Physics, 11.04.2020 03:47 aubreystechschu11331

A 2.00-kg textbook rests on a frictionless, horizontal surface. A cord attached to the book passes over a pulley whose diameter is 0.150 m, to a hanging book with mass 3.00 kg. The system is released from rest, and the books are observed to move 1.20 m in 0.800 s. (a) What is the tension in each part of the cord? (b) What is the moment of inertia of the pulley about its rotation axis? Can you explain why the tensions are not the same even though it is the same cord.

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Answer from: onna172001

a) T = 7.5 N

T' = 18.15 N

b) I = 0.016 kgm²

Explanation:

Given that:

Mass of the textbook m = 2 kg

Diameter of the pulley d = 0.150 m

Hanged mass m' = 3 kg

Displacement s = 1.2 m

Time t = 0.800 s

According to kinematics equation

Displacement s can be derived from the second equation of motion:

$s = ut + \frac{1}{2} at ^2$

where u = 0

$s = \frac{1}{2} at ^2$

making acceleration a the subject of the formula; we have:

$a = \frac{2s}{t^2}$

$a = \frac{2*1.2}{0.8^2}$

a = 3.75 m/s^2

Now; taking into account of mass m;

The tension in the cord attached to the book on the horizontal surfacce can be calculated as:

T = ma

T = 2 × 3.75

T = 7.5 N

For the mass m; the tension is calculated as :

m'g - T' = m' a

T' = m'(g-a)

T' = 3 × (9.8 - 3.75)

T' = 18.15 N

Considering the pulley:

$(T'-T) r = I\alpha$

where;

$\alpha = \frac{a}{r}$

Then

$(T'-T) r = I* \frac{a}{r}$

Then the moment of inertia I/ can be re-written as :

$I = (T'-T)\frac{r^2}{a}$

$I = (18.15-7.5)\frac{0.75^2}{3.75}$

I = 0.016 kgm²

Answer from: Quest

find the sum of the two horizontal vectors and the sum of the two vertical vectors

isoceles right angled triangle

Answer from: Quest

Answer: a charge of -1.0 × 10⁻⁴ c must be placed at x = 26 m.the sign of the charge will be the same sign as the -4.0 µc charge.======electricstatic force between two point charges is given by $f = \dfrac{kq_1q_2}{r^2}$ where k is coulomb's constant, 8.99×10⁹ n·m²/c², q₁ and q₂ are the point charges, and r is the distance between the two point charges.assign positive to q₁let q₁ = 6.0 µc, q₂ = -4.0 µc, q₀ represent the charge at the origin, and q₃ represent this new charge to place.the charge at the origin q₀ would experience the following force f₀₁ from q₁, with r = 6.0 m (note that 1 µc = 10⁻⁶ c) $f_{01} = \dfrac{kq_0q_1}{(\text{6.0 m})^2}$ the force from q₂ $f_{02} = \dfrac{kq_0q_2}{(\text{15 m})^2}$ the force from q₃ $f_{03} = \dfrac{kq_0q_1}{(\text{26 m})^2}$ the net force on the origin charge, q₀, has to be zero for no electrostatic forcehaving an equation where k and q₀ can be divided out of the equation will allow us to solve for q₃ $\begin{aligned} f_{net} & = 0\\ f_{01} + f_{02} + f_{03} & = 0 \\ \dfrac{kq_0q_1}{(\text{6.0 m})^2} + \dfrac{kq_0q_2}{(\text{15 m})^2} + \dfrac{kq_0q_3}{(\text{26 m})^2} & = 0 \\ kq_0 \left( \dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2} + \dfrac{q_3}{(\text{26 m})^2}\right) & = 0 \\ \dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2} + \dfrac{q_3}{(\text{26 m})^2} & = 0 \end{aligned}$ isolating: $\begin{aligned} \dfrac{q_3}{(\text{26 m})^2} & = -\dfrac{q_1}{(\text{6.0 m})^2} - \dfrac{q_2}{(\text{15 m})^2} \\ q_3 & = -(\text{26 m})^2\left(\dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2}\right) \\ q_3 & = -(\text{26 m})^2\left(\dfrac{6.0\times10^{-6} \text{ c}}{(\text{6.0 m})^2} + \dfrac{-4.0\times10^{-6} \text{ c}}{(\text{15 m})^2}\right) \\ q_3 & = -1.0064888 \times 10^{-4}\text{ c} \\ q_3 & = -1.0\times 10^{-4}\text{ c} \end{aligned}$ a charge of -1.0 × 10⁻⁴ c must be placed at x = 26 m.the sign of the charge will be the same as the -4.0 µc charge

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