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kondaur [170]
3 years ago
14

You are sitting on a merry-go-round at a distance of 2m from its center. It spins 15 times in 3 min. What distance do you move a

s you make one revolution? What is your angular speed in RPM? What is your linear speed in meters per SECOND?
Physics
1 answer:
soldier1979 [14.2K]3 years ago
8 0

Answer:

A) 12.57 m

B) 5 RPM

C) 3.142 m/s

Explanation:

A) Distance covered in 1 Revolution:

The formula that gives the relationship between the arc length or distance covered during circular motion to the angle subtended or the revolutions, is given as follows:

s = rθ

where,

s = distance covered = ?

r = radius of circle = 2 m

θ = Angle = 2π radians  (For 1 complete Revolution)

Therefore,

s = (2 m)(2π radians)

<u>s = 12.57 m</u>

B) Angular Speed:

The formula for angular speed is given as:

ω = θ/t

where,

ω = angular speed = ?

θ = angular distance covered = 15 revolutions

t = time taken = 3 min

Therefore,

ω = 15 rev/3 min

<u>ω = 5 RPM</u>

C) Linear Speed:

The formula that gives the the linear speed of an object moving in a circular path is given as:

v = rω

where,

v = linear speed = ?

r = radius = 2 m

ω = Angular Speed in rad/s = (15 rev/min)(2π rad/1 rev)(1 min/60 s) = 1.571 rad/s

Therefore,

v = (2 m)(1.571 rad/s)

<u>v = 3.142 m/s</u>

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Photons of wavelength 65.0 pm are Compton-scattered from a free electron which picks up a kinetic energy of 0.75 keV from the co
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0.6764*10^-10m

Explanation:

Using E= hc/wavelength

(4.14x10^-15)x(3.0x10^8)/(65x10^-12)=0.1911x10^5 eV=19.1 keV

So subtract the calculated energy from the given energy of scattered photons

9.11-0.75=18.36 keV

To find wavelength

Wavelength= hc/ E

[(4.14x 10^-15)x (3.0x10^8)]/(18.36*10^3) =0.6764^-10 m

8 0
3 years ago
A solid object has a mass of 104 kg and a volume of 1,278 m3. What is its density? 0.081 kg/m3 12.29 g/cm3 132912.00 g/cm3 canno
REY [17]
The density is 81.4 g/m3. Before you start plugging numbers into the density formula (D=M/V), you should convert 104 kg to grams, which ends up being 104,000 grams. Then you can plug in the 104,000 grams and 1,278 m3 into the formula. When you divide the mass by the volume, you get a really long decimal, which you can round to 81.4 g/m3, or whatever place your teacher wants you to round to.
4 0
3 years ago
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How far from a converging lens with a focal length of 16 cm should an object be placed to produce a real image which is the same
Feliz [49]

Answer:

32 cm

Explanation:

f = focal length of the converging lens = 16 cm

Since the lens produce the image with same size as object, magnification is given as

m = magnification = - 1

p = distance of the object from the lens

q = distance of the image from the lens

magnification is given as

m = - q/p

- 1 = - q/p

q = p                                    eq-1

Using the lens equation, we get

1/p + 1/q = 1/f

using eq-1

1/p + 1/p = 1/16

p = 32 cm

4 0
3 years ago
1) On the way to the moon, the Apollo astro-
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(1) You must find the point of equilibrium between the two forces,

<span>G * <span><span><span>MT</span><span>ms / </span></span><span>(R−x)^2 </span></span>= G * <span><span><span>ML</span><span>ms / </span></span><span>x^2
MT / (R-x)^2 = ML / x^2

So,

x = R * sqrt(ML * MT) - ML / (MT - ML)
R = is the distance between Earth and Moon.

</span></span></span>The result should be,
x = 3.83 * 10^7m
from the center of the Moon, and 

R - x = 3.46*10^8 m
from the center of the Earth.


(2) As the distance from the center of the Earth is the number we found before,
d = R - x = 3.46*10^8m
The acceleration at this point is
g = G * MT / d^2
g = 3.33*10^-3 m/s^2
6 0
3 years ago
What does this picture mean?
ivann1987 [24]
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be aware of large trucks on the roadway
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