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Aleksandr-060686 [28]

3 years ago

6

8. A solid wooden door, 90 cm wide by 2.0 m tall, has a mass of 35 kg. It is open and at rest. A small 500-g ball is thrown perp

endicular to the door with a speed of 20 m/s and hits the door 60 cm from the hinged side, causing it to begin turning. The ball rebounds along the same line with a speed of 16.0 m/s relative to the ground. If the momentum of inertia of the door around the hinge is I=1/3 Ma 2 , where a is the width of the door, how much energy was lost during this collision?

1 answer:

insens350 [35]3 years ago

7 0

Answer:

Explanation:

Kinetic energy of ball

= .5 x .5 x 20²

= 100 J

Original kinetic energy of door = 0

Total kinetic energy before ball hitting the door

= 100 J

We shall apply law of conservation of momentum to calculate angular velocity of the door after ball hitting it.

change in angular momentum of ball

= mvr - mur , u is initial velocity and v is final velocity of ball

= .5 ( 20 + 16 ) x .06

= 1.08

Change in angular momentum of door

= I x ω - 0

1/3 x 35 x .09² x ω

= .0945 x ω

so

.0945 x ω = 1.08

ω = 11.43

rotational K E of door after collision

= 1/2 I ω²

= .5 x .0945 x 11.43 ²

= 6.17 J

Kinetic energy of ball after collision

= 1/2 x .5 x 16²

= 64

Total KE of door and ball

= 64 + 6.17

= 70.17 J

LOSS OF ENERGY

= 100 - 70.17 J

= 29.83 J

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Please help Math Phys

Lorico [155]

Answer:

269 m

45 m/s

-58.6 m/s

Explanation:

Part 1

First, find the time it takes for the package to land. Take the upward direction to be positive.

Given (in the y direction):

Δy = -175 m

v₀ = 0 m/s

a = -9.8 m/s²

Find: t

Δy = v₀ t + ½ at²

(-175 m) = (0 m/s) t + ½ (-9.8 m/s²) t²

t = 5.98 s

Next, find the horizontal distance traveled in that time:

Given (in the x direction):

v₀ = 45 m/s

a = 0 m/s²

t = 5.98 s

Find: Δx

Δx = v₀ t + ½ at²

Δx = (45 m/s) (5.98 s) + ½ (0 m/s²) (5.98 s)²

Δx = 269 m

Part 2

Given (in the x direction):

v₀ = 45 m/s

a = 0 m/s²

t = 5.98 s

Find: v

v = at + v₀

v = (0 m/s²) (5.98 s) + (45 m/s

v = 45 m/s

Part 3

Given (in the y direction):

Δy = -175 m

v₀ = 0 m/s

a = -9.8 m/s²

Find: v

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3 years ago

One string of a certain musical instrument is 70.0 cm long and has a mass of 8.79 g . It is being played in a room where the spe

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To solve this problem we will apply the concepts of linear mass density, and the expression of the wavelength with which we can find the frequency of the string. With these values it will be possible to find the voltage value. Later we will apply concepts related to harmonic waves in order to find the fundamental frequency.

The linear mass density is given as,

$\mu = \frac{m}{l}$

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$\mu = 0.01255kg/m$

The expression for the wavelength of the standing wave for the second overtone is

$\lambda = \frac{2}{3} l$

Replacing we have

$\lambda = \frac{2}{3} (70*10^{-2})$

$\lambda = 0.466m$

The frequency of the sound wave is

$f_s = \frac{v}{\lambda_s}$

$f_s = \frac{344}{0.768}$

$f_s = 448Hz$

Now the velocity of the wave would be

$v = f_s \lambda$

$v = (448)(0.466)$

$v = 208.768m/s$

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$v = \sqrt{\frac{T}{\mu}}$

$v^2 = \frac{T}{\mu}$

$T= \mu v^2$

$T = (0.01255kg/m)(208.768m/s)^2$

$T = 547N$

The tension in the string is 547N

PART B) The relation between the fundamental frequency and the $n^{th}$ harmonic frequency is

$f_n = nf_1$

Overtone is the resonant frequency above the fundamental frequency. The second overtone is the second resonant frequency after the fundamental frequency. Therefore

$n=3$

Then,

$f_3 = 3f_1$

Rearranging to find the fundamental frequency

$f_1 = \frac{f_3}{3}$

$f_1 = \frac{448Hz}{3}$

$f_1 = 149.9Hz$

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Answer:

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