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kogti [31]
2 years ago
13

A fly travels along the x-axis. His starting point is x = 16 m and his ending point 15x = - 25 m. His flight lasts 4.0 seconds H

ow far has he flown, what is his displacement , what is his speed and what is his velocity?
Physics
1 answer:
gogolik [260]2 years ago
5 0

Explanation:

Starting position at x  = 16m

Ending position at x  = -25m

Time of flight  = 4s

Unknown:

Distance flown  = ?

Displacement  = ?

Speed  = ?

Velocity  = ?

Solution:

To find the distance flown, we should understand that the body is moving on the x - plane;  

  So distance  = 16 + 25  = 41m

Displacement is 41m to the left or -x axis

Speed is the distance divided by the time taken;

      Speed  = \frac{distance}{time}   = \frac{41}{4}   = 10.25m/s

Velocity is 10.25m/s along -x axis

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6 0
3 years ago
A 2100 g block is pushed by an external force against a spring (with a 22 N/cm spring constant) until the spring is compressed b
Vilka [71]

Answer:

6.5e-4 m

Explanation:

We need to solve this question using law of conservation of energy

Energy at the bottom of the incline= energy at the point where the block will stop

Therefore, Energy at the bottom of the incline consists of the potential energy stored in spring and gravitational potential energy=\frac{1}{2} kx^{2} +PE1

Energy at the point where the block will stop consists of only gravitational potential energy=PE2

Hence from Energy at the bottom of the incline= energy at the point where the block will stop

⇒\frac{1}{2} kx^{2} +PE1=PE2

⇒PE2-PE1=\frac{1}{2} kx^{2}

Also PE2-PE2=mgh

where m is the mass of block

g is acceleration due to gravity=9.8 m/s

h is the difference in height between two positions

⇒mgh=\frac{1}{2} kx^{2}

Given m=2100kg

k=22N/cm=2200N/m

x=11cm=0.11 m

∴2100*9.8*h=\frac{1}{2}*2200*0.11^{2}

⇒20580*h=13.31

⇒h=\frac{13.31}{20580}

⇒h=0.0006467m=6.5e-4

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3 years ago
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Answer:

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2 years ago
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The distance between the objective and eyepiece lenses in a microscope is 19 cm . The objective lens has a focal length of 5.5 m
kramer

Answer:

The focal length of eyepiece is 3.68 cm.

Explanation:

Given that,

Distance = 19 cm

Focal length = 5.5 mm

Magnification = 200

Object distance = -25 cm

We need to calculate the focal length

Using formula of magnification

m=\dfrac{d}{f_{o}}+\dfrac{-25}{f_{0}f_{e}}

Put the value into the formula

f_{e}=\dfrac{19\times(-25)}{.55(-200-\dfrac{19}{.55})}

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Hence, The focal length of eyepiece is 3.68 cm.

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3 years ago
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