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professor190 [17]
3 years ago
11

A purse at radius 2.30 m and a wallet at radius 3.45 m travel in uniform circular motion on the floor of a merry-go-round as the

ride turns. They are on the same radial line. At one instant, the acceleration of the purse is . At that instant and in unit-vector notation, what is the acceleration of the wallet?
Physics
1 answer:
ivolga24 [154]3 years ago
4 0

Answer:

The acceleration of the wallet is 3\hat{i}+6\hat{j}

Explanation:

Given that,

Radius of purse r= 2.30 m

Radius of wallet r'= 3.45 m

Acceleration of the purse a=2\hat{i}+4.00\hat{j}

We need to calculate the acceleration of the wallet

Using formula of acceleration

a=r\omega^2

Both the purse and wallet have same angular velocity

\omega=\omega'

\sqrt{\dfrac{a}{r}}=\sqrt{\dfrac{a'}{r'}}

\dfrac{a}{r}=\dfrac{a'}{r'}

\dfrac{a'}{a}=\dfrac{r'}{r}

\dfrac{a'}{a}=\dfrac{3.45}{2.30}

\dfrac{a'}{a}=\dfrac{3}{2}

a'=\dfrac{3}{2}\times(2\hat{i}+4.00\hat{j})

a'=3\hat{i}+6\hat{j}

Hence, The acceleration of the wallet is 3\hat{i}+6\hat{j}

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s344n2d4d5 [400]

Answer: B

Explanation: I think

3 0
3 years ago
In a Broadway performance, an 77.0-kg actor swings from a R = 3.65-m-long cable that is horizontal when he starts. At the bottom
krek1111 [17]

Answer: h =1.22 m

Explanation:

from the question we were given the following

mass of performer ( M1 ) = 77 kg

length of cable ( R ) = 3.65 m

mass of costar ( M2 ) = 55 kg

maximum height (h) = ?

acceleration due to gravity (g) = 9.8 m/s^2  (constant value)

We first have to find the velocity of the performer. From the work energy theorem work done = change in kinetic energy

work done = 1/2 x mass x ( (final velocity)^2 - (initial velocity)^2 )

initial velocity is zero in this case because the performer was at rest before swinging, therefore

work done = 1/2 x 77 x ( v^2 - 0)

work done = 38.5 x ( v^2 ) ......equation 1

work done is also equal to m x g x distance ( the distance in this case is the length of the rope), hence equating the two equations we have

m x g x R =  38.5 x ( v^2 )

77 x 9.8 x 3.65 =  38.5 x ( v^2 )

2754.29 = 38.5 x ( v^2 )

( v^2 ) =  71.54

v = 8.4 m/s  ( velocity of the performer)

After swinging, the performer picks up his costar and they move together, therefore we can apply the conservation of momentum formula which is

initial momentum of performer (P1) + initial momentum of costar (P2) = final momentum of costar and performer after pick up (Pf)  

momentum = mass x velocity therefore the equation above now becomes

(77 x 8.4) + (55 x 0) = (77 +55) x Vf  

take note the the initial velocity of the costar is 0 before pick up because he is at rest

651.3 = 132 x Vf

Vf = 4.9 m/s

the performer and his costar is 4.9 m/s after pickup

to finally get their height we can use the energy conservation equation for from after pickup to their maximum height. Take note that their velocity at maximum height is 0

initial Kinetic energy + Initial potential energy = Final potential energy + Final Kinetic energy

where

kinetic energy = 1/2 x m x v^2

potential energy  = m x g x h

after pickup they both will have kinetic energy and no potential energy, while at maximum height they will have potential energy and no kinetic energy. Therefore the equation now becomes

initial kinetic energy = final potential energy

(1/2 x (55 + 77) x 4.9^2) + 0 = ( (55 + 77) x 9.8 x h) + 0

1584.7 = 1293 x h

h =1.22 m

3 0
3 years ago
What is the centripetal acceleration of a small laboratory centrifuge in which the tip of the test tube is moving at 19.0 meters
Artist 52 [7]

Answer:3.61(10)^{3} \frac{m}{s^{2}}

The centripetal acceleration a_{c} of an object moving in a uniform circular path is given by the following equation:

a_{c}=\frac{V^{2}}{r}

Where:

V=19m/s is the velocity

r=10cm=0.1m is the radius of the circle

a_{c}=\frac{(19m/s)^{2}}{0.1m}

a_{c}=3610m/s^{2}=3.61(10)^{3}m/s^{2}

8 0
3 years ago
You are standing on a street corner and you hear an ambulance siren. The pitch of the siren seems to be getting higher. What do
Gre4nikov [31]

B.The siren will get louder and higher as the ambulance moves towards you


Hope this helped !

4 0
3 years ago
Read 2 more answers
A truck is travelling at 30km/hr with engine delivering a driving force of 800n to the road
Pavel [41]

Power = (force) x (distance / time) = force x speed .

We know the force = 800N.
We have a speed = 30km/hr, but in order to use it in the power formula,
it has to be in meters/second, so we have some work to do first.

(30 km/hr) x (1,000 m/km) x (1 hr / 3,600 sec) = 300 / 36  m/sec .

Power = (force) x (speed) = (800 N) x (300/36 m/s) = <em>6-2/3 kilowatts </em>

Work = (power) x (time) = (6,666-2/3 joule/sec) x (25sec) = <em>166,666-2/3 joules</em>.

The figure for power is slightly weird ... 746 watts = 1 horsepower,
so the truck's engine is only delivering about 8.9 horsepower.
Very fuel-efficient, but I don't think they drive trucks that way.


4 0
3 years ago
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