Answer:
a
The speed of wave is 
b
The new speed of the two waves is 
Explanation:
From the question we are told that
The mass of the string is 
The length is 
The tension is 
Now the velocity of the first wave is mathematically represented as

Where
is the linear density which is mathematically represented as

substituting values


So


Now given that the Tension, mass and length are constant the velocity of the second wave will same as that of first wave (reference PHYS 1100 )
Answer: A Neutron.
I know this is correct. Thank's and yw :3
Answer: a) - 437.8° F, b) - 261°c.
Explanation: a) the kelvin and Fahrenheit temperature scale are related by the formulae below.
5 (°F - 32) = 9 (k - 273)
Where °F = temperature in Fahrenheit and k = temperature in kelvin.
For question A, k = 12.0, by substituting to have the value for °F, we have
5(°F - 32) = 9 ( 12 - 273)
5(°F - 32) = 9(-261)
5(°F - 32) = - 2349
°F - 32 = - 2349/5
°F - 32 = - 469.8
°F = - 469.8 + 32
°F = - 437.8
Question B
The centigrade and kelvin scale are related by the formulae below
°c = k - 273
Where °c = temperature in centigrade and k = temperature in kelvin =12
°c = 12 - 273
°c = - 261
Answer:
70713
Explanation:
Because you need to multiply the amount of water lost (2430) by the time (29.1) which will equal 70713J/g needed to counter the loss.
Hope this helps:)