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AleksAgata [21]
2 years ago
8

An object is located 25.0 cm from a convex mirror. The image distance is -50.0 cm. What is the magnification?

Physics
1 answer:
Lynna [10]2 years ago
4 0

Answer:

\boxed{\sf Magnification \ (m) = 2}

Given:

Object distance (u) = 25.0 cm

Image distance (v) = -50.0 cm

To Find:

Magnification (m)

Explanation:

\boxed{\bold{\sf Magnification  \: (m) = - \frac{Image  \: distance  \: (v)}{Object  \: distance  \: (u)}}}

Substituting values of Image distance(v) & Object distance (u) in the equation:

\sf \implies m =  -  \frac{( - 50)}{25}

-(-50) = 50:

\sf \implies m =  \frac{50}{25}

\sf \implies m =  \frac{2 \times  \cancel{25}}{ \cancel{25}}

\sf \implies m = 2

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aleksley [76]

Answer:

Potential difference though which the electron was accelerated is 2.67\times 10^{-6}\ uV\  .

Explanation:

Given :

De Broglie wavelength , \lambda=750\ nm.

Plank's constant , h=6.626\times 10^{-34}\ J.s \ .

Charge of electron , e=-1.6\times 10^{-19}\ C.

Mass of electron , m=9.11\times 10^{-31}\ kg.m=9.11\times 10^{-31}\ kg.

We know , according to de broglie equation :

\lambda=\dfrac{h}{mv}\\\\ v=\dfrac{h}{m\lambda}\\\\v=\dfrac{6.626\times 10^{-34}\ J.s \ }{9.11\times 10^{-31}\ kg\times 750\times 10^{-9}\ m }= 969.78\ m/s .

Now , we know potential energy applied on electron will be equal to its kinetic energy .

Therefore ,

qV=\dfrac{mv^2}{2}\\\\ V=\dfrac{mv^2}{2q}

Putting all values in above equation we get ,

V=2.67\times 10^{-6}\ uV .

Hence , this is the required solution.

5 0
2 years ago
Diesel engines give more miles per gallon than gasoline engines, but some of this is due to the higher energy content of diesel
Lina20 [59]

Answer:

0.106

Explanation:

For 1 liter of diesel the car can get 19 km, if it takes 0.2 MJ for each km then it would take the total energy of 19*0.2 = 3.8 MJ to move an aerodynamic car 19 km. Since 1 liter of of diesel also contains 36 MJ in internal energy, then the efficiency of the diesel engine is the ratio of its output energy over its input energy:

\frac{3.8}{36} = 0.106

5 0
2 years ago
A brick of mass 2 kg is dropped from a rest position 5 m above the ground. what is its velocity at a height of 3 m above the gro
Rina8888 [55]
We can solve the problem by using the law of conservation of energy.

Using the ground as reference point, the mechanical energy of the brick when it is at 5 m from the ground is just potential energy (because the brick is initially at rest, so it doesn't have kinetic energy):
E= U = mgh=(2 kg)((9.81 m/s^2)(5 m)=98.1 J

when the brick is at h'=3 m from the ground, its mechanical energy is now sum of kinetic energy and potential energy:
E= K+U= \frac{1}{2} mv^2 + mgh'

where v is the velocity of the brick. Since E is conserved, it must be equal to the initial energy (98.1 J), so we can solve this equation to find v:
v= \sqrt{ \frac{2(E-mgh')}{m} }=6.3 m/s
8 0
2 years ago
A boat moves through the sea.
sergiy2304 [10]

Answer:

dont you have to times it

Explanation:

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2 years ago
What visible signs indicate a precipitation reaction when two solutions are mixed?
Illusion [34]

Formation of an insoluble solid

Explanation:

One of the remarkable visible signs that indicates a precipitation reaction when two solutions are mixed is the formation of an insoluble solid. The insoluble solid formed is the precipitate.

  • Precipitates usually forms in single replacement reactions and double replacement or double decomposition reactions.
  • They form when two soluble compounds react. One of the product is an insoluble solid in the solution called the precipitate.
  • The solubility table helps to predict whether precipitates forms in a reaction.

Learn more:

precipitate brainly.com/question/8896163

#learnwithBrainly

6 0
2 years ago
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