5

Newton's cradle consists of an aligned row of identical elastic balls suspended by strings so they barely touch one another. whe

Ilia_Sergeevich [38]

Initial momentum = 2 * m v

so if 1 ball popped out with 2 v momentum would be conserved

but the initial kinetic energy of the two balls is

1/2 m v^2 + 1/2 m v^2 = m v^2

But the kinetic energy of one ball with twice the speed would be

KE = 1/2 m (2 v)^2 = 2 m v^2 thus violating conservation of energy

8 0

3 years ago

HOW DOES A GRAPH HELP YOU SEE ANOMALOUS DATA?

Ghella [55]

Answer:

Anomalies consist of one or more modifications, insertions or deletions. As was described in Section 3.1, there are only three types of changes that can be made to a graph. Therefore, anomalies that consist of structural changes to a graph must consist of one of these types. Assumption 4.

Explanation:

8 0

3 years ago

The parallel axis theorem relates Icm, the moment of inertia of an object about an axis passing through its center of mass, to I

erma4kov [3.2K]

Answer:

Part a)

$I_{end} = \frac{mL^2}{3}$

Part b)

$I_{edge} = \frac{2ma^2}{3}$

Explanation:

As we know that by parallel axis theorem we will have

$I_p = I_{cm} + Md^2$

Part a)

here we know that for a stick the moment of inertia for an axis passing through its COM is given as

$I = \frac{mL^2}{12}$

now if we need to find the inertia from its end then we will have

$I_{end} = I_{cm} + Md^2$

$I_{end} = \frac{mL^2}{12} + m(\frac{L}{2})^2$

$I_{end} = \frac{mL^2}{3}$

Part b)

here we know that for a cube the moment of inertia for an axis passing through its COM is given as

$I = \frac{ma^2}{6}$

now if we need to find the inertia about an axis passing through its edge

$I_{edge} = I_{cm} + Md^2$

$I_{edge} = \frac{ma^2}{6} + m(\frac{a}{\sqrt2})^2$

$I_{edge} = \frac{2ma^2}{3}$

7 0

3 years ago

A ball of mass m is thrown straight upward from ground level at speed v0. At the same instant, at a distance D above the ground,

n200080 [17]

Answer:

a. $t = \frac{v_{0} +/- \sqrt{v_{0} ^{2} - gD} }{g}$ b. D = v₀²/2g

Explanation:

Here is the complete question

A ball is thrown straight up from the ground with speed v₀ . At the same instant, a second ball is dropped from rest from a height D , directly above the point where the first ball was thrown upward. There is no air resistance

Find the time at which the two balls collide.

Express your answer in terms of the variables D ,v₀ , and appropriate constants..

t = ?!

Part B

Find the value of D in terms of v₀ and g so that at the instant when the balls collide, the first ball is at the highest point of its motion.

Express your answer in terms of the variables v₀ and g .

D =?!

Solution

The distance moved by the ball dropped from distance,D with velocity v₀, H₁ = D - (v₀t - gt²/2) = D + v₀t + gt²/2.

The distance moved by the ball thrown straight upward with velocity v₀ is H₂ = v₀t - gt²/2.

The two balls collide when their vertical distances are equal. That is H₁ = H₂

So, D - v₀t + gt²/2 = v₀t - gt²/2

Collecting like terms

D + gt²/2 + gt²/2 = v₀t + v₀t

D +gt² = 2v₀t

gt² - 2v₀t + D = 0.

Using the quadratic formula,

$t = \frac{-(-2v_{0} ) +/- \sqrt{(-2v_{0} )^{2} - 4 X g XD} }{2g} = \frac{2v_{0} +/- \sqrt{4v_{0} ^{2} - 4gD} }{2g} = \frac{v_{0} +/- \sqrt{v_{0} ^{2} - gD} }{g}$

B. At its highest point, the velocity of the first ball, v = 0. Using v² = u² - 2gs where s = highest point of first ball when they collide and u = v₀.

0 = v₀² - 2gs

s = v₀²/2g.

Also, the time it takes the first ball to reach its highest point is gotten from v = u - gt. At highest point, v = 0 and u = v₀. So,

0 = v₀ - gt₀

t₀ = v₀/g

Also H = s₁ + s where s₁ = distance moved by second ball in time t₀ for collision = v₀t₀ - gt₀²/2.

So, H = v₀t₀ - gt₀²/2 + v₀²/2g = v₀(v₀/g) - g(v₀/g)²/2 + v₀²/2g = v₀²/2g - v₀²/2g + v₀²/2g = v₀²/2g

6 0

3 years ago

A cat napping on a window sill demonstrates which scientific concept?

ser-zykov [4K]

The answer is inertia becuase it means nothing happening or is doing nothing

8 0

3 years ago

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