All categories

3 years ago

The set of points in a plane that are equidistant from a given point form a ____.

Physics

2 answers:

il63 [147K]3 years ago

8 0

Answer:

Circle

Explanation:

gtnhenbr [62]3 years ago

6 0

Circle

.......................

You might be interested in

A 100-kg merry-go-round in the shape of a uniform, solid, horizontal disk of radius 1.50 m is set in motion by wrapping a rope a

olchik [2.2K]

Answer:

The force required is equal to $30\pi N$ .

Explanation:

We know that

Torque = force × perpendicular distance

= F × R = I× $\alpha$

I = M× $\frac{R^{2} }{2}$

From above equation $\alpha = \frac{FR}{I}$ .............. 1

We know that

$w=w_{o} +\alpha t$ ...........................2

Given that t= 2 sec

$w_{o} = 0$

$w=0.4 rev/s = 0.8\pi rad/s$

From equation 1 and 2 , we get

F = $\frac{wI}{RT}$ = $\frac{w\times m \times r}{2t}$

Upon substituting the above values we will be getting

F = $30\pi N$

3 0

3 years ago

Express 9/15 as a percentage

Zanzabum

The answer is 60%. If you divide the 9 and 5, you will get 0.6. So you just move the decimal point to the right twice. (:

7 0

3 years ago

Read 2 more answers

PLZ HELP ON #22-26!!!! <br><br>Please explain why and how you got your answer.

AleksAgata [21]

22. a - (vf^2 - vi^2)/(2d)
a = (0 - 23^2)/(170)
a = -3.1 m/s^2

23. Find the time (t) to reach 33 m/s at 3 m/s^2
33-0/t = 3
33 = 3t
t = 11 sec to reach 33 m/s^2
Find the av velocuty: 33+0/2 = 16.5 m/s
Dist = 16.5 * 11 = 181.5 meters to each 33m/s speed. Runway has to be at least this long.

24. The sprinter starts from rest. The average acceleration is found from:
(Vf)^2 = (Vi)^2 = 2as ---> a = (Vf)^2 - (Vi)^2/2s = (11.5m/s)^2-0/2(15.0m) = 4.408m/s^2 estimated: 4.41m/s^2
The elapsed time is found by solving
Vf = Vi + at ----> t = vf-vi/a = 11.5m/s-0/4.408m/s^2 = 2.61s

25. Acceleration of car = v-u/t = 0ms^-1-21.0ms^-1/6.00s = -3.50ms^-2
S = v^2 - u^2/2a = (0ms^-1)^2-(21.0ms^-1)^2/2*-3.50ms^-2 = 63.0m

26. Assuming a constant deceleration of 7.00 m/s^2
final velocity, v = 0m/s
acceleration, a = -7.00m/s^2
displacement, s - 92m
Using v^2 = u^2 - 2as
0^2 - u^2 + 2 (-7.00) (92)
initial velocity, u = sqrt (1288) = 35.9 m/s
This is the speed pf the car just bore braking.

I hope this helps!!

5 0

3 years ago

In a double slit experiment, the distance between the slits is 0.2 mm and the distance to the screen is 100 cm. What is the phas

Anni [7]

Answer:

The phase difference is $\Delta \phi = 180^o$

Explanation:

From the question we are told that

The distance between the slits is $d = 0.2 \ mm = \frac{0.2}{1000} = 0.2 *10^{-3} \ m$

The distance to the screen is $D = 100 cm = \frac{100}{100} = 1 \ m$

The wavelength is $\lambda = 400nm$

The distance of the wave from the central maximum is $L = 5mm = 5*10^{-3} m$

Generally the path difference of this waves is mathematically represented as

$y = d sin \theta$

Here $\theta$ is the angle between the the line connecting the mid-point of the slits with the screen and the line connecting the mid-point of the slits to the central maximum