Write 3 rules that describe the possible behaviors that occur when objects with different charges are brought close together

Where is most of the energy from incoming waves focused?

zavuch27 [327]

I'd say B since headland since whenever there's a crash wave, headland receives most of the impact

3 0

3 years ago

If two identical trees are cut down, one with a hand saw, and one with an electric saw...

nataly862011 [7]

The hand saw would involve more work because it takes more time and effort.

4 0

3 years ago

A meter stick is held vertically with one end on the floor and is then allowed to fall. Find the speed of the other end when it

Tems11 [23]

Answer:

5.4 ms⁻¹

Explanation:

Here we have to use conservation of energy. Initially when the stick is held vertical, its center of mass is at some height above the ground, hence the stick has some gravitational potential energy. As the stick is allowed to fall, its rotates about one. gravitational potential energy of the stick gets converted into rotational kinetic energy.

$L$ = length of the meter stick = 1 m

$m$ = mass of the meter stick

$w$ = angular speed of the meter stick as it hits the floor

$v$ = speed of the other end of the stick

we know that, linear speed and angular speed are related as

$v = r w\\w = \frac{v}{r}$

$h$ = height of center of mass of meter stick above the floor = $\frac{L}{2} = \frac{1}{2} = 0.5 m$

$I$ = Moment of inertia of the stick about one end

For a stick, momentof inertia about one end has the formula as

$I = \frac{mL^{2} }{3}$

Using conservation of energy

Rotational kinetic energy of the stick = gravitational potential energy

$(0.5) I w^{2} = mgh\\(0.5)(\frac{mL^{2} }{3}) (\frac{v}{L} )^{2} = mgh\\(0.5)(\frac{v^{2} }{3}) = gh\\(0.5)(\frac{v^{2} }{3}) = (9.8)(0.5)\\v = 5.4 ms^{-1}$

7 0

3 years ago

the chef was careless and put the aluminum cookie sheet directly on the hot stove, which melted 0.05 kg of the aluminum. how muc

sergejj [24]

Heat required to melt 0.05 kg of aluminum is 28.7 kJ.

<h3>What is the energy required to melt 0.05 kg of aluminum?</h3>

The heat energy required to melt 0.05 kg of aluminum is obtained from the heat capacity of aluminum and the melting point of aluminum.

The formula to be used is given below:

Heat required = mass * heat capacity * temperature change

Assuming the aluminum sheet was at room temperature initially.;

Room temperature = 25 °C

Melting point of aluminum = 660.3 °C

Temperature difference = (660.3 - 25) = 635.3 903

Heat capacity of aluminum = 903 J/kg/903

Heat required = 0.05 * 903 * 635.3

Heat required = 28.7 kJ

In conclusion, the heat required is obtained from the heat change aluminum and the mass of the aluminum melted.

Learn more about heat capacity at: brainly.com/question/21406849

#SPJ1

3 0

2 years ago

Fill in the blank. Consider the inverse square law: When light leaves a light bulb, it spreads out over more and more space as i

aliya0001 [1]

Answer:

Explanation:

Intensity of light is inversely proportional to distance from source

I ∝ 1 /r² where I is intensity and r is distance from source . If I₁ and I₂ be intensity at distance r₁ and r₂ .

I₁ /I₂ = r₂² /r₁²

If r₂ = 4r₁ ( given )

I₁ / I₂ = (4r₁ )² / r₁²

= 16 r₁² / r₁²

I₁ / I₂ = 16

I₂ = I₁ / 16

So intensity will become 16 times less bright .

"16 times " is the answer .

8 0

3 years ago