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WITCHER [35]
3 years ago
9

Which of the following statements about this experiment is FALSE? Before each trial one should reshape the bob into something li

ke a ball. You may assume the collision between the bob and the box is completely inelastic. The initial position for the box should be just touching the pendulum bob when it is hanging straight down. To make the box move, the pendulum bob should hit close to the bottom of the box during the collision
Physics
1 answer:
Troyanec [42]3 years ago
7 0

Which of the following statements about this experiment is FALSE? Before each trial one should reshape the bob into something like a ball. You may assume the collision between the bob and the box is completely inelastic. The initial position for the box should be just touching the pendulum bob when it is hanging straight down. To make the box move, the pendulum bob should hit close to the bottom of the box during the collision is given below

Explanation:

So, The bob is held at angle theta from initial position, P. Find the displacement from P.  

displacement= L sin  Θ. I hope you get it here. The pendulum forms a triangle with length L, the initial position and horizontal displacement.

So,the point is at displacement point the kinetic energy of bob is 0 and when it is released it will have maximum kinetic energy at P. It will hit the box and all he kinetic energy of bob will get transferred into driving force(F.D) of box. Now kinetic energy = Force * displacement.

K.E= 0.5 m V^2= F.D * Lsin Θ.

Find F.D

F.D = (0.5 m V^2) /  Lsin Θ.

Now for the box, F.D - Friction = m(box) a.

Friction = F.D + m(box) a.

Find the accelaration of box from an equation of motion. ( u=0, find displacement of box,s,time taken and so on)

U get the friction. Now,

coeff of friction = µ = friction/ reaction.

Note that reaction here = weight of box= m(box) g.   g= acc of free fall= 9.81.

So here u go.. U get the coeff of friction.. I hope am right here.. and made no mistake.. Anyway try it with the values to confirm! ;)

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PS Final Exam
tatuchka [14]

Answer:

rolling friction

Explanation:

5 0
2 years ago
The resultant of 2 forces at right angles is 100 lbs. If one of the forces makes an angle of 30 degress with the resultant, comp
Ipatiy [6.2K]

Answer:

86.6 lbs

Explanation:

Let the force is X.

Resultant force, R = 100 lbs

Other force is Y. Angle between resultant force and force X is 30°.

According to the diagram

Cos30=\frac{X}{R}

0.866=\frac{X}{100}

X = 86.6 lbs

Other force Y

Sin30=\frac{X}{R}

0.5=\frac{Y}{100}

Y = 50 lbs

5 0
3 years ago
In a double-slit experiment, the third-order maximum for light of wavelength 510 nm is located 17 mm from the central bright spo
Marysya12 [62]

Answer:

14.9 mm

Explanation:

We know dsinθ = mλ where d = separating of slit, m = order of maximum = 3 and λ = wavelength = 510nm = 510 × 10⁻⁹ m

Also tanθ = L/D where L = distance of m order fringe from central bright spot = 17 mm = 0.017 m and D = distance of screen from slit = 1.6 m

So, sinθ = mλ/d

Since θ is small, sinθ ≅ tanθ

So,

mλ/d = L/D

d = mλD/L

Substituting the values of the variables into the equation, we have

d = 3 × 510 × 10⁻⁹ m × 1.6 m/0.017 m

d = 2448 × 10⁻⁹ m²/0.017 m

d = 144000 × 10⁻⁹ m

d = 1.44 × 10⁻⁴ m

d = 0.144 × 10⁻³ m

d = 0.144 mm

Now, for the second-order maximum, m' of the 670 nm wavelength of light,

m'λ'/d = L'/D where m' = order of maximum = 2, λ' = wavelength of light = 670 nm = 670 × 10⁻⁹ m, d = slit separation = 0.144 mm = 0.144 × 10⁻³ m, L' = distance of second order maximum from central bright spot and D = distance of screen from slit = 1.6m

So, L' = m'λ'D/d

So, substituting the values of the variables into the equation, we have

L' = 2 × 670 × 10⁻⁹ m × 1.6 m/0.144 × 10⁻³ m

L' = 2144  × 10⁻⁹ m²/0.144 × 10⁻³ m

L' = 14888.89 × 10⁻⁶ m

L' = 0.01488 m

L' ≅ 0.0149 m

L' = 14.9 mm

4 0
2 years ago
the end of the tracks, 8.8 m lower vertically, is a horizontally situated spring with constant 5 × 105 N/m. The acceleration of
devlian [24]

Answer

Assuming the mass of the car, m = 43000 kg

initial speed u = 0

vertical distance moved, h = 8.8 m

spring constant k = 5 x  10⁵ N / m

acceleration of gravity = 9.8 m/s²

From law of conservation of energy ,

Gravitational potential energy at starting position =potential energy of the spring at maximum compression

                m g h =\dfrac{1}{2}k x^2

                x = \sqrt{\dfrac{2mgh}{k}}

                x = \sqrt{\dfrac{2\times 43000\times 9.8\times 8.8}{5\times 10^5}}

                    x = 14.83 m

If the mass of the car is equal to 43000 Kg the spring is compressed to 14.83 m

6 0
3 years ago
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