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Hunter-Best [27]
2 years ago
11

A thin rod of length 1.4 m and mass 140 g is suspended freely from one end. It is pulled to one side and then allowed to swing l

ike a pendulum, passing through its lowest position with angular speed 1.09 rad/s. Neglecting friction and air resistance, find (a) the rod's kinetic energy at its lowest position and (b) how far above that position the center of mass rises.
Physics
2 answers:
m_a_m_a [10]2 years ago
5 0

Answer:

a The kinetic energy is  KE = 0.0543 J

b The height of the center of mass above that position is  h = 1.372 \ m    

Explanation:

From the question we are told that

  The length of the rod is  L = 1.4m

   The mass of the rod m = 140 = \frac{140}{1000} = 0.140 \ kg  

   The angular speed at the lowest point is w = 1.09 \ rad/s

Generally moment of inertia of the rod about an axis that passes through its one end is

                   I = \frac{mL^2}{3}  

Substituting values

               I = \frac{(0.140) (1.4)^2}{3}

               I = 0.0915 \ kg \cdot m^2

Generally the  kinetic energy rod is mathematically represented as

             KE = \frac{1}{2} Iw^2

                    KE = \frac{1}{2} (0.0915) (1.09)^2

                           KE = 0.0543 J

From the law of conservation of energy

The kinetic energy of the rod during motion =  The potential energy of the rod at the highest point

   Therefore

                   KE = PE = mgh

                        0.0543 = mgh

                             h = \frac{0.0543}{9.8 * 0.140}

                                h = 1.372 \ m    

                 

Lynna [10]2 years ago
3 0

Answer:

a) Kr = 0.0543 J

b) Δy = 0.0396 m

Explanation:

a) Given

L = 1.4 m

m = 140 g = 0.14 kg

ω = 1.09 rad/s

Kr = ?

We have to get the rotational inertia as follows

I = Icm + m*d²

⇒ I = (m*L²/12) + (m*(L/2)²)

⇒ I = (0.14 kg*(1.4 m)²/12) + (0.14 kg*(1.4 m/2)²)

⇒ I = 0.09146 kg*m²

Then, we apply the formula

Kr = 0.5*I*ω²

⇒ Kr = 0.5*(0.09146 kg*m²)*(1.09 rad/s)²

⇒ Kr = 0.0543 J

b) We apply the following principle

Ei = Ef

Where the initial point is the lowest position and the final point is at the maximum height that its center of mass can achieve, then we have

Ki + Ui = Kf + Uf

we know that ωf = 0 ⇒ Kf = 0

⇒ Ki + Ui = Uf

⇒ Uf - Ui = Ki

⇒ m*g*yf - m*g*yi = Ki

⇒ m*g*(yf - yi) = Ki

⇒ m*g*Δy = Ki

⇒ Δy = Ki/(m*g)

where

Ki = Kr = 0.0543 J

g = 9.81 m/s²

⇒ Δy = (0.0543 J)/(0.14 kg*9.81 m/s²)

⇒ Δy = 0.0396 m

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shusha [124]

Answer:

- 3.72 Ns.

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Explanation:

mass of racket, M = 1000 g = 1 kg

mass of ball, m = 60 g = 0.06 kg

initial velocity of racket, U = 11 m/s

initial velocity of ball, u = 18 m/s

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Let the final velocity of the racket is V.

(a) Momentum is defined as the product of mass and velocity of the ball.

initial momentum of the ball = m x u = 0.06 x 18 = 1.08 Ns

Final momentum of the ball = m x v = 0.06 x (- 44) = - 2.64 Ns

Change in momentum of the ball = final momentum - initial momentum

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Thus, the change in momentum of the ball is - 3.72 Ns.

(b) By use of conservation of momentum

initial momentum of racket and ball = final momentum of racket and ball

1 x 11 + 0.06 x 18 =  1 x V - 0.06 x 44

12.08 = V - 2.64

V = 9.44 m/s

Thus, the final velocity of the racket afetr the impact is 9.44 m/s .

3 0
3 years ago
Two capacitors of capacitances 25 µF and 50 µF are connected in series with a 33-V battery. How much energy is stored in the 25-
torisob [31]

Answer:

6.05×10⁻³ J

Explanation:

Note: Two capacitors connected in series behaves like two resistors connected in parallel.

Using

1/Ct = 1/C1+1/C2

Ct = (C1×C2)/(C1+C2)............................ Equation 1

Where Ct = combined capacitance of the two capacitor, C1 = Capacitance of the first capacitor, C2 = capacitance of the second capacitor.

Given: C1 = 25 µF, C2 = 50 µF

Substitute into equation 1

Ct = (25×50)/(25+50)

Ct = 1250/75

Ct = 16.67 µF.

Using

Q = CV.................... Equation 2

Where Q = Charge, V = Voltage.

Given: V = 33 V, C = 16.67 µF = 16.67×10⁻⁶ F

Substitute into equation 2

Q = 33(16.67×10⁻⁶)

Q = 5.5×10⁻⁴ C.

Since both capacitors are connected in series, the same amount of charge flows through them.

Using,

E = 1/2Q²/C.................. Equation 3

Where E = Energy stored in the 25-µF capacitor

Given: Q =5.5×10⁻⁴ C, C = 25 µF = 25×10⁻⁶ F

Substitute into equation 3

E = 1/2(5.5×10⁻⁴)²/ 25×10⁻⁶

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5 0
3 years ago
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ankoles [38]

Answer:

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Rama09 [41]
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3 0
2 years ago
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