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3 years ago

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A concave mirror of radius of curvature 40 cm is to be used to obtain a real image of an object. The image is to be one-third as

large as the object. Where should the object be placed and where is the image found

1 answer:

vagabundo [1.1K]3 years ago

7 0

answer.

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It is known that birds can detect the earth's magnetic field, but the mechanism of how they do this is not known. It has been su

Lubov Fominskaja [6]

Answer:

A) 0.50 mV

Explanation:

In this problem, we can think the wings of the bird as a metal rod moving across a magnetic field. So, and emf will be induced into the wings of the bird, according to the formula:

$\epsilon = BvL sin \theta$

where

$B=5\cdot 10^{-5} T$ is the strength of the magnetic field

v = 13 m/s is the speed of the bird

L = 1.2 m is the wingspan of the bird

$\theta=40^{\circ}$ is the angle between the direction of motion and the direction of the magnetic field

Substituting numbers into the formula, we find

$\epsilon = (5.0\cdot 10^{-5} T)(13 m/s)(1.2 m) sin 40^{\circ}=0.00050 V = 0.50 mV$

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3 years ago

The Back Saver Sit and Reach from the Fitnessgram measure this fitness component.

MAVERICK [17]

The correc answer is B.

7 0

3 years ago

What services do plants provide? Select the three that apply.

Irina18 [472]

The answer to this is B, C, and D. hope this helped

6 0

3 years ago

(a) If a proton with a kinetic energy of 6.2 MeV is traveling in a particle accelerator in a circular orbit with a radius of 0.5

Tju [1.3M]

Answer:

The fraction of its energy that it radiates every second is $3.02\times10^{-11}$ .

Explanation:

Suppose Electromagnetic radiation is emitted by accelerating charges. The rate at which energy is emitted from an accelerating charge that has charge q and acceleration a is given by

$\dfrac{dE}{dt}=\dfrac{q^2a^2}{6\pi\epsilon_{0}c^3}$

Given that,

Kinetic energy = 6.2 MeV

Radius = 0.500 m

We need to calculate the acceleration

Using formula of acceleration

$a=\dfrac{v^2}{r}$

Put the value into the formula

$a=\dfrac{\dfrac{1}{2}mv^2}{\dfrac{1}{2}mr}$

Put the value into the formula

$a=\dfrac{6.2\times10^{6}\times1.6\times10^{-19}}{\dfrac{1}{2}\times1.67\times10^{-27}\times0.51}$

$a=2.32\times10^{15}\ m/s^2$

We need to calculate the rate at which it emits energy because of its acceleration is

$\dfrac{dE}{dt}=\dfrac{q^2a^2}{6\pi\epsilon_{0}c^3}$

Put the value into the formula

$\dfrac{dE}{dt}=\dfrac{(1.6\times10^{-19})^2\times(2.3\times10^{15})^2}{6\pi\times8.85\times10^{-12}\times(3\times10^{8})^3}$

$\dfrac{dE}{dt}=3.00\times10^{-23}\ J/s$

The energy in ev/s

$\dfrac{dE}{dt}=\dfrac{3.00\times10^{-23}}{1.6\times10^{-19}}\ J/s$

$\dfrac{dE}{dt}=1.875\times10^{-4}\ ev/s$

We need to calculate the fraction of its energy that it radiates every second

$\dfrac{\dfrac{dE}{dt}}{E}=\dfrac{1.875\times10^{-4}}{6.2\times10^{6}}$

$\dfrac{\dfrac{dE}{dt}}{E}=3.02\times10^{-11}$

Hence, The fraction of its energy that it radiates every second is $3.02\times10^{-11}$ .

5 0

3 years ago

An electron is confined to a one dimensional infinite potential well 150 pm. How much energy must it absorb if it is to jump to

igor_vitrenko [27]

Answer:

\Delta E=1.22\times 10^{-22}J

Explanation:

The energy of electron in any state is given by $E=\frac{n^2h^2}{8mL^2}$ here h is planck's constant n is state of electron L is the infinte potential well m is the mass of electron

We know that $h=6.67\times 10^{-34}$

Potential well dimension = $150pm=150\times 10^{-12}m$

Mass of electron $=9.1\times 10^{-31}kg$

So energy required to electron to jump from ground state to 3rd state

$\Delta E=\frac{h^2}{8mL^2}\left ( 3^2-1^2 \right )$

$\Delta E=\frac{\left ( 6.67\times 10^{-34} \right )^2}{8\times 9.1\times 10^{-31}(150\times 10^{-12})^2}\left ( 9-1 \right )$

$\Delta E=1.22\times 10^{-22}J$

7 0

3 years ago