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Elis [28]
3 years ago
8

I will give 50 pts and brainliest

Physics
1 answer:
ipn [44]3 years ago
8 0

Answer:

Limiting the use of fossil fuels such as oil, carbon and natural gas and replacing them with renewable and cleaner sources of energy, all while increasing energy efficiency

Explanation:

You might be interested in
When a condenser discharges electricity, the instantaneous rate of change of the voltage is proportional to the voltage in the c
e-lub [12.9K]

Answer:

460.52 s

Explanation:

Since the instantaneous rate of change of the voltage is proportional to the voltage in the condenser, we have that

dV/dt ∝ V

dV/dt = kV

separating the variables, we have

dV/V = kdt

integrating both sides, we have

∫dV/V = ∫kdt

㏑(V/V₀) = kt

V/V₀ = e^{kt}

Since the instantaneous rate of change of the voltage is -0.01 of the voltage dV/dt = -0.01V

Since dV/dt = kV

-0.01V = kV

k = -0.01

So, V/V₀ = e^{-0.01t}

V = V₀e^{-0.01t}

Given that the voltage decreases by 90 %, we have that the remaining voltage (100 % - 90%)V₀ = 10%V₀ = 0.1V₀

So, V = 0.1V₀

Thus

V = V₀e^{-0.01t}

0.1V₀ = V₀e^{-0.01t}

0.1V₀/V₀ = e^{-0.01t}

0.1 = e^{-0.01t}

to find the time, t it takes the voltage to decrease by 90%, we taking natural logarithm of both sides, we have

㏑(0.01) = -0.01t

So, t = ㏑(0.01)/-0.01

t = -4.6052/-0.01

t = 460.52 s

3 0
3 years ago
An object has a mass of 10 kilograms and is accelerating at 5 m/s/s, what force pushed the object?
Katen [24]

Answer:

force pushed the object = 10 × 5 = 50N

8 0
3 years ago
a radio antenna broadcasts a 1.0 MHz radio wav e with 21 kW of power. Assume that the radiation is emitted uniformly in all dire
My name is Ann [436]

Answer:

I=2.67\times 10^{-6}\ W/m

Explanation:

Given that,

Frequency of a radio antenna is 1 MHz

Power, P = 21 kW

We need to find the the waves intensity 25 km from the antenna . The object emits intenisty evenly in all direction. It can be given by :

I=\dfrac{P}{4\pi r^2}\\\\I=\dfrac{21\times 10^3}{4\pi (25000)^2}\\\\I=2.67\times 10^{-6}\ W/m

So, the wave intensity 25 km from the antenna is 2.67\times 10^{-6}\ W/m.

5 0
3 years ago
Se apunta un rifle horizontalmente con mira a un blanco pequeño que está a 200m en el suelo. La velocidad inicial de la bala es
vfiekz [6]

Answer:

Lo importante a tener en cuenta sobre esta pregunta es que la velocidad horizontal de la bala no hace ninguna diferencia en cuanto al tiempo que tarda en caer al suelo.

Debido a que el arma no ha aplicado ninguna fuerza vertical a la bala, la única fuerza que afecta la bala es la gravedad. Esto significa que la bala tarda tanto en caer al suelo como lo haría si se cayera, a pesar de que ahora viaja una gran distancia horizontal en la duración.

Para encontrar el tiempo de viaje antes de tocar el suelo, tenemos 3 valores:

-El desplazamiento desde el suelo que la bala debe viajar, s = 1.5m

-La aceleración que experimenta la bala. Como la gravedad está acelerando la bala hacia abajo, a = g = ~ 9.81m / s ^ 2

-La velocidad inicial de la bala verticalmente. Como la bala es estacionaria verticalmente (solo viaja horizontalmente al inicio), u = 0m

Examinamos nuestras ecuaciones de movimiento, comúnmente conocidas como ecuaciones SUVAT. Es posible que necesite aprender estos para su examen, pero algunas tablas de examen los proporcionan.

Debido a que tenemos s, u y a, y estamos buscando el tiempo t, la ecuación relevante es

s = ut + 0.5 (en ^ 2)

Completando nuestros valores tenemos:

1.5 = 0t + 0.5 (9.81 x t ^ 2)

1.5 = 4.905 x t ^ 2

Divide 1.5 entre 4.905 para encontrar t ^ 2

t ^ 2 = 0.3058 ...

Simplemente encontramos la raíz cuadrada de t ^ 2 para encontrar t, el tiempo que tarda la bala en llegar al suelo:

t = 0.553s (3 cifras significativas)

Para encontrar la distancia horizontal, d, que la bala ha viajado antes de tocar el suelo, podemos usar la ecuación que vincula el desplazamiento s con cierta velocidad v durante un tiempo t:

s = vt

La velocidad horizontal de la bala, v = 430

El tiempo antes de que la bala toque el suelo, t = 0.553

Entonces d = vt = 430 * 0.553 = 238m (3 cifras significativas)

3 0
3 years ago
A force of 200 N stretches a spring 30 cm. What is the spring constant of the spring? How far would this spring stretch with a f
bija089 [108]

Hooke's Law

F = k. Δx

Δx = 30 cm = 0.3 m

200 = k . 0.3

\tt k =\dfrac{200}{0.3}= 666.6

the spring stretch for 100 N:

\tt \Delta x=\dfrac{100}{666.6}=0.15=15\:cm

4 0
2 years ago
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