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3 years ago

Mass of object is 50g moves in a circular path of radius 10cm find work done

Physics

1 answer:

topjm [15]3 years ago

3 0

Answer:

Work done = 0.3142 Nm

Explanation:

Mass of Object is 50 g

Circular path of radius is 10 cm ⇒ 0.1 m

Work done = Force × Distance = ?

*Distance moved (circular path) ⇒ Circumference of the circular path

2πr = 2 × 3.142 × 0.1 ⇒ 0.6284 m

*Force that is enough to move a 50 g must be equal or more than its weight.

therefore convert 50 grams to newton = 0.5 N

Recall that; work done is force times distance

∴ 0.5 N × 0.6284 m

Work done = 0.3142 Nm

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3 years ago

Read 2 more answers

A cylindrical capacitor has an inner conductor of radius 2.7 mmmm and an outer conductor of radius 3.1 mmmm. The two conductors

Mars2501 [29]

Answer:

(A) Capacitance per unit length = $4.02 \times 10^{-10}$

(B) The magnitude of charge on both conductor is $Q = 4.22 \times 10^{-19}$ C and the sign of charge on inner conductor is $+Q$ and the sign on outer conductor is $-Q$

Explanation:

Given :

Radius of inner part of conductor $(R_{1})$ = $2.7 \times 10^{-3}$ m

Radius of outer part of conductor $(R_{2})$ = $3.1 \times 10^{-3}$ m

The length of the capacitor $(l)$ = $3 \times 10^{-3}$ m

(A)

Capacitance is purely geometrical property. It depends only on length, radius of conductor.

From the formula of cylindrical capacitor,

$C = \frac{2\pi\epsilon_{o} l }{ln\frac{R_{2} }{R_{1} } }$

Where, $\epsilon_{o} = 8.85 \times 10^{-12}$

But we need capacitance per unit length so,

$\frac{C}{l} = \frac{2\pi\epsilon_{o} }{ln\frac{R_{2} }{R_{1} } }$

capacitance per unit length = $\frac{6.28 \times 8.85 \times 10^{-12} }{ln(1.148)} = 4.02 \times 10^{-10}$

(B)

The charge on both conductors is given by,

$Q = C \Delta V$

Where, C = capacitance of cylindrical capacitor and value of $C = 12.06 \times 10^{-13}$ F, $\Delta V = 350 \times 10^{-3}$ V

∴ $Q = 4.22 \times 10^{-19}$ C

The magnitude of charge on both conductor is same as above but the sign of charge is different.

Charge on inner conductor is $+Q$ and Charge on outer conductor is $-Q$ .

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4 years ago

If a resultant vector is 12 m/s and the horizontal component is 9 m/s, what is the value of the vertical component? A. 5.8 m/s B

horrorfan [7]

The vertical component is B) 7.9 m/s

Explanation:

A vector can be resolved into its horizontal and vertical components. The horizontal and the vertical components form the sides of a right triangle, in which the resultant corresponds to the hypothenuse, so we can use Pythagorean's theorem:

$R=\sqrt{R_x^2+R_y^2}$

where

R is the resultant vector

Rx is the horizontal component

Ry is the vertical component

In this problem, we have:

R = 12 m/s is the resultant vector

Rx = 9 m/s is the horizontal component

Solving the formula for Ry, we find the vertical component:

$R_y = \sqrt{R^2-R_x^2}=\sqrt{12^2-9^2}=7.9 m/s$

Learn more about vector components:

brainly.com/question/2678571

#LearnwithBrainly

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3 years ago