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s2008m [1.1K]
3 years ago
5

A bowling pin is thrown vertically upward such that it rotates as it moves through the air, as shown in the figure. Initially, t

he center of mass of the bowling pin is moving upward with a speed vi of
10 The maximum height of the center of mass of the bowling pin is most nearly

Physics
1 answer:
zheka24 [161]3 years ago
3 0

Answer:

Explanation:

In projectile, the formula for calculating the maximum height reached by the pin is expressed as;

H = u²/2g

u is the speed/velocity = 10m/s

g is the acceleration due to gravity = 9.81m/s²

Substitute

H = 10²/2(9.81)

H = 100/19.62

H = 5.097m

Hence the maximum height of the center of mass of the bowling pin is most nearly 5.0m

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A meter stick is held vertically with one end on the floor and is then allowed to fall. Find the speed of the other end when it
Tems11 [23]

Answer:

5.4 ms⁻¹

Explanation:

Here we have to use conservation of energy. Initially when the stick is held vertical, its center of mass is at some height above the ground, hence the stick has some gravitational potential energy. As the stick is allowed to fall, its rotates about one. gravitational potential energy of the stick gets converted into rotational kinetic energy.

L = length of the meter stick = 1 m

m = mass of the meter stick

w = angular speed of the meter stick as it hits the floor

v = speed of the other end of the stick

we know that, linear speed and angular speed are related as

v = r w\\w = \frac{v}{r}

h = height of center of mass of meter stick above the floor = \frac{L}{2} = \frac{1}{2} = 0.5 m

I = Moment of inertia of the stick about one end

For a stick, momentof inertia about one end has the formula as

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

Using conservation of energy

Rotational kinetic energy of the stick = gravitational potential energy

(0.5) I w^{2} = mgh\\(0.5)(\frac{mL^{2} }{3}) (\frac{v}{L} )^{2} = mgh\\(0.5)(\frac{v^{2} }{3}) = gh\\(0.5)(\frac{v^{2} }{3}) = (9.8)(0.5)\\v = 5.4 ms^{-1}

7 0
3 years ago
Coherent light of wavelength 540 nm passes through a pair of thin slits that are 3.4 × 10-5 m apart. At what angle away from the
Scrat [10]

Answer: 1.8\°

Explanation:

The diffraction angles \theta_{n} when we have a slit divided into n parts are obtained by the following equation:

dsin\theta_{n}=n\lambda (1)

Where:

d=3.4(10)^{-5}m is the width of the slit

\lambda=540 nm=540(10)^{-9}m is the wavelength of the light  

n is an integer different from zero.

Now, the second-order diffraction angle is given when n=2, hence equation (1) becomes:

dsin\theta_{2}=2\lambda (2)

Now we have to find the value of \theta_{2}:

sin\theta_{2}=\frac{2\lambda}{d} (3)

Then:

\theta_{2}=arcsin(\frac{2\lambda}{d})   (4)

\theta_{2}=arcsin(\frac{2(540(10)^{-9}m)}{3.4(10)^{-5}m})   (5)

Finally:

\theta_{2}=1.8\°   (6)

5 0
3 years ago
Meteorology is BEST defined as
alexgriva [62]
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5 0
3 years ago
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3 0
2 years ago
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Two identical 0.400 kg masses are pressed against opposite ends of a light spring of force constant 1.75 N/cm, compressing the s
nlexa [21]

Answer:

0.853 m/s

Explanation:

Total energy stored in the spring = Total kinetic energy of the masses.

1/2ke² = 1/2m'v².................... Equation 1

Where k = spring constant of the spring, e = extension, m' = total mass, v = speed of the masses.

make v the subject of the equation,

v = e[√(k/m')].................... Equation 2

Given: e = 39 cm = 0.39 m, m' = 0.4+0.4 = 0.8 kg, k = 1.75 N/cm = 175 N/m.

Substitute into equation 2

v = 0.39[√(1.75/0.8)

v = 0.39[2.1875]

v = 0.853 m/s

Hence the speed of each mass = 0.853 m/s

7 0
3 years ago
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