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madam [21]
3 years ago
14

An object is placed 50.0 cm in front of a convex mirror. where can be the image located if the focal length is 40 cm from the mi

rror
Need now plz and thank you! ​
Physics
1 answer:
sergiy2304 [10]3 years ago
6 0

Answer:

Image will form at distance 22.22 cm behind the mirror

Explanation:

As we know that the mirror formula is given as

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

now we know that

object distance from mirror is

d_o = -50 cm

Focal length of the mirror is given as

f = 40 cm

now we have

\frac{1}{-50} + \frac{1}{d_i} = \frac{1}{40}

\frac{1}{d_i} = \frac{1}{50} + \frac{1}{40}

d_i = 22.22 cm

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A fast-food restaurant uses a conveyor belt to send the burgers through a grilling machine. if the grilling machine is 1.2 m lon
Vinil7 [7]

length of the grilling machine is 1.2 m

time taken to cook the burger is 2.7 min = 162 s

so the speed of the machine should be like this that if must have to cook till it cross the machine

v = \frac{d}{t}

v = \frac{1.2}{162}

v = 7.41* 10^{-3} m/s

now in one minute the total length of the machine that is covered is given by

L = v*t

L = 7.41*10^{-3}* 60 = 44.4 cm

now distance between the burgers is 15 cm

so total production rate will be

N = \frac{44.4}{15} = 3 burger/min

so it will produce 3 burger per minute

8 0
3 years ago
Which planet has a moon with many sulfur volcanoes?
Murljashka [212]

Jupiter i hope it is right answer

6 0
2 years ago
Birdman is flying horizontally at a
zavuch27 [327]

Answer:

X=92.49 m

Explanation:

Given that

u= 21 m/s

h= 97 m

Time taken to cover vertical distance h

h= 1/2 g t²

By putting the values

97 = 1/2 x 10 x  t²          ( g = 10 m/s²)

t= 4.4 s

The horizontal distance

X= u .t

X= 21 x 4.4

X=92.49 m

3 0
3 years ago
An Atwood's machine consists of two different masses, both hanging vertically and connected by an ideal string which passes over
Tasya [4]

Answer:

V₁ = √ (gy / 3)

Explanation:

For this exercise we will use the concepts of mechanical energy, for which we define energy n the initial point and the point of average height and / 2

Starting point

    Em₀ = U₁ + U₂

    Em₀ = m₁ g y₁ + m₂ g y₂

Let's place the reference system at the point where the mass m1 is

     y₁ = 0

    y₂ = y

    Em₀ = m₂ g y = 2 m₁ g y

End point, at height yf = y / 2

    E_{mf} = K₁ + U₁ + K₂ + U₂

    E_{mf} = ½ m₁ v₁² + ½ m₂ v₂² + m₁ g y_{f} + m₂ g y_{f}

Since the masses are joined by a rope, they must have the same speed

     E_{mf} = ½ (m₁ + m₂) v₁² + (m₁ + m₂) g y_{f}

   E_{mf}= ½ (m₁ + 2m₁) v₁² + (m₁ + 2m₁) g y_{f}

How energy is conserved

   Em₀ =  E_{mf}

   2 m₁ g y = ½ (m₁ + 2m₁) v₁² + (m₁ + 2m₁) g y_{f}

   2 m₁ g y = ½ (3m₁) v₁² + (3m₁) g y / 2

   3/2 v₁² = 2 g y -3/2 g y

   3/2 v₁² = ½ g y

   V₁ = √ (gy / 3)

5 0
3 years ago
What is the displacement of the runner in 4 laps? *<br> 2 point
Serhud [2]

Answer:

There is no displacement.

Explanation:

Because the runner is running laps and returning to the original place, there is no displacement as displacement is relative to the change in location from the original position.

Hope this helps. . .

ly UwU

3 0
2 years ago
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