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ratelena [41]
3 years ago
14

1) Andrea and Chuck are riding on a merry-go-round. Andrea rides on a horse at the outer rim of the circular platform, twice as

far from the center of the circular platform as Chuck, who rides on an inner horse. When the merry-go-round is rotating at a constant angular speed, Andrea's tangential speed is which of the following?a) twice Chuck's.b) the same as Chuck's.c) half of Chuck's.d) impossible to determine.2) When the merry go round is rotating at a constant angular speed, Andrea's tangential speed is:______.a) twice Chuck's.b) the same as Chuck's.c) half of Chuck's.d) impossible to determine.
Physics
1 answer:
telo118 [61]3 years ago
8 0

Explanation:

The tangential speed of Andrea is given by :

v=r\omega

Where

r is radius of the circular path

ω is angular speed

The merry-go-round is rotating at a constant angular speed. Let the new distance from the center of the circular platform is r'

r' = 2r

New angular speed,

v'=r'\omega'\\\\v'=(2r)\omega\\\\v'=2r\omega\\\\v'=2v

New angular speed is twice that of the Chuck's speed.

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Two kids are playing on a newly installed slide, which is 3 m long. John, whose mass is 30 kg, slides down into William (20 kg),
yuradex [85]

Answer:

v=3.564\ m.s^{-1}

\Delta v =2.16\ m.s^{-1}

Explanation:

Given:

  • mass of John, m_J=30\ kg
  • mass of William, m_W=30\ kg
  • length of slide, l=3\ m

(A)

height between John and William, h=1.8\ m

<u>Using the equation of motion:</u>

v_J^2=u_J^2+2 (g.sin\theta).l

where:

v_J = final velocity of John at the end of the slide

u_J = initial velocity of John at the top of the slide = 0

Now putting respective :

v_J^2=0^2+2\times (9.8\times \frac{1.8}{3})\times 3

v_J=5.94\ m.s^{-1}

<u>Now using the law of conservation of momentum at the bottom of the slide:</u>

<em>Sum of initial momentum of kids before & after collision must be equal.</em>

m_J.v_J+m_w.v_w=(m_J+m_w).v

where: v = velocity with which they move together after collision

30\times 5.94+0=(30+20)v

v=3.564\ m.s^{-1} is the velocity with which they leave the slide.

(B)

  • frictional force due to mud, f=105\ N

<u>Now we find the force along the slide due to the body weight:</u>

F=m_J.g.sin\theta

F=30\times 9.8\times \frac{1.8}{3}

F=176.4\ N

<em><u>Hence the net force along the slide:</u></em>

F_R=71.4\ N

<em>Now the acceleration of John:</em>

a_j=\frac{F_R}{m_J}

a_j=\frac{71.4}{30}

a_j=2.38\ m.s^{-2}

<u>Now the new velocity:</u>

v_J_n^2=u_J^2+2.(a_j).l

v_J_n^2=0^2+2\times 2.38\times 3

v_J_n=3.78\ m.s^{-1}

Hence the new velocity is slower by

\Delta v =(v_J-v_J_n)

\Delta v =5.94-3.78= 2.16\ m.s^{-1}

8 0
3 years ago
1. You place an object 63 cm in front of a converging lens, with a 40 cm focal length.
NikAS [45]

Answer:

1.

109.6 cm ,  - 1.74 , real

2.

1.5

Explanation:

1.

d₀ = object distance = 63 cm

f = focal length of the lens = 40 cm

d = image distance = ?

using the lens equation

\frac{1}{f} = \frac{1}{d_{o}} + \frac{1}{d}

\frac{1}{40} = \frac{1}{63} + \frac{1}{d}

d = 109.6 cm

magnification is given as

m = \frac{-d}{d_{o}}

m = \frac{-109.6}{63}

m = - 1.74

The image is real

2

d₀ = object distance = a

d = image distance = - (a + 5)

f = focal length of lens = 30 cm

using the lens equation

\frac{1}{f} = \frac{1}{d_{o}} + \frac{1}{d}

\frac{1}{30} = \frac{1}{a} + \frac{1}{- (a + 5)}

a = 10

magnification is given as

m = \frac{-d}{d_{o}}

m = \frac{- (- (a +5))}{a}

m = \frac{(5 + 10)}{10}

m = 1.5

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OverLord2011 [107]
<h2>Greetings!</h2>

To find this value, you need to remember the speed formula:

3 = 6 / 2

Speed = distance ÷ time

Rearrange to make distance the subject:

Distance = speed * time

Simply plug these values into this:

5.6 * 8.25 = 46.2

<h3>So the player will travel 46.2 metres!</h3>
<h2>Hope this helps!</h2>
7 0
3 years ago
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