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umka21 [38]
3 years ago
11

Two identical gliders on an air track are held together by a piece of string, compressing a spring between the gliders. While th

ey are moving to the right at a common speed of 0.325 m/s, someone holds a match under the string and burns it, letting the spring force the gliders apart. One glider is then observed to be moving to the right at 1.33 m/s. What velocity does the other glider have
Physics
1 answer:
baherus [9]3 years ago
3 0

Answer:

Explanation:

Both the gliders have same mass and same speed of .325 m /s .

The detachment of spring will create equal and opposite force on both of them , one force adding momentum to one and the other force reducing the momentum of the other . Since mass of both the gliders are same , change in velocity too will be same . Let change in velocity be v .

for forward moving glider

.325 + v = 1.33

v = 1.005 m /s

The other glider will have final velocity

= .325 - 1.005

= - .68 m /s

So other glider will go in opposite direction with velocity of .68 m /s .

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A physics professor demonstrates the Doppler effect by tying a 600 Hz sound generator to a 1.0-m-long rope and whirling it aroun
nalin [4]

Answer:

Therefore the the highest frequency is 620Hz and lowest frequency is 580Hz

Explanation:

Given data

Source Frequency fs=600Hz

Length r=1.0m

RPM=100 rpm

The speed of the generator is calculated as:

v_{s}=rw\\v_{s}=r(2\pi f)

Substitute the given values

v_{s}=(1.0m)2\pi (\frac{100}{60}rev/s )\\v_{s}=10.47m/s

For approaching generator the frequency is calculated as:

f_{+}=\frac{f_{s}}{1-\frac{v_{s}}{v} }\\f_{+}=\frac{600Hz}{1-\frac{10.47m/s}{343m/s} } \\f_{+}=620Hz

On the other hand,for the receding generator,Doppler's effect is expressed as:

 f_{-}=\frac{f_{s}}{1+\frac{v_{s}}{v} }\\f_{-}=\frac{600Hz}{1+\frac{10.47m/s}{343m/s} } \\f_{-}=580Hz

Therefore the the highest frequency is 620Hz and lowest frequency is 580Hz

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4 years ago
I need help with number 6
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7 0
3 years ago
You are holding a shopping basket at the grocery store with two 0.55-kg cartons of cereal at the left end of the basket. the bas
stepan [7]
Refer to the diagram shown below.

The basket is represented by a weightless rigid beam of length 0.78 m.
The x-coordinate is measured from the left end of the basket.

The mass at x=0 is 2*0.55 = 1.1 kg.
The weight acting at x = 0 is W₁ = 1.1*9.8 = 10.78 N

The mass near the right end is 1.8 kg.
Its weight is W₂ = 1.8*9.8 = 17.64 N

The fulcrum is in the middle of the basket, therefore its location is
x = 0.78/2 = 0.39 m.

For equilibrium, the sum of moments about the fulcrum is zero.
Therefore
(10.78 N)*(0.39 m) - (17.64 N)*(x-0.39 m) = 0
4.2042 - 17.64x + 6.8796 = 0
-17.64x = -11.0838
x = 0.6283 m

Answer:  0.63 m from the left end.

5 0
3 years ago
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3 years ago
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In the Bohr model of the hydrogen atom, an electron({rm mass};m=9.1; times 10^{ - 31;}{rm kg}) orbits a proton at a distance of
max2010maxim [7]

Answer:

n=6.56×10¹⁵Hz

Explanation:

Given Data

Mass=9.1×10⁻³¹ kg

Radius distance=5.3×10⁻¹¹m

Electric Force=8.2×10⁻⁸N

To find

Revolutions per second

Solution

Let F be the force of attraction

let n  be the number of revolutions per sec made by the electron around the nucleus then the centripetal force is given by

F=mω²r......................where ω=2π  n

F=m4π²n²r...............eq(i)

as the values given where

Mass=9.1×10⁻³¹ kg

Radius distance=5.3×10⁻¹¹m

Electric Force=8.2×10⁻⁸N

we have to find n from eq(i)

n²=F/(m4π²r)

n^{2} =\frac{8.2*10^{-8} }{9.11*10^{-31}* 4\pi^{2} *5.3*10^{-11}  }\\ n^{2}=4.31*10^{31}\\ n=\sqrt{4.31*10^{31}}\\ n=6.56*10^{15}Hz

8 0
3 years ago
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