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Dahasolnce [82]
3 years ago
13

A real object with height of 3.20 cm is placed to the left of a converging lens whose focal length is 90cm. The image is on the

right of the lens and 4.50cm tall and inverted. Where is the object? Where is the image? Is the image real or virtual?
Physics
1 answer:
pickupchik [31]3 years ago
3 0

Answer:

d_{o} = 154 cm

d_{i} = 216.6 cm

The image is real

Explanation:

h_{o} = height of the object = 3.20 cm

h_{i} = height of the image = 4.50 cm

f = focal length of the converging lens = 90 cm

d_{o} = object distance from the lens = ?

d_{i} = image distance from the lens = ?

using the equation for magnification

\frac{h_{i}}{h_{o}}= \frac{ d_{i}}{d_{o}}

\frac{4.50}{3.20}= \frac{d_{i}}{d_{o}}

d_{i} = 1.40625 d_{o}                        eq-1

using the lens equation

\frac{1}{d_{i}} + \frac{1}{d_{o}} = \frac{1}{f}

using eq-1

\frac{1}{( 1.40625)d_{o}} + \frac{1}{d_{o}} = \frac{1}{90}

d_{o} = 154 cm

Using eq-1

d_{i} = 1.40625 d_{o}  

d_{i} = 1.40625 (154)

d_{i} = 216.6 cm

The image is real

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The particle is an electron. The field slows down the electron without deflecting it. The direction of the electric field is <u>right.</u>

In physics, the motion of electrically charged particles gives rise to a field called an electric field. It is measured in force per unit charge.

This field applies force on other charged particles.

Particles bearing opposite charges attract each other while particles having similar charges repel each other in the field.

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2 years ago
A 23 kg body is moving through space in the positive direction of an x axis with a speed of 130 m/s when, due to an internal exp
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Answer:

a) Vx = 1088m/s

b) Vy = -162.93m/s

c) 5246745J

Explanation:

Mass of unbroken body = 23kg

Its velocity along +ve X-axis = 130m/s

Mass of first broken body, m1= 9.4kg

Its velocity along +ve X-axis = 130m/s

Nass of 2nd broken body, m2 = 6.1kg

Its velocity long-lived X - axis = -550m/s

Mass of 3rd broken body = ?

m3 = (23 - 9.4 - 6.1)kg

m3 = 7.5kg

Let velocity along the x-axis = Vx

Let the velocity along the x-axis = Vy

Applying law of conservation of momentum along x-axis

a) m1×0 + m2×(-550) + m3×(Vx) =M × 130

9.4 × 0 + 6.1× (-550) + 7.5(Vx) = 23 ×130

0 + (-5170) + 7.5Vx = 2990

2990 + 5170 = 7.5Vx

8160 = 7.5Vx

Vx = 8160/7.5

Vx = 1088m/s

b) Aplying conservation of momentum along the x-axis

(m1×130) + (m2 × 0) + (m3× Vy) = 0

(9.4 × 130) + (6.1 ×550) + 7.5Vy = 0

1222 + 0 + 7.5Vy = 0

1222 = -7.5Vy

Vy = 1222/(-7.5)

Vy = -262.93m/s

c) The energy released or change in KE is given by:

1/2[(m1v1^2) + (m2v2^2) +(m3Vx^2) ]= MV^2

Change in KE = 1/2[ 9.4× 130^2 + 6.1 × 550^2 + 7.5 × 1088^2 ] - 1/2(23 × 130^2)

Change in KE = 1/2[158860 + 1845250 + 8878080] - 1/2[388700]

Change in KE = 5441095 - 194350

Change in KE = 5246745J

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If the substance doesn't change chemically, it is a physical reaction.
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Answer:

150.6 km

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