How is the wave speed calculated?

A bicyclist bikes the 90 mi to a city averaging a certain speed. The return trip is made at a speed that is 1 mph slower. Total

Lynna [10]

Answer:

his speeds while going to city is 10 mph and while his round trip the speed will be 9 mph

Explanation:

Let say the speed of the bicycle while he moves towards the city is "v"

now the speed of the round trip must be smaller by 1 mph

so its speed for round trip will be

$v_2 = v - 1$

now we know that total time of the motion is 19 hr

so we will have

$t_1 = \frac{90}{v}$

$t_2 = \frac{90}{v - 1}$

so we will have

$t_1 + t_2 = 19 hr$

$\frac{90}{v} + \frac{90}{v-1} = 19$

$90(2v - 1) = 19(v^2 - v)$

$19 v^2 - 199 v + 90 = 0$

by solving above equation we have

$v = 10 mph$

so his speeds while going to city is 10 mph and while his round trip the speed will be 9 mph

5 0

3 years ago

Which statement is true of equinoxes? They occur in June and December. Days and nights are equal in length everywhere. The lengt

Zanzabum

<span>Days and nights are equal in length everywhere.(gradpoint)</span>

8 0

3 years ago

Read 2 more answers

A car slows down from speed of 72

Nastasia [14]

Explanation:

Given parameters:

Initial velocity = 72km/hr

Final velocity = 0km/hr

Time taken = 25s

Unknown:

Acceleration = ?

Solution:

To solve this problem, convert km/hr to m/s;

1000m = 1km

3600s = 1hr

72km/hr;

1km/hr = 0.278m/s

72km/hr = 0.278 x 72 = 20.02m/s

Acceleration is the change in velocity divided by the time taken;

Acceleration = $\frac{final velocity - initial velocity }{time}$

Acceleration = $\frac{0 - 20.02}{25}$ = -0.8m/s

The car is actually decelerating at a rate of 0.8m/s

5 0

2 years ago

A pair of slits separated by 1 mm, are illuminated with monochromatic light of wavelength 411 nm. The light falls on a screen 1.

Ilya [14]

Answer:

$t = 0.192 \mu m$

Explanation:

Path difference due to a transparent slab is given as

$\Delta x = (\mu - 1) t$

here we know that

$\mu = 1.79$

now total shift in the bright fringe is given as

$Shift = \frac{D(\mu - 1)t}{d}$

Also we know that the fringe width of maximum intensity is given as

$\delta x = \frac{\lambda D}{d}$

now we have

$\frac{D}{d} = \frac{\delta x}{\lambda}$

now the shift is given as

$Shift = \frac{(\mu - 1) t \delta x}{\lambda}$

given that the shift is

$Shift = 0.37 \delta x$

here we have

$0.37 \delta x = \frac{(\mu - 1) t \delta x}{\lambda}$

now plug in all values in it

$0.37 = \frac{(1.79 - 1) t}{411 \times 10^{-9}}$

$t = 0.192 \times 10^{-6} m$

$t = 0.192 \mu m$

3 0

3 years ago

A small spinning asteroid is in a circular orbit around a star, much like the earth's motion around our sun. The asteroid has a

Fudgin [204]

Answer:

Temperature will be 305 K

Explanation:

We have given The asteroid has a surface area $A=7.70m^2$

Power absorbed P = 3800 watt

Boltzmann constant $\sigma =5.67\times 10^{-8}Wm/K^4$

According to Boltzmann rule power radiated is given by

$P=\sigma AT^4$

$3800=5.67\times 10^{-8}\times 7.70\times T^4$

$T^4=87.0381\times 10^8$

$T=305K$

So temperature will be 305 K

8 0

3 years ago