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Paladinen [302]
2 years ago
15

The experimental apparatus shown in the figure above contains a pendulum consisting of a 0.66 kg ball attached to a string of le

ngth 0,60 m. The pendulum is released from rest at an angle of
60 degrees and collides with a ball of mass 0.22 kg initially at rest at the edge of a table. The 0.22 kg ball hits the floor a distance of 1.4 m from the edge of the table.
a. Calculate the speed of the 0.66 kg ball just before the collision
b. Calculate the speed of the 0.22 kg ball immediately after the collision
c. Calculate the speed of the 0.66 kg ball immediately after the collision
d. Indicate the direction of motion of the 0.66 kg ball immediately after the collision
To the left
to the right
e. Calculate the height to which the 0.66 kg ball rises after the collision
f. Based on your data, is the collision elastic?
Yes
No
Justify your answer.

Physics
1 answer:
lara31 [8.8K]2 years ago
8 0

The problem is solved and the questions are answered below.

Explanation:

a. To calculate the speed of the 0.66 kg ball just before the collision

V₀ + K₀ = V₁ + K₁

= mgh₀ = 1/2 mv₁²

where, h= r - r cosθ

V = \sqrt{2gh}

 V = 2.42 m/s

b. Calculate the speed of the 0.22 kg ball immediately after the collision

y = y₀ + Vy₀t - 1/2 gt²

0 = 1.2 - 1/2 gt²

t = 0.495 s

x = x₀ + Vx₀t

1.4 = 0 + vx₀ (0.495)

Vx₀ = 2.83 m/s

C. To Calculate the speed of the 0.66 kg ball immediately after the collision

m₁ v₁ = m₁ v₃ + m₂ v₄

(0.66)(2.42) = (0.66) v₃ + (0.22)(2.83)

V₃ = 1.48 m/s

D. To Indicate the direction of motion of the 0.66 kg ball immediately after the collision is to the right.

E. To Calculate the height to which the 0.66 kg ball rises after the collision

V₀ + k₀ = V₁ + k₁

1/2 mv₀² = mgh₁

h₁ = v₀²/2 g

  = 0.112 m

F. Based on your data, No the collision is not elastic.

Δk = 1/2 m₁v₃² =1/2 m₂v₄² - 1/2 m₁v₁²

     = 1/2 (0.66)(1.48)² + 1/2 (0.22)(2.83)² - 1/2 (0.66)(2.42)²

    = - 0.329 J

Hence, kinetic energy is not conserved.

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Answer:

Bulb 1 has more resistance.

Explanation:

Given that,

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The power of circuit 1, P₁ = 50 W

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We need to find the bulb that has a higher resistance.

The power of the bulb is given by :

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For bulb 1,

R_1=\dfrac{V^2}{P_1}\\\\R_1=\dfrac{(120)^2}{50}\\\\R_1=288\ \Omega

For bulb 2,

R_2=\dfrac{V^2}{P_2}\\\\R_2=\dfrac{(120)^2}{100}\\\\R_2=144\ \Omega

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8 0
2 years ago
A long line of charge with uniform linear charge density λ1 is located on the x-axis and another long line of charge with unifor
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Answer:

A.The positive z-direction

Explanation:

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We have to find the direction of electric field at z=a on the positive z-axis if \lambda_1 and \lambda_2 are positive.

The direction of electric field  at z=a on the positive z-axis  is positive z-direction .

Because \lambda_1 and \lambda_2 are positive and the electric field is  applied away from the positive charge.

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6 0
3 years ago
A sealed tank containing seawater to a height of 10.5 mm also contains air above the water at a gauge pressure of 2.95 atmatm. W
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Answer:

The water is flowing at the rate of 28.04 m/s.

Explanation:

Given;

Height of sea water, z₁ = 10.5 m

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To determine the speed of the water, apply Bernoulli's equation;

P_1 + \rho gz_1 + \frac{1}{2}\rho v_1^2 = P_2 + \rho gz_2 + \frac{1}{2}\rho v_2^2

where;

P₁ = P_{gauge \ pressure} + P_{atm \ pressure}

P₂ = P_{atm}

v₁ = 0

z₂ = 0

Substitute in these values and the Bernoulli's equation will reduce to;

P_1 + \rho gz_1 + \frac{1}{2}\rho v_1^2 =  P_2 + \rho gz_2 + \frac{1}{2}\rho v_2^2\\\\P_1 + \rho gz_1 + \frac{1}{2}\rho (0)^2 =  P_2 + \rho g(0) + \frac{1}{2}\rho v_2^2\\\\P_1 + \rho gz_1 =  P_2 + \frac{1}{2}\rho v_2^2\\\\P_{gauge} + P_{atm} + \rho gz_1 = P_{atm} + \frac{1}{2}\rho v_2^2\\\\P_{gauge} +  \rho gz_1 =  \frac{1}{2}\rho v_2^2\\\\v_2^2 = \frac{2(P_{gauge} +  \rho gz_1)}{\rho} \\\\v_2 = \sqrt{ \frac{2(P_{gauge} +  \rho gz_1)}{\rho} }

where;

\rho is the density of seawater = 1030 kg/m³

v_2 = \sqrt{ \frac{2(2.95*101325 \ + \  1030*9.8*10.5 )}{1030} }\\\\v_2 = 28.04 \ m/s

Therefore, the water is flowing at the rate of 28.04 m/s.

7 0
3 years ago
An ideal monatomic gas at 275 K expands adiabatically and reversibly to six times its volume. What is its final temperature (in
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The final temperature is 83 K.

<u>Explanation</u>:

For an adiabatic process,

T {V}^{\gamma - 1} = \text{constant}

\cfrac{{T}_{2}}{{T}_{1}} = {\left( \cfrac{{V}_{1}}{{V}_{2}} \right)}^{\gamma - 1}

Given:-

{T}_{1} = 275 \; K  

{T}_{2} = T \left( \text{say} \right)

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{V}_{2} = 6V

\gamma = \cfrac{5}{3} \;    (the gas is monoatomic)

\therefore \cfrac{T}{275} = {\left( \cfrac{V}{6V} \right)}^{\frac{5}{3} - 1}

 

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T  =  275 \times 0.30

T  =  83 K.

3 0
3 years ago
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lord [1]

Answer:

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Explanation:

From the question we are told that

    The  magnitude of the horizontal force is F  =  92.7 \  N

     The mass of the crate is  m  =  40.5 \  kg

     The acceleration of the crate is  a =  1.13 \ m/s

Generally the net force acting on the crate is mathematically represented as

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Here F_f is force of kinetic friction (in N) acting on the crate

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            92.7  -  F_f =  40.5 * 1.13

=>         F_f =  46.935 \  N

5 0
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