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2 years ago

1. A 3.0 kg mass is tied to a rope and swung in a horizontal circle. If the velocity of the mass is 4.0 ms and

Physics

1 answer:

saul85 [17]2 years ago

4 0

10.67m/s²

32N

Explanation:

Given parameters:

Mass of the body = 3kg

velocity of the mass = 4m/s

radius of circle = 0.75m

Unknown:

centripetal acceleration = ?

centripetal force = ?

Solution:

The centripetal force is the force that keeps a radial body in its circular motion. It is directed inward:

Centripetal acceleration = $\frac{v^{2} }{r}$

v is the velocity of the body

r is the radius of the circle

putting in the parameters:

Centripetal acceleration = $\frac{4^{2} }{0.75}$

Centripetal acceleration = 10.67m/s²

Centripetal force = m $\frac{v^{2} }{r}$

m is the mass

Centripetal force = mass x centripetal acceleration

= 3 x 10.67

= 32N

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Acceleration brainly.com/question/3820012

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Answer:

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2 years ago

A 5.00μF parallel-plate capacitor is connected to a 12.0 V battery. After the capacitor is fully charged, the battery is disconn

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In this problem, the capacitor is connected to the 12.0 V, until it is fully charged. Considering the capacity of the capacitor, $C=5.00 \mu F$ , the charged stored on the capacitor at the end of the process is

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When the battery is disconnected, the charge on the capacitor remains unchanged. But the capacitance, C, also remains unchanged, since it only depends on the properties of the capacitor (area and distance between the plates), which do not change. Therefore, given the relationship

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(b) (i) 24.0 V

In this case, the plate separation is doubled. Let's remind the formula for the capacitance of a parallel-plate capacitor:

$C=\frac{\epsilon_0 \epsilon_r A}{d}$

where:

$\epsilon_0$ is the permittivity of free space

$\epsilon_r$ is the relative permittivity of the material inside the capacitor

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As we said, in this case the plate separation is doubled: d'=2d. This means that the capacitance is halved: $C'=\frac{C}{2}$ . The new voltage across the plate is given by

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and since Q (the charge) does not change (the capacitor is now isolated, so the charge cannot flow anywhere), the new voltage is

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While the separation between the plate was unchanged (d); so, the new capacitance will be

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So, the capacitance has increased by a factor 4; therefore, the new voltage is

$V'=\frac{Q}{C'}=\frac{Q}{4C}=\frac{1}{4} \frac{Q}{C}=\frac{V}{4}$

which means

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However, the displacement of the object is zero (because the object returns to the starting point), and so the average velocity is also zero:

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