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Mumz [18]
3 years ago
7

A diverging lens of focal length 18.0m is used to view a shark that is 90.0m away from the lens. If the image formed is 1.0m lon

g, calculate the: (i) image distance; (ii) length of the shark.
Physics
1 answer:
Reil [10]3 years ago
7 0

Answer:

i. + 22.5 m ii. 4.0 m

Explanation:

i. Image distance

Using the lens formula

1/u + 1/v = 1/f where f = focal length = + 18.0 m, u = object distance = distance of shark away from lens = + 90.0 m and v = image distance from lens = unknown

So, we find v

1/v = 1/f - 1/u

= 1/+18 - 1/+90

= (5 - 1)/90

= 4/90

v = 90/4

= + 22.5 m

So the image is real and formed 22.5 m away on the other side of the lens.

ii Length of Shark

Using the magnification formula, m = image height/object height = image distance/object distance. image height = 1.0 m where object height = length of shark.

m = image distance/object distance

= v/u

= +22.5/+90

= 0.25

0.25 = image height/object height

So,

object height = image height/0.25

= 1.0 m/0.25

= 4.0 m

So, the length of the shark is 4.0 m

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I would most likely go with B. but im not 100% sure

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4 years ago
The opening to a cave is a tall, 30.0-cm-wide crack. A bat is preparing to leave the cave emits a 30.0 kHz ultrasonic chirp. How
Vlada [557]

Answer:

The value is  w =  7.54 \  m        

Explanation:

From the question we are told that

     The length of the crack is  a =  0.3 \  m

     The  frequency is  f =  30.0 \ kHz =  30 *10^{3} \  Hz

      The distance outside the cave that is being consider is  D =  100 \  m

      The speed of sound is v_s =  340 \  m/s

Generally the wavelength of the wave is mathematically represented as

        \lambda =  \frac{v}f}

=>     \lambda =  \frac{340 }{30*10^{3}}

=>     \lambda = 0.0113 \ m/s

Generally for a  single slit the path difference between the interference patterns of the sound wave and the center  is mathematically represented as  

          y =  \frac{ n *  \lambda * D}{a}

=>     y =  \frac{ 1  *  0.0113 * 100}{0.3}

=>     y = 3.77 \  m

Generally the width of the sound beam is mathematically represented as

         w =  2 *  y

=>      w =  2 *  3.77

=>      w =  7.54 \  m        

4 0
3 years ago
If the sum of the external forces on an object is zero, then the sum of the external torques on it
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Answer:

True.

Explanation:

If the sum of the external forces on an object is zero, then the sum of the external torques on it  must also be zero.

The net external force and the net external torque acting on the object have to be zero for an object to be in mechanical equilibrium.

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How do you calculate energy lost due to friction in an experiment?
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Explanation:

6 0
4 years ago
Suppose that the height of the incline is h = 14.7 m. Find the speed at the bottom for each of the following objects. 1.solid sp
tensa zangetsu [6.8K]

Answer:

1. 14.4 m/s  2. 13.2 m/s 3. 12.0 m/s 4. 13.9 m/s

Explanation:  

Assuming no friction present, the different objects roll without slipping, so there is a constant relationship between linear and angular velocity, as follows:

ω= v/r

If no friction exists, the change in total kinetic energy must be equal in magnitude to the change in the gravitational potential energy:

∆K = -∆U

 ½ *m*v² + ½* I* ω²  = m*g*h

Simplifying and replacing the value of the angular velocity:

½ * v² + ½ I *(v/r)² = g*h (1)

In order to answer the question, we just need to replace h by the value given, and I (moment of inertia) for the value for each different object, as follows:

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                Replacing in (1):

                ½ * v² + ½ (2/5 *m*r²) *(v/r)² = g*h

                Replacing by the value given for h, and solving for v:

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  • Spherical shell I=2/3*m*r²

                Replacing in (1):

                ½ * v² + ½ (2/3 *m*r²) *(v/r)² = g*h

                Replacing by the value given for h, and solving for v:

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  • Hoop   I= m*r²

                Replacing in (1):

                ½ * v² + ½ (m*r²) *(v/r)² = g*h

               Replacing by the value given for h, and solving for v:

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                 Replacing in (1):

                ½ * v² + ½ (1/2 *m*r²) *(v/r)²= g*h

                 Replacing by the value given for h, and solving for v:

                v = 2*√(1/3*9.8 m/s2*14.7 m)  = 13.9 m/s

5 0
3 years ago
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