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natita [175]
3 years ago
6

Sphere 1 has surface area A₁ and volume V₁, sphere 2 has surface area A₂ and volume V₂. If the radius of sphere 2 is six times t

he radius of sphere 1, what is the ratio \frac{A2}{A1} of the areas? What is the ratio \frac{V2}{V1} of the volumes?
Physics
1 answer:
Sveta_85 [38]3 years ago
8 0

Let , radius of sphere 1 is r .

So , radius of sphere 2 is 6r .

Surface area of sphere is given by :

A=4\pi r^2

So ,

\dfrac{A_2}{A_1}=\dfrac{4\pi(6r)^2}{4\pi r^2}\\\\\dfrac{A_2}{A_1}=36

Volume is given by :

V=\dfrac{4}{3}\pi r^3

Ratio of sphere 2 by sphere 1 is given by :

\dfrac{V_2}{V_1}=\dfrac{\dfrac{4}{3}\pi (6r)^3}{\dfrac{4}{3}\pi r^3}\\\\\dfrac{V_2}{V_1}=216

Therefore ,  the ratio of area and volume is 36 and 216 respectively .

Hence , this is the required solution .

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At time t=0, a particle is located at the point (3,6,9). It travels in a straight line to the point (5,2,7), has speed 8 at (3,6
Elis [28]

The particle has constant acceleration according to

\vec a(t)=2\,\vec\imath-4\,\vec\jmath-2\,\vec k

Its velocity at time t is

\displaystyle\vec v(t)=\vec v(0)+\int_0^t\vec a(u)\,\mathrm du

\vec v(t)=\vec v(0)+(2\,\vec\imath-4\,\vec\jmath-2\,\vec k)t

\vec v(t)=(v_{0x}+2t)\,\vec\imath+(v_{0y}-4t)\,\vec\jmath+(v_{0z}-2t)\,\vec k

Then the particle has position at time t according to

\displaystyle\vec r(t)=\vec r(0)+\int_0^t\vec v(u)\,\mathrm du

\vec r(t)=(3+v_{0x}t+t^2)\,\vec\imath+(6+v_{0y}t-2t^2)\,\vec\jmath+(9+v_{0z}t-t^2)\,\vec k

At at the point (3, 6, 9), i.e. when t=0, it has speed 8, so that

\|\vec v(0)\|=8\iff{v_{0x}}^2+{v_{0y}}^2+{v_{0z}}^2=64

We know that at some time t=T, the particle is at the point (5, 2, 7), which tells us

\begin{cases}3+v_{0x}T+T^2=5\\6+v_{0y}T-2T^2=2\\9+v_{0z}T-T^2=7\end{cases}\implies\begin{cases}v_{0x}=\dfrac{2-T^2}T\\\\v_{0y}=\dfrac{2T^2-4}T\\\\v_{0z}=\dfrac{T^2-2}T\end{cases}

and in particular we see that

v_{0y}=-2v_{0x}

and

v_{0z}=-v_{0x}

Then

{v_{0x}}^2+(-2v_{0x})^2+(-v_{0x})^2=6{v_{0x}}^2=64\implies v_{0x}=\pm\dfrac{4\sqrt6}3

\implies v_{0y}=\mp\dfrac{8\sqrt6}3

\implies v_{0z}=\mp\dfrac{4\sqrt6}3

That is, there are two possible initial velocities for which the particle can travel between (3, 6, 9) and (5, 2, 7) with the given acceleration vector and given that it starts with a speed of 8. Then there are two possible solutions for its position vector; one of them is

\vec r(t)=\left(3+\dfrac{4\sqrt6}3t+t^2\right)\,\vec\imath+\left(6-\dfrac{8\sqrt6}3t-2t^2\right)\,\vec\jmath+\left(9-\dfrac{4\sqrt6}3t-t^2\right)\,\vec k

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An fm radio station broadcasts electromagnetic radiation at a frequency of 100.6 mhz. the wavelength of this radiation is ______
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The wavelength would be 2.980044314115. Reduced it would be 2.980 or just 2.98.
5 0
3 years ago
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an electric device is plugged into a 110v wall socket. if the device consumes 500 w of power, what is the resistance of the devi
andreev551 [17]

Answer: R=24.2Ω

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P=V.i

P=R.i²

P=\frac{V^{2}}{R}

The resistance of the system is:

P=\frac{V^{2}}{R}

R=\frac{V^{2}}{P}

R=\frac{110^{2}}{500}

R = 24.2Ω

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4 0
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Suppose you increase your walking speed from 5 m/s to 14 m/s in a period of 3 s. What is your acceleration
kirza4 [7]
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3 years ago
A 5 kgkg sphere having a charge of ++ 8 μCμC is placed on a scale, which measures its weight in newtons. A second sphere having
Mrac [35]

Answer:

 F_Balance = 46.6 N    ,m' = 4,755 kg

Explanation:

In this exercise, when the sphere is placed on the balance, it indicates the weight of the sphere, when another sphere of opposite charge is placed, they are attracted so that the balance reading decreases, resulting in

          ∑ F = 0

          Fe –W + F_Balance = 0

         F_Balance = - Fe + W

           

The electric force is given by Coulomb's law

          Fe = k q₁ q₂ / r₂

The weight is

          W = mg

Let's replace

           F_Balance = mg - k q₁q₂ / r₂

Let's reduce the magnitudes to the SI system

          q₁ = + 8 μC = +8 10⁻⁶ C

          q₂ = - 3 μC = - 3 10⁻⁶ C

          r = 0.3 m = 0.3 m

Let's calculate

         F_Balance = 5 9.8 - 8.99 10⁹  8 10⁻⁶ 3 10⁻⁶ / (0.3)²

         F_Balance = 49 - 2,397

         F_Balance = 46.6 N

This is the balance reading, if it is calibrated in kg, it must be divided by the value of the gravity acceleration.

Mass reading is

          m' = F_Balance / g

          m' = 46.6 /9.8

          m' = 4,755 kg

6 0
3 years ago
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