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2 years ago

This homework is meant to be in for today first lesson, please help!!!

Physics

1 answer:

Black_prince [1.1K]2 years ago

4 0

The agriculture land needed to grow the willow is 1500000 Hectares

<h3>What is Power?</h3>

Power can be defined as the rate at which work is done by an object. The S.I unit of power is Watt(W).

A coal burning power station burns 6 million tonnes of coal per kg. Coal has a average energy value of 29.25 MJ per year. Wood chip from willow trees has an energy value of 13 MJ per kg. A hectare of agricultural land can produce 9 tonnes of drt willow wood per year.

The coal consumption per year = 29.25 x 1000 x 6000000

=175500000000 MJ

If this power station burned dry willow wood instead of coal, agriculture land would be needed to grow the willow,

1 hectare of willow will produce 9 x 13x 1000 = 117000 MJ.

The agriculture land needed to grow the willow is

= 175500000000 MJ / 117000 MJ.

= 1500000 Hectares.

Thus, agriculture land needed is 1500000 Hectares.

Learn more about power.

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Both speed and velocity measure how fast something is moving. However, since speed is not a _________ it does not require a(n) _

Jet001 [13]

Both speed and velocity measure how fast something is moving. Hence, since speed is not a vector it does not require direction.

Speed only depends on the distance covered by the object and how much time it took to cover that distance. It does not care about the direction, it just depend on the length of the entire path covered. Whereas, velocity depends on magnitude as well as the direction of the object's motion for example: When an airplane is traveling, It is better to deal the motion with velocity and when we are moving in a car inside the city, we should deal the motion with speed.

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4 years ago

2.) The speed of light in a vacuum is c = 300,000 km / s. The light of Sol takes 8 minutes and 19 seconds to reach Earth. Calcul

Vinvika [58]

Answer:

149,700,000 km (= 1.50 x 10⁸ km)

Explanation:

Given,

Speed of light, c = 300,000 km/s

Time Taken = 8 min 19 s = 499 seconds

Recall, Distance = Speed x Time

= 300,000 km/s x 499 s

= 149,700,000 km

= 1.50 x 10⁸ km

3 0

3 years ago

Read 2 more answers

Where is the sun's energy most concentrated

Maru [420]

B. At the equator

Explanation:

The energy coming from the Sun hits the Earth's surface at different angles, depending on the latitude of the place. The more perpendicular the ray of lights hit the surface, the more the energy transmitted to the Earth's surface, the warmer the location.

The angle at which the ray of lights hit the Earth is related to the latitude: in particular, the ray of lights arrive perpendicular at the equator ( $0^{\circ}$ ), they arrive at larger angle in the United States (which is located at intermediate latitudes) and they arrive at the largest angles at the poles. For this reason, the sun's most energy is concentrated at the equator.

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3 years ago

Please help me Calculate the total energy stored in the capacitor

Neko [114]

La fórmula: P = W/t ó W = P x t. donde:

P = potencia

W = trabajo

t = tiempo

Otra fórmula de potencia es: P= I x V

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Una fórmula muy importante que también hay que tener en cuenta es: V = q/C que indica que el voltaje es proporcional a la carga que hay en un condensador.

De la fórmula de potencia P= I x V y considerando que la corriente es constante (corriente continua), entonces la potencia es proporcional al voltaje. Si el voltaje aumenta en forma lineal, la potencia aumentará igual. Ver el siguiente diagrama.

Como la potencia varía en función del tiempo, no se puede aplicar la fórmula W = P x t, para calcular la energía transferida. Pero observando el gráfico, se ve que esta energía se puede determinar midiendo el área bajo la curva de la figura.

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El área bajo la curva es igual a la mitad de la potencia en el momento “t”, multiplicada por “t”.

Entonces: W = (P x t) / 2. Pero se sabe que P = V x I. Si se reemplaza esta última fórmula en la anterior se obtiene: W = (V x I x t) / 2, y como I x t = CV = Q, entonces para saber cuanta energía (W) hay en un condensador usamos una de las siguientes fórmulas:

W = (CV2/2) julios

W = (QV/2) julios

W = (Q2/2C) julios

, donde:

W = Trabajo (Energía) en julios

C = Capacidad en faradios

V = voltaje en voltios en los extremos del condensador

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3 0

3 years ago

Read 2 more answers

A crate of mass 9.2 kg is pulled up a rough incline with an initial speed of 1.58 m/s. The pulling force is 110 N parallel to th

kari74 [83]

Given that,

Mass = 9.2 kg

Force = 110 N

Angle = 20.2°

Distance = 5.10 m

Speed = 1.58 m/s

(A). We need to calculate the work done by the gravitational force

Using formula of work done

$W_{g}=mgd\sin\theta$

Where, w = work

m = mass

g = acceleration due to gravity

d = distance

Put the value into the formula

$W_{g}=9.2\times(-9.8)\times5.10\sin20.2$

$W_{g}=-158.8\ J$

(B). We need to calculate the increase in internal energy of the crate-incline system owing to friction

Using formula of potential energy

$\Delta U=-W$

Put the value into the formula

$\Delta U=-(-158.8)\ J$

$\Delta U=158.8\ J$

(C). We need to calculate the work done by 100 N force on the crate

Using formula of work done

$W=F\times d$

Put the value into the formula

$W=100\times5.10$

$W=510\ J$

We need to calculate the work done by frictional force

Using formula of work done

$W=-f\times d$

$W=-\mu mg\cos\theta\times d$

Put the value into the formula

$W=-0.4\times9.2\times9.8\cos20.2\times5.10$

$W=-172.5\ J$

We need to calculate the change in kinetic energy of the crate

Using formula for change in kinetic energy

$\Delta k=W_{g}+W_{f}+W_{F}$

Put the value into the formula

$\Delta k=-158.8-172.5+510$

$\Delta k=178.7\ J$

(E). We need to calculate the speed of the crate after being pulled 5.00m

Using formula of change in kinetic energy

$\Delta k=\dfrac{1}{2}m(v_{2}^2-v_{1}^{2})$

$v_{2}^2=\dfrac{2\times\Delta k}{m}+v_{1}^2$

Put the value into the formula

$v_{2}^2=\dfrac{2\times178.7}{9.2}+1.58$

$v_{2}=\sqrt{\dfrac{2\times178.7}{9.2}+1.58}$

$v_{2}=6.35\ m/s$

Hence, (A). The work done by the gravitational force is -158.8 J.

(B). The increase in internal energy of the crate-incline system owing to friction is 158.8 J.

(C). The work done by 100 N force on the crate is 510 J.

(D). The change in kinetic energy of the crate is 178.7 J.

(E). The speed of the crate after being pulled 5.00m is 6.35 m/s

8 0

3 years ago