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sukhopar [10]
2 years ago
15

Identify a form of potential energy.

Physics
1 answer:
brilliants [131]2 years ago
5 0

Answer:

Gravitational potential energy (ex. height)

Elastic potential energy (ex. stretched rubber band)

Electrical potential energy (ex. negative charges close together want to repel each other)

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Use the drop-down menus to complete the
bagirrra123 [75]

Answer:

Cocoa mix is the: Solute

Water is the: Solvent

The solution has reached: Saturation

Explanation:

5 0
4 years ago
How much energy has 4×10^10m^3 of water collected in a reservoir at a hight of 100 m from the power house ?What kind of energy i
sveticcg [70]

Answer:

PE = 3.92x10^16J

potential energy

Explanation:

PE = m*g*h

mass of water = 1000kg/m³

(4*10^10m³)*1000kg = 4*10^13kg

PE = (4*10^13kg)*(9.81m/s²)*(100m)

PE = 3.92x10^16J

7 0
3 years ago
A nonconducting spherical shell, with an inner radius of 4 cm and an outer radius of 6 cm, has charge spread non uniformly throu
Aloiza [94]
In other words a infinitesimal segment dV caries the charge 
<span>dQ = ρ dV </span>

<span>Let dV be a spherical shell between between r and (r + dr): </span>
<span>dV = (4π/3)·( (r + dr)² - r³ ) </span>
<span>= (4π/3)·( r³ + 3·r²·dr + 3·r·(dr)² + /dr)³ - r³ ) </span>
<span>= (4π/3)·( 3·r²·dr + 3·r·(dr)² + /dr)³ ) </span>
<span>drop higher order terms </span>
<span>= 4·π·r²·dr </span>

<span>To get total charge integrate over the whole volume of your object, i.e. </span>
<span>from ri to ra: </span>
<span>Q = ∫ dQ = ∫ ρ dV </span>
<span>= ∫ri→ra { (b/r)·4·π·r² } dr </span>
<span>= ∫ri→ra { 4·π·b·r } dr </span>
<span>= 2·π·b·( ra² - ri² ) </span>

<span>With given parameters: </span>
<span>Q = 2·π · 3µC/m²·( (6cm)² - (4cm)² ) </span>
<span>= 2·π · 3×10⁻⁶C/m²·( (6×10⁻²m)² - (4×10⁻²m)² ) </span>
<span>= 3.77×10⁻⁸C </span>
<span>= 37.7nC</span>
6 0
4 years ago
Dinah is playing cards with her friends, and it is her turn to deal. A card with a mass of 2.3 g slides 0.35 m along the table b
Inga [223]

Answer:

soln,

=2.3+0.35-0.24

=2.41 answer

4 0
3 years ago
A billiard ball is moving in the x-direction at 30.0 cm/s and strikes another billiard ball moving in the y-direction at 40.0 cm
ipn [44]

To solve this problem it is necessary to apply the trigonometric ratios of the given velocity components.

If we make a graph of the velocity vectors in their respective velocities according to the given description we will have something similar to the attached graph.

The angle could be obtained from the components of the opposite leg and the adjacent leg so that

tan\theta = \frac{x}{y}

\theta = tan^{-1}(\frac{x}{y})

The opposite leg value (y) is 40cm / s and the adjacent leg (x) is 30cm / s

\theta = tan^{-1}(\frac{30}{40})

\theta = 36.87\°

Therefore the final direction that does the first ball is 36.87°

6 0
3 years ago
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