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Maru [420]
3 years ago
8

Which is not a way to conserve existing energy resources?

Physics
2 answers:
zimovet [89]3 years ago
8 0
For the answer to the question above, the answer is simple.<span> among</span> the choices given the only way on not to conserve energy is by using the available fossil fuel.
I hope my answer helped you. feel free to ask more questions. Have a nice day!
Ghella [55]3 years ago
5 0
Use available fossil fuel
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photo one- refraction

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What scale ranks minerals from softest to hardest
rodikova [14]

Answer: Mohs Scale

Explanation: Mohs Scale of Hardness

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Un móvil con velocidad inicial de 19,8km/h adquiere una aceleración constante de 2,4m/s^2. Determine la velocidad y el espacio r
nata0808 [166]

Responder:

Velocidad = 41.5m / s

Espacio recorrida = 352.5 metros

Explicación:

Dado lo siguiente:

Velocidad inicial (u) = 19.8 km / h

Aceleración (a) = 2.4m / s ^ 2

Tiempo de viaje (t) = 15 s

A.) velocidad después de 15 s

Velocidad inicial = (19.8 × 1000) m / 3600s Velocidad inicial = 19800m / 3600 = 5.5m / s

Usando la ecuación: v = u + at, donde v es la velocidad

v = 5.5 + 2.4 (15)

v = 5.5 + 36

v = 41.5m / s

Espacio recorrida:

v ^ 2 = u ^ 2 + 2aS; donde S es la distancia recorrida

41.5 ^ 2 = 5.5 ^ 2 + 2 × (2.4) × S

1722.25 = 30.25 + 4.8S

1722.25 - 30.25 = 4.8S

1692 = 4.8S S = 1692 / 4.8 S = 352.5m

8 0
3 years ago
In the real word what is the terminal velocity
olga55 [171]
Based on wind resistance?
5 0
4 years ago
A magnetic field of 0.276 T exists in the region enclosed by a solenoid that has 517 turns and a diameter of 10.5 cm. Within wha
Advocard [28]

Answer:

The period the field must be reduced to zero is 9.81 x 10⁻⁵ s

Explanation:

Given;

initial value of the magnetic field, B₁ = 0.276 T

number of turns of the solenoid, N = 517 turns

diameter of the solenoid, d = 10.5 cm = 0.105 m

induced emf, = 12.6 kV = 12,600 V

when the field becomes zero, then the final magnetic field value, B₂ = 0

The induced emf is given by Faraday's law;

emf = -\frac{NA\Delta B}{t} \\\\emf = -\frac{NA (B_2 -B_1)}{t} \\\\t = -\frac{NA (B_2 -B_1)}{emf}\\\\t = \frac{NA (B_1 -B_2)}{emf}\\\\where;\\\\t \ is \ the \ time \ when \ B = 0 \ \ (i.e\ B_2 = 0)\\\\A \ is \ the \ area \ of \ the \ coil\\\\A = \frac{\pi d^2}{4} = \frac{\pi (0.105)^2}{4} = 0.00866 \ m^2\\\\t= \frac{(517) \times (0.00866)\times  (0.276 -0)}{12,600}\\\\t = 9.81 \times 10^{-5} \ s

Therefore, the period the field must be reduced to zero is 9.81 x 10⁻⁵ s

4 0
4 years ago
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