What two factors are multiplied to determine wave speed

Mr. Stephenson drives his car from Dallas to Town X, a distance of 437 km. The trip takes 7.27 hours. What is the average speed

klasskru [66]

Average speed = (distance covered) / (time to cover the distance).

Distance covered = 437 km = 437,000 meters

Time to cover the distance = (7.27 hrs) x (3600 sec/hr) = 26,172 seconds

Average speed = (437,000 meters) / (26,172 seconds)

Average speed = 16.7 m/s

8 0

3 years ago

A person lowers a bucket into a well by turning the hand crank, as the drawing illustrates. The crank handle moves with a consta

dimulka [17.4K]

Answer:

0.453 m/s

Explanation:

Assuming the handle has diameter of 0.4 m while inner part diameter is 0.1 m then the circumference of outer part is $\pi d_h$ where d is diameter and subscript h denote handle. By substituting 0.4 for the handle's diameter then cirxumference of outer part is $\pi\times 0.4\approx 1.256 m$

The rate of rotation will then be 1.81/1.256=1.441 rev/s

Similarly, circumference of inner part will be $\pi d_i$ where subscript i represent inner. Substituting 0.1 for inner diameter then

$\pi\times 0.1\approx 0.3142 m$

The rate of rotation found for outer handle applies for inner hence speed will be 0.3142*1.441=0.453 m/s

7 0

2 years ago

So far in your life, you may have assumed that as you are sitting in your chair right now, you are not accelerating. However, th

tia_tia [17]

Answer:

a) $a=33.73mm/s^{2}$

b) mg>N

c) $\%_{change}=0.343\%$

d) $a=24.07mm/s^{2}$

Explanation:

In order to solve part a) of the problem, we can start by drawing a free body diagram of the presented situation. (see attached picture).

In this case, we know the centripetal acceleration is given by the following formula:

$a_{c}=\omega ^{2}r$

where:

$\omega=\frac{2\pi}{T}$

we know the period of rotation of the earth is about 24 hours, so:

$T=24hr*\frac{3600s}{1hr}=86400s$

so we can now find the angular speed:

$\omega=\frac{2\pi}{86400s}$

$\omega=72.72x10^{-6} rad/s^{2}$

So the centripetal acceleration will be:

$a_{c} =(72.72x10^{-6} rad/s^{2})^{2}(6478x10^{3}m)$

which yields:

$a_{c}=33.73mm/s^{2}$

b)

In order to answer part b, we must draw a free body diagram of us sitting on a chair. (See attached picture.)

So we can do a sum of forces in equilibrium:

$\sum F=0$

so we get that:

$N-mg+ma_{c} = 0$

and solve for the normal force:

$N=mg-ma_{c}$

In this case, we can clearly see that:

$mg>mg-ma_{c}$

therefore mg>N

This is because the centripetal acceleration is pulling us upwards, that will make the magnitude of the normal force smaller than the product of the mass times the acceleration of gravity.

c)

So let's calculate our weight and normal force:

Let's say we weight a total of 60kg, so:

$mg=(60kg)(9.81m/s^{2})=588.6N$