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3 years ago

An object is spun around in circular motion such that its frequency is 12 Hz.

Physics

1 answer:

vichka [17]3 years ago

4 0

Explanation:

Given parameters:

Frequency = 12Hz

Unknown:

a. Period of its rotation

b. Time required to complete 86 rotations

Solution:

Frequency is the number of times of rotation per unit of time.

Frequency = $\frac{n}{t}$

Period is the opposite of frequency;

Period = $\frac{1}{12}$ = 0.083s

b. Time for 86 rotations;

Since frequency = $\frac{n}{t}$

n is the number of rotation

t is the time

t = $\frac{n}{Frequency}$ = $\frac{86}{12}$ = 7.2s

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Two charges are located in the x–y plane. If q1 = -2.90 nC and is located at x = 0.00 m, y = 0.840 m and the second charge has m

Lunna [17]

Answer:

Epx= - 21.4N/C

Epy= 19.84N/C

Explanation:

Electric field theory

The electric field at a point P due to a point charge is calculated as follows:

E= k*q/r²

E= Electric field in N/C

q = charge in Newtons (N)

k= electric constant in N*m²/C²

r= distance from load q to point P in meters (m)

Equivalences

1nC= 10⁻⁹C

known data

q₁=-2.9nC=-2.9 *10⁻⁹C

q₂=5nC=5 *10⁻⁹C

r₁=0.840m

$r_{2} =\sqrt{1^{2} +0.8^{2} } =\sqrt{1.64}$

$sin\beta =\frac{0.8}{\sqrt{1.64} } =0.6246$

$cos\beta =\frac{1}{\sqrt{1.64} } =0.7808$

Calculation of the electric field at point P due to q1

Ep₁x=0

$Ep_{1y} =\frac{k*q_{1} }{r_{1}^{2} } =\frac{8.99*10^{9}*2.9*10^{-9} }{0.84^{2} } =36.95\frac{N}{C}$

Calculation of the electric field at point P due to q2

$Ep_{2x} =-\frac{k*q_{2} *cos\beta }{r_{2}^{2} } =-\frac{8.99*10^{9}*5*10^{-9} *0.7808 }{(\sqrt{1.64})^{2} } =-21.4\frac{N}{C}$

$Ep_{2y} =-\frac{k*q_{2} *sin\beta }{r_{2}^{2} } =-\frac{8.99*10^{9}*5*10^{-9} *0.6242 }{(\sqrt{1.64})^{2} } =-17.11\frac{N}{C}$

Calculation of the electric field at point P(0,0) due to q1 and q2

Epx= Ep₁x+ Ep₂x==0 - 21.4N/C =- 21.4N/C

Epy= Ep₁y+ Ep₂y=36.95 N/C-17.11N =19.84N/C

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3 years ago

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iron β is a solid phase of iron still unknown to science. The only difference between it and ordinary iron is that Iron β forms

saw5 [17]

Answer:

8.60 g/cm³

Explanation:

In the lattice structure of iron, there are two atoms per unit cell. So:

$\frac{2}{a^{3} } = \frac{N_{A} }{V_{molar} }$ where $V_{molar} = \frac{A}{\rho }$ an and A is the atomic mass of iron.

Therefore:

$\frac{2}{a^{3} } = \frac{N_{A} * p }{A}$

This implies that:

$A = (\frac{2A}{N_{A} * p)^{\frac{1}{3} } }$

= $\frac{4}{\sqrt{3} }r$

Assuming that there is no phase change gives:

$\rho = \frac{4A}{N_{A}(2\sqrt{2r})^{3} }$

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4 years ago

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The answer to ur question is B

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3 years ago

How much force is needed to stop a 4000 kg truck moving at 8 m/s in 0 2 seconds?

maksim [4K]

Answer:

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Explanation:

Force Calculater

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