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3 years ago

If an object is placed between a convex lens and its focal point, which type of image will be produced?

Physics

2 answers:

Andreas93 [3]3 years ago

8 0

<span>virtual, upright, and magnified</span>

Nastasia [14]3 years ago

4 0

Answer:

virtual, upright and magnified

Explanation:

As we know by lens formula

$\frac{1}{d_i} + \frac{1}{d_o} = \frac{1}{f}$

now we know that

$d_o < f$

now we have

$\frac{1}{d_i} = \frac{1}{f} - \frac{1}{d_o}$

now we have

$d_i = -\frac{d_o f}{f - d_o}$

now the magnification is given as

$m = \frac{d_i}{d_o}$

$m = -\frac{f}{f - d_o}$

Since here magnification is negative

so image is virtual and upright

also we can see from equation

m > 1

so here we have

virtual, upright and magnified

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A laser with a power of 1.0 mW has a beam radius of 1.0 mm. What is the peak value of the electric field in that beam

Fynjy0 [20]

Answer:

The peak value of the electric field is 489.64 V/m

Explanation:

Given;

power of the laser, P = 1.0 mW = 1 x 10⁻³ W

Radius of the beam, R = 1.0 mm = 1 x 10⁻³ m

Area of the beam = πr² = π(1 x 10⁻³ )² = 3.142 x 10⁻⁶ m²

The average intensity of the light = P / A

The average intensity of the light = ( 1 x 10⁻³) / (3.142 x 10⁻⁶)

The average intensity of the light = 318.27 W/m²

The peak value of the electric field is given by;

$E_o = \sqrt{\frac{2I_{avg}}{c\epsilon_o}}\\\\E_o = \sqrt{\frac{2(318.27)}{(3*10^8)(8.85*10^{-12})}}\\\\E_o = 489.64 \ V/m$

Therefore, the peak value of the electric field is 489.64 V/m.

4 0

4 years ago

Two tugboats pull a disabled supertanker. Each tug exerts a constant force of 1.20×10^6 N, one at an angle 14.0∘ west of north,

laila [671]

Answer:

1.45544 J

Explanation:

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5 0

3 years ago

Uest<br>1. State Newton's law of cooling.

garik1379 [7]

Answer:

Newton's law of cooling states that the rate of heat loss of a body is directly proportional to the difference in the temperatures between the body and its surroundings. The law is frequently qualified to include the condition that the temperature difference is small and the nature of heat transfer mechanism remains the same. As such, it is equivalent to a statement that the heat transfer coefficient, which mediates between heat losses and temperature differences, is a constant. This condition is generally met in heat conduction (where it is guaranteed by Fourier's law) as the thermal conductivity of most materials is only weakly dependent on temperature. In convective heat transfer, Newton's Law is followed for forced air or pumped fluid cooling, where the properties of the fluid do not vary strongly with temperature, but it is only approximately true for buoyancy-driven convection, where the velocity of the flow increases with temperature difference. Finally, in the case of heat transfer by thermal radiation, Newton's law of cooling holds only for very small temperature differences.

When stated in terms of temperature differences, Newton's law (with several further simplifying assumptions, such as a low Biot number and a temperature-independent heat capacity) results in a simple differential equation expressing temperature-difference as a function of time. The solution to that equation describes an exponential decrease of temperature-difference over time. This characteristic decay of the temperature-difference is also associated with Newton's law of cooling

6 0

3 years ago

What is the kinetic energy of a 150 gram object moving at a velocity of 100 m/s?

Alex Ar [27]

Answer:

750 J

Explanation:

lets convert mass into kg first , 150 /1000 = 0.15 kg

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3 0

3 years ago

Four charges with equal magnitudes of 10.6 × 10-12 C are placed at the corners of a rectangle. The lengths of the sides of the r

cricket20 [7]

Answer:

Figure a. E_net = 99.518 N/C

Figure b. E_net = 177.151 N / C

Explanation:

Given:

- Attachment for figures missing in the question.

- The dimensions for rectangle are = 7.79 x 3.99 cm

- All four charges have equal magnitude Q = 10.6*10^-12 C

Find:

Find the magnitude of the electric field at the center of the rectangle in Figures a and b.

Solution:

- The Electric field generated by an charged particle Q at a distance r is given by:

E = k*Q / r^2

- Where, k is the coulomb's constant = 8.99 * 10^9

Part a)

- First we see that the charges +Q_1 and +Q_3 produce and electric field equal but opposite in nature. So the sum of Electric fields:

E_1 + E_3 = 0

- For Charges -Q_2 and +Q_4, they are equal in nature but act in the same direction towards the negative charge -Q_2. Hence, the net Electric Field at center of the rectangle can be given as:

E_net = E_2 + E_4

E_2 = E_4

E_net = 2*E = 2*k*Q / r^2

- The distance r from each corner to mid-point of the rectangle is constant. It can be evaluated by Pythagoras Theorem as follows:

r = sqrt ( (7.79/200)^2 + (3.99/200)^2 )

r = sqrt ( 1.9151*10^-3 ) = 0.043762 m

- Plug the values in the E_net expression developed above:

E_net = 2*(8.99*10^9)*(10.6*10^-12) / 1.9151*10^-3

E_net = 99.518 N/C

Part b)

- Similarly for Figure b, for Charges -Q_2 and +Q_4, they are equal in nature but act in the same direction towards the negative charge -Q_2. Also, Charges -Q_1 and +Q_3, they are equal in nature but act in the same direction towards the negative charge -Q_1. These Electric fields are equal in magnitude to what we calculated in part a).

- To find the vector sum of two Electric Fields E_1,3 and E_2,4 we see the horizontal components of each cancels each other out. While the vertical components E_1,3 and E_2,4 are equal in magnitude and direction.

Hence,

E_net = 2*E_part(a)*cos(Q)

- Where, Q is the angle between resultant, vertical in direction, and each of the electric field. We can calculate Q using trigonometry as follows:

Q = arctan ( 3.99 / 7.79 ) = 27.12 degrees.

- Now, compute the net electric field E_net:

E_net = 2*(99.518)*cos(27.12)

E_net = 177.151 N / C

5 0

3 years ago