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2 years ago

6

A record spins at 33 rpm (revolutions per minute), which is an angular velocity of about 3.46 radians per second. What is the ap

proximate linear velocity of a fly that sits on the record, 12 cm from the center

1 answer:

lubasha [3.4K]2 years ago

6 0

Hi there!

We can use the following equation to relate angular velocity to linear velocity.

$v = \omega r$

v = linear velocity (m/s)

ω = angular velocity (3.46 rad/sec)

r = distance from axis of rotation (.12 m)

Plug in the given values.

$v = (3.46)(.12) = \boxed{.415 \frac{m}{s}}$

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Vector A with arrow, which is directed along an x axis, is to be added to vector B with arrow, which has a magnitude of 5.5 m. T

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Answer:

Magnitude of vector A = 0.904

Explanation:

Vector A , which is directed along an x axis, that is

$\vec{A}=x_A\hat{i}$

Vector B , which has a magnitude of 5.5 m

$\vec{B}=x_B\hat{i}+y_B\hat{j}$

$\sqrt{x_{B}^{2}+y_{B}^{2}}=5.5\\\\x_{B}^{2}+y_{B}^{2}=30.25$

The sum is a third vector that is directed along the y axis, with a magnitude that is 6.0 times that of vector A $\vec{A}+\vec{B}=6x_A\hat{j}\\\\x_A\hat{i}+x_B\hat{i}+y_B\hat{j}=6x_A\hat{j}$

Comparing we will get

$x_A=-x_B\\\\y_B=6x_A$

Substituting in $x_{B}^{2}+y_{B}^{2}=30.25$

$\left (-x_{A} \right )^{2}+\left (6x_{A} \right )^{2}=30.25\\\\37x_{A}^2=30.25\\\\x_{A}=0.904$

So we have

$\vec{A}=0.904\hat{i}$

Magnitude of vector A = 0.904

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3 years ago

A space vehicle accelerates uniformly from 85 m/s at t = 0 to 164 m/s at t = 10.0 s .How far did it move between t = 2.0 s and t

creativ13 [48]

First, we have a change in the velocity from 85 to 164 m/s in 10 sec.

Then, we calculate the <u>acceleration </u>as:

$a=\frac{v_{f}-v_{i} }{t} =\frac{164-85}{10}=7.9 m/s^2$

Hence we need to calculate the velocity of the space vehicle at t = 2 sec using the first equation of motion:

$v_{f}=v_{i}+at=85+7.9*2=100.8m/s$

Then, using the second equation of motion to calculate the distance:

$d=v_{i} t+\frac{1}{2}at^2$

$d=100.8*2+\frac{1}{2}*7.9*(2)^2=217.4m$

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3 years ago

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An uncharged, nonconducting, hollow sphere of radius 10.0cm surrounds a 10.0-μC charge located at the origin of a Cartesian coor

MrMuchimi

The electric flux through the hole is $56.45\ webber$ .

Electric flux is the number of electric field lines cutting through the surface and is measured as surface intregal of electric field over that surface
Mathematically it is given by $\phi_E=E.A \ Nm^2/C$ where E is the electric field and A is the area.
Gauss's law states that electric flux through closed surface is equal to the 1 / ε₀ times the charge enclosed by that surface which is given by Ф = q / ε₀ where q is the central charge and ε₀ is the permittivity of the medium.

It is given , hollow sphere of radius 10.0cm surrounds a 10.0-μC charge.

The whole surface of hollow sphere $= 4\pi r^2$

$= 4\times 3.14\times (10 \times 10^{-2})^2 \\\\= 12.56\times 10^{-2} m^2$

Area of the hole ( both side ) $= 2\times \pi r^2$

$= 2\times 3.14 \times (10^-^3)^2\\= 6.28 \times 10^-^6 m^2$

According to Gauss's theorem, the flow from a particular charge in the center is given by

$\phi= \frac{10\times10^-^6}{8.85\times 10^-^1^2}\\\\\phi=1.13\times10^6$

This flux flows through the surface of the sphere, so the flux per unit area which is given by

$= \frac{ 1.13\times 10^6 }{ 12.56\times 10^-^2} \\\\= 8.99 \times 10^6 \ weber / m^2$

Flux through area of hole is given by :

$= 8.99\times10^6 \times6.28 \times 10^-^6\\ = 56.45 \ weber$

Learn about more electric flux here :

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A laundry detergent
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