A385-g tile hangs from one end of a string that goes over a pulley with a moment of inertia of and a radius of 15.0 cm. a mass of 710 g hangs from the other end of the string. when the tiles are released, the larger one accelerates downward while the lighter one accelerates upward. the pulley has no friction in its axle and turns without the string slipping. what is the tension in the string on the side of the 710-g tile?

T1=5.588N

T2=4.543N

Explanation:

A 385-g tile hangs from one end of a string that goes over a pulley with a moment of inertia of and a radius of 15.0 cm. A mass of 710 g hangs from the other end of the string. When the tiles are released, the larger one accelerates downward while the lighter one accelerates upward. The pulley has no friction in its axle and turns without the string slipping. What is the tension in the string on the side of the 710-g tile?

convert grams to kilograms

Let

M₁ = mass 1 = 0.710 kg

M₂ = mass 2 = 0.385 kg

I = moment of inertia of the pulley = 0.0125 kgm²

R = radius of the pulley = 0.15 m

T₁ = tension in the string in connection to M₁

T₂ = tension in the string in connection to M₂

A = acceleration of the system

lets ave a deep dive into the formula

M₁×g - T₁ = M₁×A

(0.710 kg)×(9.8 m/s²) - T₁ = (710 kg)×A

(6.958 N) - T₁ = (0.710 kg)×A

T₁ = (6.958N) - (0.710 kg)×A . . . . . . . equation 1

 second equation: as follows

T₂ - M₂×g = M₂×A

T₂ - (0.385 kg)×(9.8 m/s²) = (0.385 kg)×A

T₂ - (3.773N) = (0.385 kg)×A

T₂ = (0.385 kg)×A + (3.773 N) . . . . . . . 2

third equation:

T₁×R - T₂×R = I×A/R

T₁×(0.15 m) - T₂×(0.15 m) = (0.0125 kgm²)×A/(0.15 m)

T₁×(0.15 m) - T₂×(0.15 m) = (0.0833 kgm)×A . . . . . . . 3

substitute the values f 1 and 2 in equation 3

T₁ = (6.958N) - (0.710 kg)×A

T₂ = (0.385 kg)×A + (3.773 N)

(6.958N) - (0.710 kg×A)×(0.15 m) - (0.385 kg)×A + (3.773 N)×(0.15 m) = (0.0833 )×A . . . . . . . 3

1.0437-0.106A-(0.05A+.565)=0.0833A

0.477=0.0833A+0.05A+0.106A

0.477=0.2393A

A=1.993m^2

A=2m/s^2

from equation 1

T₁ = (6.958N) - (0.710 kg)×2

T1=5.588N

from equation 2

substituting the value of A, which is the acceleration

T₂ = (0.385 kg)×2 + (3.773 N)

T2=4.543N

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2