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kicyunya [14]
3 years ago
8

A 290 gg bird flying along at 6.2 m/sm/s sees a 9.0 gg insect heading straight toward it with a speed of 34 m/sm/s (as measured

by an observer on the ground, not by the bird). The bird opens its mouth wide and enjoys a nice lunch.
What is the bird's speed immediately after swallowing?
Physics
1 answer:
Murrr4er [49]3 years ago
6 0

Answer:

The bird's speed immediately after swallowing is 4.98 m/s.

Explanation:

Given that,

Mass of bird = 290 g

Speed = 6.2 m/s

Mass of sees = 9.0 g

Speed = 34 m/s

We need to calculate the bird's speed immediately after swallowing

Using conservation of momentum

m_{1}v_{1}+m_{2}v_{2}=(m_{1}+m_{2})v_{3}

Put the value into the formula

0.290\times6.2+0.009\times(-34)=(0.290+0.009)\times v_{3}

v_{3}=\dfrac{0.290\times6.2-0.009\times34}{(0.290+0.009)}

v_{3}=4.98\ m/s

Hence, The bird's speed immediately after swallowing is 4.98 m/s.

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slamgirl [31]

Answer:

18.89cm

Explanation:

As we know that the person is standing 5m in front of the camera

d_0=5m=500cm

The focal length of the lens =50cm

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By Lens formula we have:

\dfrac{1}{f} = \dfrac{1}{d_i} + \dfrac{1}{d_o}\\\dfrac{1}{50} = \dfrac{1}{d_i} + \dfrac{1}{500}\\\dfrac{1}{d_i} =\dfrac{1}{50}-\dfrac{1}{500}\\\dfrac{1}{d_i}=0.018\\d_i=55.56cm

By the formula of magnification

\dfrac{h_i}{h_o} = \dfrac{55.56}{500}\\\\h_i = \dfrac{55.56}{500} \times h_o\\\\ h_o=1.70m=170cm\\\\Therefore: h_i=\dfrac{55.56}{500} \times$ 170 cm\\\\h_i =18.89 cm

The height of the image formed is 18.89cm.

5 0
3 years ago
Choose the situation below in which the force applied is the greatest.
Gnesinka [82]

Answer:

D

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We know the formula for Work to be:

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d is the distance

A)

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Distance = 50

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Force = 50/50 = 1

B)

Work = 400

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Force = 400/80 = 5

C)

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Distance = 73

Force = 365/73 = 5

D)

Work = 144

Distance = 16

Force = 144/16 = 9

Hence, D is the situation in which the force applied is the greatest.

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A. Ears <br> B. Air <br> C. Water <br> D. Rocks
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This potential energy is then converted to kinetic energy. Halfway, the kinetic energy is equal to KE1.

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KE = \frac{1}{2}mv^{2}

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