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marusya05 [52]
3 years ago
9

A very thin circular disk of radius R = 20.00 cm has charge Q = 30.00 mC uniformly distributed on its surface. The disk rotates

at a constant angular velocity ω = 5.00 rad/s around the z-axis through its center. Calculate the magnitude of the magnetic field strength on the z axis at a distance d = 2.000 × 103 cm from the center. Note that d >> R.
Physics
1 answer:
lutik1710 [3]3 years ago
3 0

Answer:

B= 7.5*10^{-15}T

Explanation:

The magnetic field strenght on the z-axis at a distance d from the center is,

B= \frac{\mu_0 Q\omega R^2}{8\pi d^3}

Our values are:

R=20cm\\Q=30mc\\w=5rad/s\\d=2*10^3cm=20m

Replacing,

B= \frac{(4\pi*10^{}-7)(30*10^{-3})(5)(0.2)^2}{8\pi(20)}

B= 7.5*10^{-15}T

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In an unweathered sample of igneous rock, the ratio of an unstable isotope to its stable daughter isotope is 1:15. If no daughte
svlad2 [7]

Answer:

200 million years

Explanation:

The equation that describes the decay of a radioactive isotope is

N(t)=N_0 (\frac{1}{2})^{\frac{t}{t_{1/2}}}

where

N(t) is the amount of radioactive isotope left at time t

N_0 is the initial amount of isotope

t_{1/2} is the half-life of the sample

In this problem, the ratio between unstable isotope and daughter isotope is 1:15; this means that

\frac{N(t)}{N_0}=\frac{1}{16}

Because the "total proportion" of original sample was 1+15=16.

Also we know that the half-life is

t_{1/2}=50\cdot 10^6 y

So we can re-arrange the equation to find t, the age of the rock:

t=t_{1/2} log_{0.5}(\frac{N}{N_0})=(50\cdot 10^6)log_{0.5}(\frac{1}{16})=200\cdot 10^6 y

So, 200 million years.

6 0
4 years ago
An electron in a one-dimensional box of width L = .75 nm decays from the n = 3 state to the ground state by emitting two photons
Vlad1618 [11]

Answer: 232 nm

Explanation: As it is well known the energy for the one dimentional box is given by:

E=\frac{h^2}{8mx L^2} *n^2

where h is the Planck constant, m the electron mass and L the width of the box. n is the energy level

For n=3 we a energy equal :9.62 * 10^-19 J= 6.01 eV

For n=1 the energy is: 1.069 *10^-19 J=0.66 eV

The energy difference Is 5.35 eV so by using this relationship

λ=1240/Energy = 232 nm

7 0
3 years ago
10. An object of a mass 300 KG is observed to accelerate at the rate of 4M/S2. Calculate the force required to produce this acce
zheka24 [161]

Answer:

1200 Newton

Explanation:

Force is equal to mass times acceleration

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4 0
3 years ago
Question 20
oksian1 [2.3K]

Answer:

The image distance is 17.56 cm

Explanation:

We have,

Height of light bulb is 3 cm.

The light bulb is placed at a distance of 50 cm. It means object distance is, u =-50 cm

Focal length of the lens, f = +13 cm

Let v is distance between image and the lens. Using lens formula :

\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{13}+\dfrac{1}{(-50)}\\\\v=17.56\ cm

So, the image distance is 17.56 cm.

5 0
3 years ago
An example of an object in projectile motion is
Usimov [2.4K]

Answer: a. a leaping frog

Explanation: In a projectile motion the object must have a certain path or trajectory with respect to a certain angle. For the leaping frog its velocity can be resolve into components having the final velocity at the highest point to be zero.

4 0
3 years ago
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