Arock with density 1900 kg/m3 is suspended from the lower end of a light string. when the rock is in air, the tension in the string is 48.0 nwhat is the tension in the string when the rock is totally immersed in a liquid with density 750 kg/m3?

the tension T2 when the rock is completely immersed is T2 =  29.05 N

Explanation:

from Newton's second law

F= m*a

where F= force , m= mass , a= acceleration

when the rock is suspended ,a=0 since it is at rest. Then

T1 - m*g = 0 , T1= tension when suspended in air , g= gravity

assuming constant density of the rock

m= ρ rock *V , where  ρ rock = density of the rock , V= volume

thus

T1= m*g = ρ rock *g*V

V=  T1/(ρ rock *g)

when the rock is submerged in oil , it receives an upward force that equals the weight of the volume of displaced oil (V displaced). Since it is completely submerged the volume displaced is the volume of the rock V=Vdisplaced  

When the rock is at rest , then

F= m*a=0

T2 + ρ oil *g*V displaced - ρ rock *g*V  =0

T2 = ρ rock *g*V - ρ oil *g*V = g*V (ρ rock - ρ oil)

T2 = g*V (ρ rock - ρ oil) = T1/(ρ rock *g) *g * (ρ rock - ρ oil)

T2 = T1 * (ρ rock - ρ oil)/ρ rock

replacing values

T2 = 48 N * (1900 kg/m3- 750 kg/m3)/ 1900 kg/m3 = 29.05 N

T2 =  29.05 N

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west

pretty sure it's false