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jolli1 [7]
3 years ago
12

A steel ball is whirled on the end of a chain in a horizontal circle of radius R with a constant period T. If the radius of the

circle is then reduced to 0.75R, while the period remains T, what happens to the centripetal acceleration of the ball?
Physics
1 answer:
CaHeK987 [17]3 years ago
4 0

Answer:

<em>The centripetal acceleration will be increased to 1.33 of its initial state. </em>

Explanation:

<h3>Centripetal acceleration</h3>

The Centripetal acceleration of  an object is the acceleration of the object along a circular path  moving towards the center of the circular path. The centripetal acceleration is represented in the equation bellow

a_{c}  = \frac{V^{2} }{r} ...................................... 1

where a_{c} is the centripetal acceleration  

          v is the tangential velocity

          and r is the radius.

<h3>How the Change of Radius Affects the Centripetal Acceleration</h3>

Reference to equation 1 the centripetal acceleration ( a_{c}) is inversely proportional (y = \frac{k}{x}) to the radius of the circle or path. this means that when the radius increases the centripetal acceleration reduces and when the radius reduces the centripetal acceleration increases. The radius was reduced to 0.75R in the question that will amount to 1.33a_{c} increase in the centripetal acceleration. This can be obtained by multiplying the centripetal acceleration by the inverse of 0.75 which is 1.33.

Therefore, when the radius is reduced by 0.75R , the centripetal acceleration of the steel ball will increase by 1.33a_{c}. since the period is kept constant

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Pavlova-9 [17]

Answer:1.71 m/s

Explanation:

Given

mass of Susan m=12 kg

Inclination \theta =30^{\circ}

Tension T=29 N

coefficient of Friction \mu =0.18

Resolving Forces Along x axis

F_x=T\cos \theta -f_r

where f_r=friction\ Force  

F_y=mg-N-T\sin \theta

since there is no movement in Y direction therefore

N=mg-T\sin \theta

and f_r=\mu N

Thus F_x=T\cos \theta -\mu N

F_x=29\cos (30)-\0.18\times (12\times 9.8-29\sin (30))                

F_x=25.114-18.558

F_x=6.556 N

Work done by applied Force is equal to change to kinetic Energy

F_x\cdot x=\frac{1}{2}\cdot mv_f^2-\frac{1}{2}\cdot mv_i^2

6.556\times 2.7=\frac{1}{2}\cdot 12\times v_f^2

v_f^2=\frac{6.556\times 2.7\times 2}{12}

v_f^2=2.95

v_f=1.717 m/s        

8 0
3 years ago
How does temperature increase?
vovikov84 [41]
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6 0
3 years ago
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The iron nail’s mass is 16 grams and its temperature drops 650 C when dropped into the water. How much heat energy did the iron
Mice21 [21]

The heat energy transferred by the iron nail is 4680 J

Explanation:

The thermal energy transferred by a substance to another substance is given by the equation

Q=mC\Delta T

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m is the mass of the substance

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For the iron nail in this problem, we have:

m = 16 g

C=0.450 J/g^{\circ}C

\Delta T = -650^{\circ}C

So, the amount of heat energy given off by the nail is

Q=(16)(0.450)(-650)=-4680 J

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Learn more about specific heat capacity:

brainly.com/question/3032746

brainly.com/question/4759369

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5 0
3 years ago
Which layer of the earths atmosphere contains the ozone layer?
Burka [1]
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8 0
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Helpp pleaseeeeee …..
Lilit [14]

Answer:

300 cos 30 = 40 a + 40 * .2 * 10

Total force = mass * acceleration + frictional force

260 = 40 a + 80

a = 180 / 40 = 4.5 m/s^2

Check:

15 a + 15 * 10 * .2 = T    acceleration of 15 kg block (assuming a = 4.5)

T = 15 (4.5) + 30 = 97.5     force required to accelerate 15 kg block

260 - 97.5 = 162.5     net force on 25 kg block

162.5 = 4.5 (25) + 25 * 10 * .2

162.5 = 112.5 + 50 = 162.5

4.5 m/s^2 checks out as correct

8 0
2 years ago
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