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3 years ago

Cold water flows to the solar panels at 15°C. During the day, the panels supply 3.8 kg of hot water

Physics

1 answer:

Yuliya22 [10]3 years ago

5 0

Answer:

The average energy that the solar panel delivers to the water in one hour is 798,000 Joules

Explanation:

The given parameters are;

The temperature of the water that flows to the solar panels, T₁ = 15°

The mass of the hot water the panel supplies during the day, m = 3.5 kg

The temperature of the hot water supplied by the panel, T₂ = 65 °C

The heat capacity of the water, c = 4,200 J/(kg·°C)

The heat delivered to the solar panels in one hour, ΔQ, is given by the following formula

ΔQ = m·c·ΔT

m = 3.8 kg, c = 4,200 J/(kg·°C), and ΔT = T₂ - T₁ = 65 °C - 15 °C = 50 °C

∴ ΔQ = 3.8 kg × 4,200 J/(kg·°C) × 50°C = 798,000 joules

∴ΔQ = 798,000 joules

The average energy that the solar panel delivers to the water in one hour, ΔQ = 798,000 joules.

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A uniform cylinder of radius 25 cm and mass 27 kg is mounted so as to rotate freely about a horizontal axis that is parallel to

Alisiya [41]

Answer:

Explanation:

STEP 1

Given

Radius of cylinder = r = 25cm, 2.5m

mass = 27kg

cylinder is mounted so as to rotate freely about a horizontal axis that is parallel to and 60cm to the central logitudinal axis of the cylinder

height = 0.6m

part 1

The cylinder is mounted so as to rotate freely about a horizontal axis tha is paralle to 60cm from the central longitudinal axis of then cylinder. The rotational inertia of the cylinder about the axis of rotation is given by

I = Icm + mh²

∴ I = 1/2mr² + mh² = 1/2x27x (0.5)² + 20 x (0.6)²

I=13.09kg.m²

where

Icm is the rotational inertia of the cylinder about its central axis

m is the mass of the cylinder

h is the distance between the axis of the rotation and the central axis of the cylinder

r is the radius of the cylinder

 I=13.09kg.m²

part2

from the conservation of the total mechanical energy of the meter stick, the change in gravitational potential energyof the meter stick plus the change in kinetic energy must be zero

Δk + Δu = 0

1/2 I(w²-w²) = Ui-Uf

1/2 x 13.09w² = mgh

∴w=√20 x 9.8 x 0.6/(1/2 x 13.09) =117.6/6.5

w=18.09rad/s

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3 years ago

Un pez llamado PARGO ROJO vive a grandes profundidades. Si se pesca, al salir a la superficie puede tomar el aspecto de la foto

zalisa [80]

Answer:

Hay diversas leyes que podemos usar acá.

Acá sabemos que la vejiga aumenta su tamaño al reducir la presión, esto tiene sentido, pues al haber menos presión, hay menos fuerza que comprime la vejiga, lo que le permite aumentar su volumen.

Acá tenemos una relación inversa de la forma: V = K/P

Una relación inversa donde la presión esta en el denominador y K es un termino que no depende ni del volumen ni de la presión.

Entonces, a medida que aumenta P, el denominador aumenta, por lo que el valor del volumen decrece.

Un ejemplo de una ecuación similar es la del gas ideal, por ejemplo, para un gas ideal dentro de un globo de volumen V para una dada presión P:

V = nRT/P

donde n es el numero de moles, R es la constante termodinámica y T es la temperatura, acá podemos ver que esta ecuación tiene la misma forma fundamental que la escrita arriba.

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3 years ago

Two hoops, starting from rest, roll down identical inclined planes. The work done by nonconservative forces, such as air resiste

Maurinko [17]

Answer: both hoops have the same kinetic energy at the bottom of the incline.

Explanation:

If we assume no work done by non conservative forces (like friction) , the total mechanical energy must be conserved.

K1 + U1 = K2 + U2

If both hoops start from rest, and we choose the bottom of the incline to be the the zero reference level for gravitational potential energy, then

K1 = 0 and U2 = 0

⇒ ΔK = ΔU = m g. h

If both inclines have the same height, and both hoops have the same mass m, the change in kinetic energy, must be the same for both hoops.

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3 years ago

A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.6 m/s at ground level.

kow [346]

Before the engines fail, the rocket's altitude at time t is given by

$y_1(t)=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2$

and its velocity is

$v_1(t)=80.6\dfrac{\rm m}{\rm s}+\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t$

The rocket then reaches an altitude of 1150 m at time t such that

$1150\,\mathrm m=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2$

Solve for t to find this time to be

$t=11.2\,\mathrm s$

At this time, the rocket attains a velocity of

$v_1(11.2\,\mathrm s)=124\dfrac{\rm m}{\rm s}$

When it's in freefall, the rocket's altitude is given by

$y_2(t)=1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2$

where $g=9.80\frac{\rm m}{\mathrm s^2}$ is the acceleration due to gravity, and its velocity is

$v_2(t)=124\dfrac{\rm m}{\rm s}-gt$

(a) After the first 11.2 s of flight, the rocket is in the air for as long as it takes for $y_2(t)$ to reach 0:

$1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2=0\implies t=32.6\,\mathrm s$

So the rocket is in motion for a total of 11.2 s + 32.6 s = 43.4 s.

(b) Recall that

${v_f}^2-{v_i}^2=2a\Delta y$

where $v_f$ and $v_i$ denote final and initial velocities, respecitively, $a$ denotes acceleration, and $\Delta y$ the difference in altitudes over some time interval. At its maximum height, the rocket has zero velocity. After the engines fail, the rocket will keep moving upward for a little while before it starts to fall to the ground, which means $y_2$ will contain the information we need to find the maximum height.

$-\left(124\dfrac{\rm m}{\rm s}\right)^2=-2g(y_{\rm max}-1150\,\mathrm m)$

Solve for $y_{\rm max}$ and we find that the rocket reaches a maximum altitude of about 1930 m.

(c) In part (a), we found the time it takes for the rocket to hit the ground (relative to $y_2(t)$ ) to be about 32.6 s. Plug this into $v_2(t)$ to find the velocity before it crashes: