A string has a mass of 0.8Kg. The length of the string is 2m. The tension in the string is 5N. The string is stretched and fixed at both ends such that when struck it forms the second harmonic. Calculate the frequency of the wave which results on the string.

1.25Hz

Explanation:

For waves on a string, the second harmonic is obtained from;

2fo = 1/l √T/M

Where;

l = length of the string

M= mass in kilograms

T = tension in the string

2fo = 1/2√5/0.8

2f0 = 1.25Hz

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft

the answer is false. this is because for ice to melt, it requires latent heat of melting. latent heat means heat energy required to change phase without any increase in temperatures. therefore, the ice will take energy from the environment to melt hence atmospheric temperatures will reduce.