3. Control variables are those that are changed in an experiment to test

You are standing a distance of 17.0 meters from the center of a merry-go-round. The merry-go-round takes 9.50 seconds to go comp

GenaCL600 [577]

Answer:

(a)11.24 m/s

(b)7.44 m/s

(c)409 N

(d) $539.55\mu$

(e) 0

Explanation:

The period for 1 circle $2\pi$ of the merry go around is 9.5s. It means the angular speed is:

$\omega = \theta / t = 2\pi / 9.5 \approx 0.661 rad/s$

(a)The speed is

$v = \omega * R = 0.661 * 17 = 11.24 m/s$

(b) Centripetal acceleration:

$a = \frac{v^2}{R} = \frac{11.24^2}{17} = 7.44 m/s^2$

(c) Magnitude of the force that keeps you go around at this acceleration

$F = ma = 55 * 7.44 = 409 N$

(d) let the coefficient of friction by $\mu$ . The frictional force shall be this coefficient multiplied by normal force reverting gravity of the man

$F_f = mg\mu = 55*9.81\mu = 539.55\mu$

3 0

3 years ago

4. The blades on a fan have a frequency of 15 Hz.

vichka [17]

Answer:

a) 4500 cycles b) 0.0667s c) 6.67s

Explanation:

a) 15 Hz= 15 cycles/ s

5 mins= 300s

15 cycles/s * 300s= 4500 cycles

b) Period= 1/ frequency

Period= 1/ 15 cycles/s

Period= 0.0667s

c) Period * number of revolutions= time

0.0667 * 100= 6.67s

6 0

3 years ago

Sonar signals and infrared light are are used to send messages to submarines deep under water. If we compare the two signals, wh

WARRIOR [948]

What are the statement choices?

6 0

3 years ago

As the object's particles move faster, the thermal energy increases, causing the _________ to also increase. *

ryzh [129]

Temperature that will be my answer number 1

7 0

3 years ago

Read 2 more answers

A double-slit experiment uses light of wavelength 650 nm with a slit separation of 0.100 mm and a screen placed 4.0 m away. a) W

dezoksy [38]

Answer:

Explanation:

a ) Slit separation d = .1 x 10⁻³ m

Screen distance D = 4 m

wave length of light λ = 650 x 10⁻⁹ m

Width of central fringe = λ D / d

= $\frac{650\times10^{-9}\times4}{.1\times10^{-3}}$

= 26 mm

b ) Distance between 1 st and 2 nd bright fringe will be equal to width of dark fringe which will also be equal to 26 mm

c ) Angular separation between the central maximum and 1 st order maximum will be equal to angular width of fringe which is equal to

λ / d

= $\frac{650\times10^{-9}}{.1\times10^{-3}}$

= 6.5 x 10⁻³ radian.

8 0

3 years ago