Which describes why people on earth can see light from the stars in the sky that are so far away?

A sound source is moving at 80 m/s toward a stationary listener that is standing in still air (a) Find the wavelength of the sou

Setler [38]

Answer:

a. wavelength of the sound, $\vartheta = 1.315\vartheta_{o}$

b. observed frequecy, $\lambda = 0.7604\lambda_{o}$

Given:

speed of sound source, $v_{s}$ = 80 m/s

speed of sound in air or vacuum, $v_{a}$ = 343 m/s

speed of sound observed, $v_{o}$ = 0 m/s

Solution:

From the relation:

v = $\vartheta \lambda$ (1)

where

v = velocity of sound

$\vartheta$ = observed frequency of sound

$\lambda$ = wavelength

(a) The wavelength of the sound between source and the listener is given by:

$\lambda = \frac{v_{a}}{\vartheta }$ (2)

(b) The observed frequency is given by:

$\vartheta = \frac{v_{a}}{v_{a} - v_{s}}\vartheta_{o}$

$\vartheta = \frac{334}{334 - 80}\vartheta_{o}$

$\vartheta = 1.315\vartheta_{o}$ (3)

Using eqn (2) and (3):

$\lambda = \frac{334}{1.315} = \frac{1}{1.315}\frac{v_{a}}{\vartheta_{o}}$

$\lambda = 0.7604\lambda_{o}$

4 0

3 years ago

Please help if you can.

VMariaS [17]

Initially there were 10 bulbs of 60 Watt power

So total power of all bulbs = 60 * 10 = 600 W

now each bulb used for 4 hours daily

so total energy consumed daily

$E = P * t$

$E = 600 * 4 = 2400 Wh$

$E = 2.4 kWh$

now we have total power consumed in 1 year

$E = 365 * 2.4 = 876 kWh$

cost of electricity = 10 cents/ kWh

so total cost of energy for one year

$P_1 = 876 * 10 = 8760 cents = $87.60$

Now if all 60 Watt bulbs are replaced by 30 Watt bulbs

So total power of all bulbs = 30 * 10 = 300 W

now each bulb used for 4 hours daily

so total energy consumed daily

$E = P * t$

$E = 300 * 4 = 1200 Wh$

$E = 1.2 kWh$

now we have total power consumed in 1 year

$E = 365 * 1.2 = 438 kWh$

cost of electricity = 10 cents/ kWh

so total cost of energy for one year

$P_1 = 438 * 10 = 4380 cents = $43.80$

total money saved in 1 year

$Savings = 87.6 - 43.8 = $43.80$

3 0

3 years ago

A ball is kicked horizontally from a 60 meter tall cliff at 10 m/s. How far from

o-na [289]

Hi there!

We can begin by deriving the equation for how long the ball takes to reach the bottom of the cliff.

$\large\boxed{\Delta d = v_it+ \frac{1}{2}at^2}}$

There is NO initial vertical velocity, so:

$\large\boxed{\Delta d= \frac{1}{2}at^2}}$

Rearrange to solve for time:

$2\Delta d = at^2\\\\t = \sqrt{\frac{2\Delta d}{g}}$

Plug in the given height and acceleration due to gravity (g ≈ 9.8 m/s²)

$t = \sqrt{\frac{2(60)}{(9.8)}} = 3.5 s$

Now, use the following for finding the HORIZONTAL distance using its horizontal velocity:

$\large\boxed{d_x = vt}\\\\d_x = 10(3.5) = \karge\boxed{35 m}$

6 0

3 years ago

A truck moving at 13.3 m/s hits a concrete wall. As a result of the collision, a 6-kg wrench moves forwards and strikes the wall

Naddik [55]

Answer:

Explanation:

The velocity of the wrench must be equal to the velocity of the truck . So momentum of the wrench before it hits the wall

= mv = 6 x 13.3 = 79.8 kg m /s

If resisting force of wall be F , impulse on the wrench = F x time

= F x .07

Impulse = change in momentum of the wrench = mv - 0 = mv = 79.8 kgm/s

So F x .07 = 79.8

F = 1140 N .

8 0

3 years ago

Coasting due west on your bicycle at 8 m/s, you encounter a sandy patch of road 7.2m

jeka57 [31]

Answer:

V = (v1 + v2) / 2 = (8 + 6.5) / 2 = 7.25 m/s average speed

t = 7.2 / 7.25 = .993 sec time to cross patch

a = (v2 - v1) / t = (6.5 - 8) / .993 = -1.51 m/s^2 or 1.5 m/s^2

8 0

2 years ago