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1 year ago

What is time taken by the stone ?

Physics

1 answer:

SSSSS [86.1K]1 year ago

8 0

The time taken by the stone to hit the ground would be 5.12 seconds.

<h3>What are the three equations of motion?</h3>

There are three equations of motion given by Newton

The first equation is given as follows

v = u + at

the second equation is given as follows

S = ut + 1/2×a×t²

the third equation is given as follows

v² - u² = 2×a×s

Keep in mind that these calculations only apply to uniform acceleration.

As given in the problem, a stone is dropped from the helicopter which is ascending at the speed of 19.6 m/s

height(S) = 156.8 meters

initial velocity(u) = -19.6 m/s

acceleration(a) = 9.81 m/s²

By using the second equation of motion given by newton

S = ut + 1/2at²

S = 156.8m ,u= -19.6 m/s , a= 9.81 m/s² and t =? seconds

156.8= -19.6t + 9.81t²

t = 5.12 seconds

Thus, the time taken by the stone to hit the ground would be 5.12 seconds.

Learn more about equations of motion from here,

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Answer:

The angular speed of the new system is $3\,\frac{rad}{s}$ .

Explanation:

Due to the absence of external forces between both disks, the Principle of Angular Momentum Conservation is observed. Since axes of rotation of each disk coincide with each other, the principle can be simplified into its scalar form. The magnitude of the Angular Momentum is equal to the product of the moment of inertial and angular speed. When both disks begin to rotate, moment of inertia is doubled and angular speed halved. That is:

$I\cdot \omega_{o} = 2\cdot I \cdot \omega_{f}$

Where:

$I$ - Moment of inertia of a disk, measured in kilogram-square meter.

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

$\omega_{f}$ - Final angular speed, measured in radians per second.

This relationship is simplified and final angular speed can be determined in terms of initial angular speed:

$\omega_{f} = \frac{1}{2}\cdot \omega_{o}$

Given that $\omega_{o} = 6\,\frac{rad}{s}$ , the angular speed of the new system is:

$\omega_{f} = \frac{1}{2}\cdot \left(6\,\frac{rad}{s} \right)$

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The angular speed of the new system is $3\,\frac{rad}{s}$ .

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2 years ago

A solid nonconducting sphere of radius R has a charge Q uniformly distributed throughout its volume. A Gaussian surface of radiu

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Answer:

1. E x 4πr² = ( Q x r³) / ( R³ x ε₀ )

Explanation:

According to the problem, Q is the charge on the non conducting sphere of radius R. Let ρ be the volume charge density of the non conducting sphere.

As shown in the figure, let r be the radius of the sphere inside the bigger non conducting sphere. Hence, the charge on the sphere of radius r is :

Q₁ = ∫ ρ dV

Here dV is the volume element of sphere of radius r.

Q₁ = ρ x 4π x ∫ r² dr

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Q₁ = (4π x ρ x r³ )/3

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Applying Gauss law of electrostatics ;

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Substitute the value of Q₁ in the above equation. Hence,

E x 4πr² = ( Q x r³) / ( R³ x ε₀ )

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3 years ago

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3 years ago

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3 years ago

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