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daser333 [38]
3 years ago
10

A string is attached to a ball that has a mass of 0.11 kg. A student pulls up on the string so that the ball accelerates upward

at 0.5 m/s2. What is the tension in the string? (Take the magnitude of acceleration due to gravity to be 9.8 m/s2.)
(Hint: How do all the forces involved interact with each other?)
Physics
1 answer:
enot [183]3 years ago
7 0

Answer:

T=+1.133N

Explanation:

Tension and weight are forces that have opposite directions

Weight is negative (downward)

W=m*g= 0.11kg*(-9.8m/s^2)

W= -1.078N

Tension is possitive (upward)

The total force will be the sum of both (the difference taking in consideration the direction)

Ft= T+W

Also the total force is the product of the mass due to acceleration:

Ft=m*a

Ft= +0.11kg*0.5m/s^2

Ft=+0.055N (upward)

Tension will be the difference between Ft and W:

T= Ft-W

T=+0.055N-(-1.078N)

T=+1.133N

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A 2.20-kg object is attached to a spring and placed on frictionless, horizontal surface. A horizontal force of 29.0 N is require
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Answer:

a. 145 N/m b. 1.29 Hz c. 1.62 m/s d.  0 m e. 13.2 m/s² f. ± 0.2 m g. 2.9 J h. 0.54 m/s i. 4.39 m/s²

Explanation:

a. The force constant of the spring

The spring force F = kx and k = F/x where k is the spring constant. F = 29.0 N and x = 0.200 m

k = 29.0 N/0.200 m = 145 N/m

b. The frequency of oscillations, f

f = 1/2π√(k/m)    m = mass = 2.20 kg

f = 1/2π√(145 N/m/2.20 kg) = 1.29 Hz

c. maximum speed of the object

The maximum elastic potential energy of the spring = maximum kinetic  energy of the object

1/2kx² = 1/2mv²

v = (√k/m)x where v is the maximum speed of the object

v = (√145/2.2)0.2 = 1.62 m/s

d Where does the maximum speed occur?

The maximum speed occurs at  0 m

e. The maximum acceleration

a = kx/m = 145 × 0.2/2.2 = 13.2 m/s²

f. The maximum acceleration occurs at x = ± 0.2 m

g. The total energy of the system is the maximum elestic potential energy of the system

E = 1/2kx² = 1/2 × 145 × 0.2² = 2.9 J

h. When x = x₀/3

1/2k(x₀/3)² = 1/2mv²

kx₀²/9 = mv²

v = 1/3(√k/m)x₀ = 1/3(√145/2.2)0.2 = 0.54 m/s

i When x = x₀/3

a = kx₀/3m =  145 × 0.2/(2.2 × 3)= 4.39 m/s²

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Find the current passing through a circuit consisting of a battery and one resistor. The resistance has a retance of 2ohms and t
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The current passing through a circuit consisting of a battery of 12 V and resistor of 2 ohms is 6 Ampere .

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