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3 years ago

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The experimenter fom the video rotates on his stool, this time holding his empty hands in his lap. You stand on a desk above him

and drop a long, heavy been bag straight down so that it lands across his lap, in his hands. what happens?

1 answer:

gtnhenbr [62]3 years ago

7 0

Answer:

Explanation:

The experimenter is rotating on his stool with angular velocity ω ( suppose )

His moment of inertia is I say

We are applying no torque from outside . therefore , the angular momentum will remain the same

Thus angular momentum L = I ω = constant

Thus we can say I₁ ω₁ = I₂ω₂ = constant

here I₁ is the initial moment of inertia and ω₁ is the initial angular velocity

Similarly I₂ is the final moment of inertia and ω₂ is the final angular velocity

When a been bag is dropped on his lap , his moment of inertia increases due to increase in mass

In the above equation, when moment of inertia increases , the angular velocity decreases . So its motion of rotation will decrease .

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A pair of slits in a double slit experiment are illuminated with monochromatic light of wavelength 480 nm. The slits are separat

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Using the appropriate approximations:

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d = slit separation

x = fringe spacing

L = distance between slits and screen

m = some integer, used to determine the distance from the central bright fringe to another bright fringe

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Given values:

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3 years ago

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3 years ago

A wheel rotating about a fixed axis with a constant angular acceleration of 2.0 rad/s2 starts from rest at t = 0. The wheel has

irga5000 [103]

Answer:

The total linear acceleration is approximately 0.246 meters per square second.

Explanation:

The total linear acceleration ( $a$ ) consist in two components, <em>radial</em> ( $a_{r}$ ) and <em>tangential</em> ( $a_{t}$ ), in meters per square second:

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$a_{t} = \alpha \cdot r$ (2)

Since both components are orthogonal to each other, the total linear acceleration is determined by Pythagorean Theorem:

$a = \sqrt{a_{r}^{2}+a_{t}^{2}}$ (3)

Where:

$r$ - Radius of the wheel, in meters.

$\omega$ - Angular speed, in radians per second.

$\alpha$ - Angular acceleration, in radians per square second.

Given that wheel accelerates uniformly, we use the following kinematic equation:

$\omega = \omega_{o}+ \alpha\cdot t$ (4)

Where:

$\omega_{o}$ - Initial angular speed, in radians per second.

$t$ - Time, in seconds.

If we know that $r = 0.1\,m$ , $\alpha = 2\,\frac{rad}{s^{2}}$ , $\omega_{o} = 0\,\frac{rad}{s}$ and $t = 0.60\,s$ , then the total linear acceleration is:

$\omega = \omega_{o}+ \alpha\cdot t$

$\omega = 1.2\,\frac{rad}{s}$

$a_{r} = \omega^{2}\cdot r$

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$a_{t} = \alpha \cdot r$

$a_{t} = 0.2\,\frac{m}{s^{2}}$

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The total linear acceleration is approximately 0.246 meters per square second.

3 0

3 years ago

A high diver of mass 51.7 kg steps off a board 10.0 m above the water and falls vertical to the water, starting from rest. If he

zmey [24]

Answer:

851.33 N

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Given: u = 0 m/s(from rest), h = 10 m g = 9.8 m/s².

Substitute into equation 1

v² = 0² + 2(9.8)(10)

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Note: As the Diver touches the water, u = 14 m/s and v = 0 m/s( stopped)

Using,

d = (v+u)t/2 .............. equation 2

Where d = distance moved by the diver in water before its motion stopped, t = time taken before it comes to rest

Given: v = 0 m/s, u = 14 m/s t = 2.10 s

Substitute into equation 2

d = (0+14)2.1/2

d = 14.7 m.

Finally

work done by the water to stop the diver = potential energy of the diver

F×d = mgh'................Equation 3

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making F the subject of the equation,

F = mgh/d ............ Equation 4

Given: m = 51.7 kg, h = 10 m, g = 9.8 m/s², d = 14.7 m.

Substitute into equation 4

F = 51.7(10+14.7)(9.8)/14.7

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8 0

3 years ago

PLEASE ANSWER, I NEED HELP

Scorpion4ik [409]

1) The gravitational force between Ellen and the moon is $1.56\cdot 10^{-3} N$

2) The two forces are equal, while the acceleration of the bus is smaller than the acceleration of the bicycle.

Explanation:

1)

The magnitude of the gravitational force between two objects is given by

$F=G\frac{m_1 m_2}{r^2}$

where

$G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}$ is the gravitational constant

$m_1, m_2$ are the masses of the two objects

r is the separation between them

In this problem, we have:

$m_1 = 47 kg$ is the mass of Ellen

$m_2 = 7.35\cdot 10^{22} kg$ is the mass of the moon

$r=3.84\cdot 10^8 m$ is the distance between Ellen and the moon

Substituting, we find the gravitational force between Ellen and the moon:

$F=(6.67\cdot 10^{-11})\frac{(47)(7.35\cdot 10^{22})}{(3.84\cdot 10^8)^2}=1.56\cdot 10^{-3} N$

2)

We can analyze the forces acting in the collision between the bus and the bicycle by using Newton's third law of motion, which states that:

"When an object A exerts a force (called action) on an object B, then object B exerts an equal and opposite force (called reaction) on object A"

Applied to our problem, this means that the force exerted by the bus on the bicycle during the collision (action force) is equal (and opposite) to the force exerted by the bicycle on the bus (reaction force).

Now let's analyze the accelerations of the two vehicles. We can find the acceleration of each vehicle by using Newton's second law:

$a=\frac{F}{m}$

where

a is the acceleration

F is the force exerted on the vehicle

m is the mass of the vehicle

As we said previously, the force F exerted on each of the two vehicles: so, the acceleration only depends on the mass. In particular, the acceleration is inversely proportional to the mass: therefore, the larger the mass of the vehicle, the smaller the acceleration. This means that the acceleration of the bus is smaller than the acceleration of the bicycle.

Learn more about gravitational force:

brainly.com/question/1724648

brainly.com/question/12785992

And about Newton's third law:

brainly.com/question/11411375

#LearnwithBrainly

6 0

3 years ago