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UkoKoshka [18]
3 years ago
15

A piece of string 2 meters and has a mass of 5g. On one end of the string hangs a 200 g mass. Find the tension of the string and

compute for the velocity of the speed of the wave generated. What is the mass constant in the string? (Answer: T=1.96N; U = 2.5 x10 3kg/m; v = 28 m/s)
Physics
1 answer:
viktelen [127]3 years ago
4 0

Explanation:

It is given that,

Length of the string, l = 2 m

Mass of the string, m=5\ g=5\times 10^{-3}\ kg

Hanged mass in the string, m'=200\ g=0.2\ kg

1. The tension in the string is given by :

T=m'g

T=0.2\times 9.8

T = 1.96 N

2. Velocity of the transverse wave in the string is given by :

v=\sqrt{\dfrac{T}{M}}

m = M/l

v=\sqrt{\dfrac{Tl}{m}}

v=\sqrt{\dfrac{1.96\times 2}{5\times 10^{-3}}}

v = 28 m/s

Hence, this is the required solution.

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Mariana [72]

Answer:

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3 years ago
Name:
Brums [2.3K]

Answer:

1.a) 1 kJ

1.b) 4 kJ

     ratio 1:4

1.c) 4 times as before

2.a)  3.33 m/s2

Explanation:

1.a) bicycle's velocity =Displacement/time

                                   =100/20 m/s

                                   =5 m/s

bicycler's KE =1/2 *mass*(velocity)^2

                      =1/2*80*5^2

                       =1000 J = 1 kJ

1.b) bicycle's new velocity =200/20 m/s

                                   =10 m/s

bicycler's new KE =1/2*80*10^2

                             =4000 J = 4 kJ

Ratio= KE 1 :KE new

        = 1 :4

1.c)  when bicycler's speed was doubled it increased the KE by 4 times (2^2). because In KE we consider the square of the speed , so the factor we increase the speed , the KE will get increased with the square value of it

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2.a) car acceleration = (20-0)/6 m/s2

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3 years ago
How are acceleration and speed related​
Gre4nikov [31]

Answer:

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Explanation:

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6 0
3 years ago
Read 2 more answers
A hot (70°C) lump of metal has a mass of 250 g and a specific heat of 0.25 cal/g⋅°C. John drops the metal into a 500-g calorimet
Gnom [1K]

Answer:

d. 37 °C

Explanation:

m_{m} = mass of lump of metal = 250 g

c_{m} = specific heat of lump of metal  = 0.25 cal/g°C

T_{mi} = Initial temperature of lump of metal = 70 °C

m_{w} = mass of water = 75 g

c_{w} = specific heat of water = 1 cal/g°C

T_{wi} = Initial temperature of water = 20 °C

m_{c} = mass of calorimeter  = 500 g

c_{c} = specific heat of calorimeter = 0.10 cal/g°C

T_{ci} = Initial temperature of calorimeter = 20 °C

T_{f} = Final equilibrium temperature

Using conservation of heat

Heat lost by lump of metal = heat gained by water + heat gained by calorimeter

m_{m} c_{m} (T_{mi} - T_{f}) = m_{w} c_{w} (T_{f} - T_{wi}) +  m_{c} c_{c} (T_{f} - T_{ci}) \\(250) (0.25) (70 - T_{f} ) = (75) (1) (T_{f} - 20) + (500) (0.10) (T_{f} - 20)\\T_{f} = 37 C

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4 years ago
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