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vodka [1.7K]
4 years ago
6

Young David who slew Goliath experimented with slings beforetackling the giant. He found that he could revolve a sling oflength

0.600 m at the rate of9.00 rev/s. If he increased the lengthto 0.900 m, he could revolve the slingonly 7.00 times per second.
(a) What is the speed of the stone for eachrate of rotation?
1 m/s at 9.00 rev/s
2 m/s at 7.00 rev/s

(b) What is the centripetal acceleration of the stone at9.00 rev/s?
3 m/s2

(c) What is the centripetal acceleration at 7.00 rev/s?
4m/s2
Physics
1 answer:
Mars2501 [29]4 years ago
3 0

Answer:

a) v1 = 33.9 m/s , v2 = 39.6 m/s, b) a = 1015.4 m/s² and c)  a = 1742.4 m/s²

Explanation:

To solve this exercise we will use the relationship between angular and linear quantities

a) The angular and linear velocity are related by

     v = w r

Let's reduce the quantities to the SI system

    w1 = 9.00 rev / s (2π rad / 1rev) = 18 π rad / s = 56.55 rad / s

    w2 = 7.00 rev / s = 14π rad / s = 43.98vrad / s

     v1 = 56.55 0.6

     v1 = 33.9 m / s

   

     v2 = 43.98 0.9

    v2 = 39.6 m / s

b) centripetal acceleration for v1

     a = v² / r

     a = 33.9² / 0.6

     a = 1015.4 m / s²

c) centripetal acceleration for v2

     a = 39.6² / 0.9

     a = 1742.4 m / s²

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Answer:

<u>In an ionic bond , an element will have to lose or gain electrons.</u>

Explanation:

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A wire 2.80 m in length carries a current of 5.20 A in a region where a uniform magnetic field has a magnitude of 0.430 T. Calcu
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Question:

A wire 2.80 m in length carries a current of 5.20 A in a region where a uniform magnetic field has a magnitude of 0.430 T. Calculate the magnitude of the magnetic force on the wire assuming the following angles between the magnetic field and the current.

(a)60 (b)90 (c)120

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Explanation:

From the question

Length of wire (L) = 2.80 m

Current in wire (I) = 5.20 A

Magnetic field (B) = 0.430 T

Angle are different in each part.

The magnetic force is given by

F=I \times B \times L \times sin(\theta)

So from data

F = 5.20 A \times 0.430 T \times 2.80 sin(\theta)\\\\F=6.2608 sin(\theta) N

Now sub parts

(a)

\theta=60^{o}\\\\Force = 6.2608 sin(60^{o}) N\\\\Force = 5.42 N

(b)

\theta=90^{o}\\\\Force = 6.2608 sin(90^{o}) N\\\\Force = 6.26 N

(c)

\theta=120^{o}\\\\Force = 6.2608 sin(120^{o}) N\\\\Force = 5.42 N

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