How does a parallel circuit differ from a series circuit?

juan aims his boat directly across a river flowing at 8mi/hr. juan's boat travels at 6 miles/hr in still water. how was does jua

son4ous [18]

6mph:

Suppose the boat is traveling on a y axis. The river flow acts on the x axis. Motion on each axes are independent. The linear speed of the boat is not changed. Furthermore the projectile motion is changing, but you're specifically asking about the linear speed of the boat which is unchanged.

4 0

3 years ago

According to Newton's Brd Law, an object's momentum depends on it's velocity and mass, 4 dog-sled teams

Kryger [21]

Answer:

Team A has the greatest momentum

Explanation:

The momentum of an object is a vector quantity given by

$p=mv$

where

m is the mass of the object

v is its velocity

In this problem, we have to compare the momenta of the different sleds.

We have:

A) Sled Team A 48 kg moving at 10m/s:

m = 48 kg

v = 10 m/s

p = (48)(10) = 480 kg m/s

B) Sled Team B 14 kg moving at 18m/s

m = 14 kg

v = 18 m/s

p = (14)(18) = 252 kg m/s

C) Sled Team C 28 kg moving at 12m/s

m = 28 kg

v = 12 m/s

p = (28)(12) = 336 kg m/s

D) Sled Team D 22 kg moving at 12m/s

m = 22 kg

v = 12 m/s

p = (22)(12) = 264 kg m/s

So, team A has the greatest momentum.

4 0

3 years ago

What kind of model is shown below?

Kay [80]

Still there ain’t no picture

7 0

3 years ago

Read 2 more answers

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:

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

and let's calculate the normal force:

$N=m(g-a_{c})$

$N=(60kg)(9.81m/s^{2}-33.73x10^{-3}m/s^{2})$

N=586.58N

so now we can calculate the percentage change:

$\%_{change} = \frac{mg-N}{mg}x100\%$

so we get:

$\%_{change} = \frac{588.6N-586.58N}{588.6N} x 100\%$

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

which is a really small change.

d) In order to find this acceleration, we need to start by calculating the radius of rotation at that point of earth. (See attached picture).

There, we can see that the radius can be found by using the cos function:

$cos \theta = \frac{AS}{h}$

In this case:

$cos \theta = \frac{r}{R_{E}}$

so we can solve for r, so we get:

$r= R_{E}cos \theta$

in this case we'll use the average radius of earch which is 6,371 km, so we get:

$r = (6371x10^{3}m)cos (44.4^{o})$

which yields:

r=4,551.91 km

and now we can calculate the acceleration at that point:

$a=\omega ^{2}r$

$a=(72.72x10^{-6} rad/s)^{2}(4,551.91x10^{3}m$

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

5 0

3 years ago

A dog can hear sounds in the range from 15 to 50,000 hz. what wavelength corresponds to the lower cut-off point of the sounds at

Ira Lisetskai [31]

The frequency corresponding to the lower cut-off point is
$f=15 Hz$
The speed of sound at a temperature of $20^{\circ}C$ is approximately
$v=343 m/s$

Therefore we can use the relationship between speed of a wave, frequency and wavelength to calculate the corresponding wavelength:
$\lambda= \frac{v}{f}= \frac{343 m/s}{15 Hz}=22.87 m$
So, the corresponding wavelength is 22.87 m.

3 0

3 years ago