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2 years ago

1) How is chemical energy transformed into potential energy?

Physics

1 answer:

tia_tia [17]2 years ago

6 0

Answer:

Chemical potential energy is a form of potential energy related to the structural arrangement of atoms or molecules. This arrangement may be the result of chemical bonds within a molecule or otherwise. Chemical energy of a chemical substance can be transformed to other forms of energy by a chemical reaction.

Explanation:

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3 years ago

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While standing on a balcony a child drops a penny. The penny lands on the ground floor 1.5 s later. How fast was the penny trave

sveticcg [70]

Answer:

14.7 m/s.

Explanation:

From the question given above, the following data were obtained:

Time (t) = 1.5 s

Acceleration due to gravity (g) = 9.8 m/s².

Height = 11.025 m

Final velocity (v) = 0 m/s

Initial velocity (u) =?

We, can obtain the initial velocity of the penny as follow:

H = ½(v + u) t

11.025 = ½ (0 + u) × 1.5

11.025 = ½ × u × 1.5

11.025 = u × 0.75

Divide both side by 0.75

u = 11.025/0.75

u = 14.7 m/s

Therefore, the penny was travelling at 14.7 m/s before hitting the ground.

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2 years ago

A small airplane flies 780 miles with an average speed of 260 miles per hour. 1.5 hours after the plane leaves, a Boeing 747 lea

Lera25 [3.4K]

Answer:

520 miles per hour

Explanation:

Let the speed of the Boeing 747 be x miles per hour.

The small airplane covers distance of 780 miles with speed 260 miles per hour.

Also,

After 1.5 hours the Boeing 747 leave the same place and reaches at same time. Both covered distance of 780 miles.

So,

Time taken by Boeing 747 + 1.5 hours = Time taken by small plane.

Also,

Time = Distance/ speed

So,

780 / x + 1.5 = 780/ 260

Solving for x, we get:

x = 520 miles per hour

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3 years ago

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2 years ago

A point charge q is located at the center of a spherical shell of radius a that has a charge −q uniformly distributed on its sur

muminat

Answer:

a) E = 0

b) $E = \dfrac{k_e \cdot q}{ r^2 }$

Explanation:

The electric field for all points outside the spherical shell is given as follows;

a) $\phi_E = \oint E \cdot dA = \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}$

From which we have;

$E \cdot A = \dfrac{{\Sigma Q}}{\varepsilon _{0}} = \dfrac{+q + (-q)}{\varepsilon _{0}} = \dfrac{0}{\varepsilon _{0}} = 0$

E = 0/A = 0

E = 0

b) $\phi_E = \oint E \cdot dA = \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}$

$E \cdot A = \dfrac{+q }{\varepsilon _{0}}$

$E = \dfrac{+q }{\varepsilon _{0} \cdot A} = \dfrac{+q }{\varepsilon _{0} \cdot 4 \cdot \pi \cdot r^2}$

By Gauss theorem, we have;

$E\oint dS = \dfrac{q}{\varepsilon _{0}}$

Therefore, we get;

$E \cdot (4 \cdot \pi \cdot r^2) = \dfrac{q}{\varepsilon _{0}}$

The electrical field outside the spherical shell

$E = \dfrac{q}{\varepsilon _{0} \cdot (4 \cdot \pi \cdot r^2) }= \dfrac{q}{4 \cdot \pi \cdot \varepsilon _{0} \cdot r^2 }= \dfrac{q}{(4 \cdot \pi \cdot \varepsilon _{0} )\cdot r^2 }$

$k_e= \dfrac{1}{(4 \cdot \pi \cdot \varepsilon _{0} ) }$

Therefore, we have;

$E = \dfrac{k_e \cdot q}{ r^2 }$

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2 years ago