Which of the following is not a type of system?

Distance = 6 km south<br><br> 60 minutes<br><br> What was the average velocity

blsea [12.9K]

v = x/t

v = average velocity, x = displacement, t = elapsed time

Given values:

x = 6km south, t = 60min

Plug in and solve for v:

v = 6/60

v = 0.1km/min south

3 0

3 years ago

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HELPPPPPP PLEASEEEEEEE-

Alex787 [66]

Answer:

One piece has a north pole only, and the other piece has à soutn pole only.

Explanation:

mark me brainliest!!

8 0

2 years ago

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The near point of an eye is 56.0 cm. A corrective lens is to be used to allow this eye to focus clearly on objects at the distan

irakobra [83]

Answer:

Explanation:

Near point = 56 cm .

near point of healthy person = 25 cm

person suffers from long sightedness

convex lens will be required .

object distance u = 25 cm

image distance v = 56 cm

both will be negative as both are in front of the lens.

lens formula

I/v - 1 / u = 1/f

- 1/56 +1/25 = 1/f

- .01785 + .04 = 1/f

1/f = .02215

f = 45.15 cm .

4 0

3 years ago

A train whistle is heard at 300 Hz as the train approaches town. The train cuts its speed in half as it nears the station, and t

spin [16.1K]

Answer:

The speed of the train before and after slowing down is 22.12 m/s and 11.06 m/s, respectively.

Explanation:

We can calculate the speed of the train using the Doppler equation:

$f = f_{0}\frac{v + v_{o}}{v - v_{s}}$

Where:

f₀: is the emitted frequency

f: is the frequency heard by the observer

v: is the speed of the sound = 343 m/s

$v_{o}$ : is the speed of the observer = 0 (it is heard in the town)

$v_{s}$ : is the speed of the source =?

The frequency of the train before slowing down is given by:

$f_{b} = f_{0}\frac{v}{v - v_{s_{b}}}$ (1)

Now, the frequency of the train after slowing down is:

$f_{a} = f_{0}\frac{v}{v - v_{s_{a}}}$ (2)

Dividing equation (1) by (2) we have:

$\frac{f_{b}}{f_{a}} = \frac{f_{0}\frac{v}{v - v_{s_{b}}}}{f_{0}\frac{v}{v - v_{s_{a}}}}$

$\frac{f_{b}}{f_{a}} = \frac{v - v_{s_{a}}}{v - v_{s_{b}}}$ (3)

Also, we know that the speed of the train when it is slowing down is half the initial speed so:

$v_{s_{b}} = 2v_{s_{a}}$ (4)

Now, by entering equation (4) into (3) we have:

$\frac{f_{b}}{f_{a}} = \frac{v - v_{s_{a}}}{v - 2v_{s_{a}}}$

$\frac{300 Hz}{290 Hz} = \frac{343 m/s - v_{s_{a}}}{343 m/s - 2v_{s_{a}}}$

By solving the above equation for $v_{s_{a}}$ we can find the speed of the train after slowing down:

$v_{s_{a}} = 11.06 m/s$

Finally, the speed of the train before slowing down is:

$v_{s_{b}} = 11.06 m/s*2 = 22.12 m/s$

Therefore, the speed of the train before and after slowing down is 22.12 m/s and 11.06 m/s, respectively.

I hope it helps you!

7 0

2 years ago

How much kinetic energy does a 50 kg object have if its moving at a velocity of 2 m/s?

bulgar [2K]

Answer:

C) 100 joules

Explanation:

The kinetic energy of an object is given by:

$K=\frac{1}{2}mv^2$

where m is the mass of the object and v its speed.

In this problem, we have an object of mass m = 50 kg and v = 2 m/s, so by using the formula we can find its kinetic energy:

$K=\frac{1}{2}(50 kg)(2 m/s)^2=100 J$

3 0

2 years ago