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Anna007 [38]
2 years ago
5

a canary sits 10 m from the edge of a 30-m-long clothesline, and a grackle sits 5 m from the other end. The rope is pulled by tw

o poles that each exerts a 200-N force on it. The mass per unit length is 0.10 kg/m. At what frequency must you vibrate the line in order to dislodge the grackle while allowing the canary to sit undisturbed?
Physics
1 answer:
rosijanka [135]2 years ago
6 0

Answer: frequency = 2.23Hz

Explanation:

1. The rope on which the birds stands is pulled by two poles that each exerts 200N on it. The mass per unit length = u = 0.1kg/m

2. In order to keep the canary undisturbed, a standing wave should exist on the rope while the position of the canary must be at one of the modes of the standing wave. If we started measuring the distance from the length side of the rope we can that x = 5m (Position of the grackle), the wave should have a node at x = 20m ( position of canary). We can deduce that ∆ = 2/3 × L

Therefore, ∆ = 2/3 × 30 = 20m

3. Therefore, speed of the wave on the rope using the relation V=√f/u

We substitute u = 0.1kg/m and f = 200N

V = √200N/0.1kg/m = 44.7m/s

Therefore, frequency f = V/∆ = 44.7m/s/20m = 2.23Hz

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MrMuchimi

Answer:

Summer begins in the Northern Hemisphere on June 20 or 21 when the North Pole is tilted a full 23.5° toward the sun. On this day, the Northern Hemisphere has the most hours of daylight, while the Southern Hemisphere has the least hours of daylight.

Explanation:

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3 years ago
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At t = 0, a car registers at 30 miles/hr. Forty seconds later, the car’s velocity is now at 50 miles/hr. Assuming constant accel
ratelena [41]
D.

50 mph - 30 mph= 20 mph net velocity
change.
20mph/3600 seconds/hour= .00555 MPS
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.0055/40= .000138

7 0
2 years ago
The most recently discovered system close to Earth is a pair of brown dwarfs known as Luhman 16. It has a distance of 6.5 light-
just olya [345]

Answer:

1.99 parsecs.

Explanation:

We have been given that the most recently discovered system close to Earth is a pair of brown dwarfs known as Luhman 16. It has a distance of 6.5 light-years.

We know that one light year equals to 0.306601 parsecs. To convert 6.5 light-years to parsecs, we will multiply 0.306601 by 6.5.

6.5\text{ Light-years}=6.5\times 0.306601\text{ Parsecs}

6.5\text{ Light-years}=1.9929065\text{ Parsecs}

6.5\text{ Light-years}\approx 1.99\text{ Parsecs}

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5 0
3 years ago
(a) what is the acceleration of two falling sky divers (mass 132 kg including parachute) when the upward force of air resistance
Komok [63]

As per the question the mass of two falling sky drivers is 132 kg.

First we have to calculate their acceleration.

Whenever a body falls freely under gravity,its acceleration is acceleration due to gravity i.e g whose value is 9.8 m/s^2.

The earth pulls the object with a force equal to the weight of the body.

Hence the force gravity  F=W= mg   [ here m is mass of the body]

Here m =132 kg.

Hence force of gravity F= mg

                                        =132 kg ×9.8 m/s^2

                                        =1293.6 kg m/s^2

                                         =1293.6 N      [ here N[newton] is the unit of force.]

As per the question the air resistance is one fourth of weight of the bodies.

Hence air resistance F' =1/4 mg

                                       =\frac{1}{4} *1293.6N

                                        =323.4 N

Here F acts in vertically downward direction while F' acts in vertically upward direction.

Hence the net force acting on the particle is F-F'.

                                                 F_{net} =1293.6N -323.4N

                                                         =970.2 N

From Newton's second law of motion we know that net force is the product of mass and acceleration i.e  

                                   F_{net} =ma  [Here a is the acceleration]

                                             a =\frac{F_{net} }{m}

                                                  = \frac{970.2}{132} m/s^2

                                                  =7.35 m/s^2

In the second question it has been told that they descend with uniform speed.hence acceleration of the two bodies will be zero.

 we know that F= ma

                           =m×0

                            =0 N

Hence they will not get any force when they will descend with a uniform speed.


4 0
2 years ago
Joe the house painter stands on a uniform oak board weighing 600 N and held up by vertical ropes at each end. Joe has been dieti
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Answer

given,

weight of the oak board = 600 N

Weight of Joe = 844 N

length of board = 4 m

Joe is standing at 1 m from left side

vertical wire is supporting at the end.

Assuming the system is in equilibrium

T₁ and T₂ be the tension at the ends of the wire

equating all the vertical force

T₁ + T₂ = 600 + 844

 T₁ + T₂ = 1444...........(1)

taking moment about T₂

 T₁ x 4 - 844 x 3 - 600 x 2 = 0

 T₁ x 4 = 3732

 T₁ = 933 N

from equation (1)

 T₂ = 1444 - 933

 T₂ = 511 N

3 0
3 years ago
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