Physics, 18.03.2020 18:02 vickReis06

You throw a 3.00-N rock vertically into the air from ground level. You observe that when it is 15.0 m above the ground, it is traveling at 25.0 m/s upward. Use the work–energy theorem to find (a) the rock’s speed just as it left the ground and (b) its maximum height.

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Answer from: AshIeigh

(a) 30.32 m/s

(b) 46.88 m

Explanation:

Given:

Weight of the rock ( F_g ) = 3.00 N

Velocity at height 15.0 m (v) = 25 m/s

Displacement of the rock (S) = 15.0 m

(a)

Let the velocity at the start be 'u'.

The mass of the rock is given as:

$m=\frac{F_g}{g}=\frac{3\ N}{9.8\ N/kg}=0.306\ kg$

Now, the work energy theorem states that the work done by the net force is equal to the change in the kinetic energy of the body.

Here, the net force is gravitational force acting on the rock vertically downward and the displacement of the rock is vertically upward. Therefore, the work done by gravity is negative and given as:

$W_{net}=-F_g\times S =-3\ N\times 15\ m= -45\ Nm$

Now, change in kinetic energy is given as:

$\Delta K=\frac{1}{2}m(v^2-u^2)$

Plug in the given values and simplify. This gives,

$\Delta K=\frac{1}{2}\times 0.306(25^2-u^2)\\\\\Delta K=0.153(625-u^2)$

From work-energy theorem:

$\Delta K=W_{net}\\\\0.153(625-u^2)=-45\\\\625-u^2=\frac{-45}{0.153}\\\\u^2=625+294.12\\\\u=\sqrt{919.12}=30.32\ m/s$

Therefore, the speed with which it leaves the ground is 30.32 m/s.

(b)

Let the maximum height reached be 'h'.

We know that, at maximum height, the speed is 0 m/s.

Therefore, final speed (v) = 0 m/s

Initial speed (u) = 30.32 m/s

The work done by gravity is given as:

$W_{net}=-3h$

The change in kinetic energy is given as:

$\Delta K=\frac{1}{2}m(v^2-u^2)\\\\\Delta K=\frac{1}{2}\times 0.306(0-(30.32)^2)\\\\\Delta K=0.153\times -919.12=-140.625\ J$

Again from work-energy theorem, we have:

$W_{net}=\Delta K$

$-3h=-140.625\\\\h=\frac{140.625}{3}\\\\h=46.88\ m$

Hence, the maximum height reached is 46.88 m.

Answer from: Quest

Liquid. good luck : )

Answer from: Quest

electromagnetic shielding is the practice of reducing the electromagnetic field in a space by blocking the field with barriers made of conductive or magnetic materials. shielding is typically applied to enclosures to isolate electrical devices from their surroundings, and to cables to isolate wires from the environment through which the cable runs. electromagnetic shielding that blocks radio frequency electromagnetic radiation is also known as rf shielding.

the shielding can reduce the coupling of radio waves, electromagnetic fields and electrostatic fields. a conductive enclosure used to block electrostatic fields is also known as a faraday cage. the amount of reduction depends very much upon the material used, its thickness, the size of the shielded volume and the frequency of the fields of interest and the size, shape and orientation of apertures in a shield to an incident electromagnetic

from wikipedia

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You throw a 3.00-N rock vertically into the air from ground level. You observe that when it is 15.0...

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