = m * 72.182

m = 36/72.182

At breaking point

Radius = 0.468 m and Velocity = 75.1 m/s

a = V²/R = 75.1²/0.468

a = 12051.3 m/s

F = Mass(m) * Acceleration (a)

F = m * 12051.3

m = F/ 12051.3

Settings the ratio of mass equal

m = m

=> 36/72.182 = F/12051.3

F = 12051.3 * 36/72.182

F = 6010.457N

3 0

3 years ago

An archer shoots an arrow with a velocity of 45.0m/s at an angle of 50.0degrees with the horizontal.An assistant standing on the

garri49 [273]

Answer:

a) u = 30.29 m/s

b) t = 2.09 s

Explanation:

given,

velocity = 45 m/s

angle (θ) = 50°

horizontal velocity = 45 cos 50°

time taken to reach 150 m.

times = $\dfrac{150}{45 cos 50^0}$

t = 5.19 s

a) height of arrow

$s = u t +\dfrac{1}{2}gt^2$

$s = v sin \theta \times t+\dfrac{1}{2}gt^2$

$s = 45 sin 50^0 \times 5.19 -\dfrac{1}{2}\times 9.81\times 5.19^2$

s = 46.78 m

v² - u² = 2 g s

u² = 2 × 9.81 × 46.78

u = 30.29 m/s

b) time taken by the apple = $\dfrac{u}{g}=\dfrac{30.29}{9.81}$

= 3.09 s

time after which it has to be thrown = 5.19-3.09 = 2.1 s

5 0

3 years ago

Which items in this image are electrically conductive? Check all that apply.

balu736 [363]

Answer: a and d

Explanation: A.) the power lines themselves

B.) the wooden pole that supports the lines

C.) the rubber soles on the worker’s boots

D.) the metal tools the worker uses

E.) the wooden ladder leaning against the lines

4 0

3 years ago