Large amounts of water that quickly infiltrate soil on a sloped surface can

A 3 om light bulb and a 6 om light bulb are connected in series, and then hooked up to a battery. Which of the two will shine th

RUDIKE [14]

Answer:

the 6 om is brighter because 6-3=3

Explanation:

3 0

3 years ago

A heat pump is to be used for heating a house in winter. The house is to be maintained at 70°F at all times. When the temperatur

Anna35 [415]

Answer:

$\dot{W_{H} } = 4244.48 Btu/h$

Explanation:

Temperature of the house, $T_{H} = 70^{0} F$

Convert to rankine, $T_{H} = 70^{0}+ 460 = 530 R$

Heat is extracted at 40°F i.e $T_{L} = 40^{0}F = 40 + 460 = 500 R$

Calculate the coefficient of performance of the heat pump, COP

$COP = \frac{T_{H} }{T_{H} - T_{L} } \\COP = \frac{530 }{530 - 500 }\\ COP = \frac{530}{30} \\COP = 17.67$

The minimum power required to run the heat pump is given by the formula:

$\dot{W_{H} } = \frac{\dot{Q_{H} }}{COP} \\$ ...............(*)

Where the heat losses from the house, $\dot{Q_{H} } = 75,000 Btu/h$

Substituting these values into * above

$\dot{W_{H} } = \frac{75000}{17.67} \\ \dot{W_{H} } = 4244.48 Btu/h$

3 0

2 years ago

Two particles are located on the x axis. particle 1 has a mass m and is at the origin. particle 2 has a mass 2m and is at x = +l

wlad13 [49]

The solution would be like this for this specific problem:

The force on m is:

GMm / x^2 + Gm(2m) / L^2 = 2[Gm (2m) / L^2] -> 1

The force on 2m is:

GM(2m) / (L - x)^2 + Gm(2m) / L^2 = 2[Gm (2m) / L^2] -> 2

From (1), you’ll get M = 2mx^2 / L^2 and from (2) you get M = m(L - x)^2 / L^2

Since the Ms are the same, then

2mx^2 / L^2 = m(L - x)^2 / L^2

2x^2 = (L - x)^2

xsqrt2 = L - x

x(1 + sqrt2) = L

x = L / (sqrt2 + 1) From here, we rationalize.

x = L(sqrt2 - 1) / (sqrt2 + 1)(sqrt2 - 1)

x = L(sqrt2 - 1) / (2 - 1)

x = L(sqrt2 - 1)

= 0.414L

Therefore, the third particle should be located the 0.414L x axis so that the magnitude of the gravitational force on both particle 1 and particle 2 doubles.

8 0

3 years ago

a person throws a ball upward into the air with an initial velocity of 20 m/s. calculate (a) how high it goes, and (b) how long

Phantasy [73]

Answer:

a) about 20.4 meters high

b) about 4.08 seconds

Explanation:

Part a)

To find the maximum height the ball reaches under the action of gravity (g = 9.8 m/s^2) use the equation that connects change in velocity over time with acceleration.

$a=\frac{Vf-Vi}{t}$

$-9.8 \frac{m}{s} =\frac{Vf-Vi}{t}$

In our case, the initial velocity of the ball as it leaves the hands of the person is Vi = 20 m/s, while thw final velocity of the ball as it reaches its maximum height is zero (0) m/s. Therefore we can solve for the time it takes the ball to reach the top:

$-9.8 =\frac{0-20}{t}\\t=\frac{20}{9.8} s = 2.04 s$

Now we use this time in the expression for the distance covered (final position Xf minus initial position Xi) under acceleration:

$Xf-Xi=Vi*t+\frac{1}{2} a t^{2} \\Xf-Xi=20*(2.04)-\frac{1}{2} 9.8*2.04^{2}\\Xf-Xi=20.408 m$

Part b) Now we use the expression for distance covered under acceleration to find the time it takes for the ball to leave the person's hand and come back to it (notice that Xf-Xi in this case will be zero - same final and initial position)

$Xf-Xi=0=20*(t)-\frac{1}{2} 9.8*t^{2$

To solve for "t" in this quadratic equation, we can factor it out as shown:

$0= t(20-\frac{9.8}{2} t)$

Therefore there are two possible solutions when each of the two factors equals zero:

1) t= 0 (which is not representative of our case) , and

2) the expression in parenthesis is zero:

$0= 20-\frac{9.8}{2} t\\t=\frac{20*2}{9.8} = 4.08 s$

7 0

3 years ago

Que unidad de medida se utiliza para expresar la capacidad o volumen de un objeto? Da un ejemplo

galina1969 [7]

Answer:

Explicado a continuación

Explanation:

Hay una pequeña diferencia conceptual entre la capacidad y el volumen de un objeto, a saber:

El volumen hace referencia al espacio que ocupa un objeto, mientras que la capacidad hace referencia al espacio que este contiene. Calcular el volumen de un cuerpo es medir cuánto ocupa mientras que calcular su capacidad es medir cuánto cabe en él.

En la práctica, ambos conceptos son usados indistintamente, ya que tienen unidades equivalentes.

El volumen tiene unidades de longitud al cubo, como por ejemplo:

$m^3,\ cm^3\ ,dm^3,\ pie^3\ , pulg^3$

y la capacidad se suele expresar en litros o unidades derivadas: litro, mililitro, centilitro, etc.

Como mencionamos, hay equivalencia engre los dos grupos de unidades. Entre las más conocidas están:

$1\ lt=1\ dm^3,\ 1\ m^3=1000\ lt$

5 0

2 years ago