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djverab [1.8K]
3 years ago
9

A completely inelastic collision occurs between two balls of wet putty that move directly toward each other along a vertical axi

s. Just before the collision, one ball of mass 3.0 kg is moving upward at 20 m/s, and the other ball, of mass 2.0 kg, is moving downward at 14 m/s. How high do the combined two balls of putty rise above the collision point? (Neglect air drag.)
Physics
1 answer:
NeX [460]3 years ago
6 0

Answer:

h = 2.087 m

Explanation:

Given

m₁ = 3 kg

v₁ = 20 m/s

m₂ = 2 kg

v₂ = - 14 m/s

In a completely inelastic collision the colliding objects stick together after the collision and move together as a single object.

In the given problem, lets assume that the balls of putty are initially moving along the  y  axis, upward direction being the positive  y  direction. And the collision occurs at the origin of the coordinate system.

We can apply the equation

vs = (m₁*v₁ + m₂*v₂) / (m₁ + m₂)  

⇒   vs = (3 kg*20 m/s + 2 kg*(- 14 m/s)) / (3 kg + 2 kg)  

⇒   vs = 6.4 m/s (↑)

To calculate the maximum height  h  attained by the combined system of two balls of putty after the the collision, we use the expression for linear motion under gravity:

vf² = vi² - 2*g*h

where

vf = 0 m/s  

g = 9.81 m/s²

vi = vs = 6.4 m/s

finally we get h:

h = vi² / (2*g)

⇒   h = (6.4 m/s)² / (2*9.81 m/s²) = 2.087 m

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Answer:

A

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E. An ocean wave moving through water is an example of a mechanical wave

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and Ans a is also correct

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2 years ago
You drop a steel ball bearing, with a radius of 2.40 mm, into a beaker of honey. Note that honey has a viscosity of 6.00 Pa/s an
Stells [14]

Answer:

The “terminal speed” of the ball bearing is 5.609 m/s

Explanation:

Radius of the steel ball R = 2.40 mm

Viscosity of honey η = 6.0 Pa/s

\text { Viscosity has Density } \sigma=1360 \mathrm{kg} / \mathrm{m}^{3}

\text { Steel has a density } \rho=7800 \mathrm{kg} / \mathrm{m}^{3}

\left.\mathrm{g}=9.8 \mathrm{m} / \mathrm{s}^{2} \text { (g is referred to as the acceleration of gravity. Its value is } 9.8 \mathrm{m} / \mathrm{s}^{2} \text { on Earth }\right)

While calculating the terminal speed in liquids where density is high the stokes law is used for viscous force and buoyant force is taken into consideration for effective weight of the object. So the expression for terminal speed (Vt)

V_{t}=\frac{2 \mathrm{R}^{2}(\rho-\sigma) \mathrm{g}}{9 \eta}

Substitute the given values to find "terminal speed"

\mathrm{V}_{\mathrm{t}}=\frac{2 \times 0.0024^{2}(7800-1360) 9.8}{9 \times 6}

\mathrm{V}_{\mathrm{t}}=\frac{0.0048 \times 6440 \times 9.8}{54}

\mathrm{V}_{\mathrm{t}}=\frac{302.9376}{54}

\mathrm{V}_{\mathrm{t}}=5.609 \mathrm{m} / \mathrm{s}

The “terminal speed” of the ball bearing is 5.609 m/s

7 0
3 years ago
Which characteristic of a sound is affected by the amount of energy used to create that sound? In what direction the sound will
olya-2409 [2.1K]

Answer:

The correct option is;

How loud or soft the sound is

Explanation:

The loudness of a sound wave is given by the amount of energy that the pressure wave carries and it is measured in decibels (dB) which is the relative intensity of the pressure wave of a sound to the standard pressure

A loud sound has a high amplitude and a soft sound has a low amplitude, such that as the amplitude of the sound is increased, due to increased energy input, the sound becomes louder, and as the amplitude of the sound is decreased due to reduced energy input, the sound becomes softer.

5 0
2 years ago
a. How many excess electrons must be distributed uniformly within the volume of an isolated plastic sphere 25.0 cm in diameter t
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Answer:

Explanation:

The electric field outside the sphere is given as,

E = k Q /r²

here Q = n x 1.6 x 10⁻¹⁹ C

where n is the number of electons

if the dimeter of sphere d= 25 cm= 0.25 m

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we get

n= E r²/ k x 1.6 x 10⁻¹⁹ C

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olga nikolaevna [1]

Answer:

The bowling ball would

Explanation:

Because it contains more weight! And that will make it fall down quicker.

Hope it helped

3 0
2 years ago
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