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zmey [24]
1 year ago
10

A bead of mass 15g is sliding on a wire. It has a speed of 2.0m/s at A, and it stops as it reaches the point C. The length of th

e wire from A to C is 250cm. How large an average friction force opposed the motion of the bead?
Physics
1 answer:
umka2103 [35]1 year ago
7 0

Answer:

.12 N

Explanation:

Velocity goes from 2 m/s    to   0  m/s   so average velocity = 1 m/s

.250 m / 1 m/s  =  .25 second to stop

change in velocity / change in time = accel

  2 m/s / .25 sec =  8 m/s^2

F = ma

  = .015 kg *  8 m/s^2  = .12 N

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Three pendulums all have the same length and start from the same height. The first pendulum is very light and has a mass of 67 g
vovikov84 [41]

Answer:

All three pendulum will attain same velocity

Explanation:

All three pendulum will attain same velocity irrespective of their mass difference in isolated system (means where air drag are negligible) and at same length

As you know when velocity is calculated we can not take mass into account.  

3 0
3 years ago
The objective lens of a microscope has a focal length of 5.5mm. Part A What eyepiece focal length will give the microscope an ov
son4ous [18]

Complete Question

The distance between the objective and eyepiece lenses in a microscope is 19 cm . The objective lens has a focal length of 5.5 mm .

What eyepiece focal length will give the microscope an overall angular magnification of 300?

Answer:

The  eyepiece focal length is  f_e  = 0.027 \ m

Explanation:

From the question we are told that

    The focal length is  f_o =  5.5 \ mm =  -0.0055 \ m

This negative sign shows the the microscope is diverging light

     The  angular magnification is m = 300

     The  distance between the objective and the eyepieces lenses is  Z =  19 \ cm  = 0.19 \ m

Generally the magnification is mathematically represented as

        m  =  [\frac{Z - f_e }{f_e}] [\frac{0.25}{f_0} ]

Where f_e is the eyepiece focal length of the microscope

  Now  making f_e the subject  of the formula

         f_e  = \frac{Z}{1 - [\frac{M  *  f_o }{0.25}] }

substituting values

        f_e  = \frac{ 0.19 }{1 - [\frac{300  *  -0.0055 }{0.25}] }

         f_e  = 0.027 \ m

     

5 0
3 years ago
A ball is thrown horizontally from the top of a 60.0-m building and lands 100.0 m from the base of thebuilding. Ignore air resis
Bumek [7]

Answer:

a)3.5s

b)28.57m/S

c)34.33m/S

d)44.66m/S

Explanation:

Hello!

we will solve this exercise numeral by numeral

a) to find the time the ball takes in the air we must consider that vertically the ball experiences a movement with constant acceleration whose value is gravity (9.81m / S ^ 2), that the initial vertical velocity is zero, we use the following equation for a body that moves with constant acceleration

Y= VoT+0.5gt^{2}

where

Vo = Initial speed =0

T = time

g=gravity=9.81m/s^2

y = height=60m

solving for time

Y=0.5gt^2\\t=\sqrt{\frac{Y}{0.5g} } \\t=\frac{60}{0.5(9.81)} \\

T=3.5s

b)The horizontal speed remains constant since there is no horizontal acceleration. with the value of the distance traveled (100m) and the time that lasts in the air (3.5s) we estimate the horizontal speed

V=\frac{x}{t} =\frac{100}{3.5}=28.57m/s

c)

to find the final vertical velocity we use the equations for motion with constant velocity as follows

Vf=Vo+g.t    

Vf=0+(9.81 )(3.5)=34.335m/S          

d)Finally, to find the resulting velocity, we add the horizontal and vertical velocities vectorially, this is achieved by finding the square root of the sum of its squares

V=\sqrt{Vx^2+Vy^2} =\sqrt{34.33^2+28.57^2} =44.67m/S

7 0
3 years ago
What’s a real life example of relative motion?
vesna_86 [32]

Answer:

let's say you're on a bus going 50 km/hr, you are moving at a velocity of zero relative to the bus. however, relative to the ground you are moving at the same velocity as the bus.

Explanation:

physics

7 0
3 years ago
Describe what happens to radiant energy once it enters the atmosphere? How does Radiant energy cause weather?
Triss [41]
Once energy from the Sun gets to Earth, several things can happen to it:

Energy can be scattered or absorbed by aerosols in the atmosphere. Aerosols are dust, soot, sulfates and nitric oxides. When aerosols absorb energy, the atmosphere becomes warmer. When aerosols scatter energy, the atmosphere is cooled.
Short wavelengths are absorbed by ozone in the stratosphere.
Clouds may act to either reflect energy out to space or absorb energy, trapping it in the atmosphere.
The land and water at Earth's surface may act to either reflect energy or absorb it. Light colored surfaces are more likely to reflect sunlight, while dark surfaces typically absorb the energy, warming the planet.
Albedo is the percentage of the Sun's energy that is reflected back by a surface. Light colored surfaces like ice have a high albedo, while dark colored surfaces tend to have a lower albedo. The buildings and pavement in cities have such a low albedo that cities have been called "heat islands" because they absorb so much energy that they warm up.
6 0
3 years ago
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