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zubka84 [21]
1 year ago
15

Consider a concave spherical mirr or that has focal length f = +19.5 cm.

Physics
1 answer:
lidiya [134]1 year ago
7 0

The distance of an object from the mirror's vertex if the image is real and has the same height as the object is 39 cm.

<h3>What is concave mirror?</h3>

A concave mirror has a reflective surface that is curved inward and away from the light source.

Concave mirrors reflect light inward to one focal point and it usually form real and virtual images.

<h3>Object distance of the concave mirror</h3>

Apply mirrors formula as shown below;

1/f = 1/v + 1/u

where;

  • f is the focal length of the mirror
  • v is the object distance
  • u is the image distance

when image height = object height, magnification = 1

u/v = 1

v = u

Substitute the given parameters and solve for the distance of the object from the mirror's vertex

1/f = 1/v + 1/v

1/f = 2/v

v = 2f

v = 2(19.5 cm)

v = 39 cm

Thus, the distance of an object from the mirror's vertex if the image is real and has the same height as the object is 39 cm.

Learn more about concave mirror here: brainly.com/question/27841226

#SPJ1

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Which of the following statements is true in the case of a collision?
cluponka [151]

Answer: D

Reduced impact time will increase the impact force

Explanation: Collision occurs when two or more bodies collide and exert forces on each other within a short time.

If a body of mass M moving with a velocity V collide with another body, the kinetic energy of the body is equal to the work done by the body.

That is, K.E = 1/2mv^2 = F × s

Where workdone = Force × distance

Make F the subject of formula

Mv^2/2s = F

But V = distance s/time t

Substitute for V

Ms^2/2t^2s = F

Ms/2t^2 = F

From the equation above, we can deduce that F is inversely proportional to the square of time.

Therefore, the reduced impact time will increase the impact force

3 0
3 years ago
a ball is thrown down at 25 m/s from a 500m tall building. how fast is it traveling when it hits the ground?
Mrrafil [7]

Answer:

The speed of the ball when it hits the ground is 102.1 m/s

Explanation:

Given;

initial velocity of ball, u = 25 m/s

distance traveled by the ball = height of the building = h = 500 m

when the ball hits the ground, the final velocity, v = ?

The final velocity of the ball is given by;

v² = u² + 2gh

where;

g is acceleration due to gravity = 9.8 m/s²

v² = (25)² + 2(9.8)(500)

v² = 10425

v = √10425

v = 102.1 m/s

Therefore, the speed of the ball when it hits the ground is 102.1 m/s

3 0
2 years ago
A 42.6kg lamp is hanging from wires as shown in figure.The ring has negligible mass. Find tensionsT1, T2,T3 if the object is in
IRISSAK [1]

Answer:

T1 = 417.48N

T2 = 361.54N

T3 = 208.74N

Explanation:

Using the sin rule to fine the tension in the strings;

Given

amass = 42.6kg

Weight = 42.6 * 9.8 = 417.48N

The third angle will be 180-(60+30)= 90 degrees

Using the sine rule

W/Sin 90 = T3/sin 30 = T2/sin 60

Get T3;

W/Sin 90 = T3/sin 30

417.48/1 = T3/sin30

T3 = 417.48sin30

T3 = 417.48(0.5)

T3 = 208.74N

Also;

W/sin90 = T2/sin 60

417.48/1 = T2/sin60

T2 = 417.48sin60

T2 = 417.48(0.8660)

T2 = 361.54N

The Tension T1 = Weight of the object = 417.48N

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How far should pencil held from a convex mirror of radius 80 cm to form an image one-half the size of the pencil?
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Answer:

10.0

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