Ask question

Ask question

All categories

2 years ago

6

The average energy per unit time per unit area that reaches the upper atmosphere of the Earth from the Sun, called the solar con

stant, is 1.47 kW/m2. Because of absorption and re- flection by the atmosphere, about 0.9 kW/m2 reaches the surface of the earth on a clear day. How much energy is collected during 1 h of daylight by a window that measures 1.1 m by 1.5 m? The window is on a mount that rotates, keeping the window facing the sun so the sun's rays remain perpendicular to the window. Answer in units of MJ.

1 answer:

Lynna [10]2 years ago

8 0

Answer: 5.34 MJ

Explanation: The window is a rectangle with area:

A = 1.1(1.5)

A = 1.65 m²

We know that only 0.9 kW/m² reaches Earth so:

P = $0.9\frac{kW}{m^{2}}(1.65)m^{2}$

P = 1.485 x 10³ W

Watts is an unit of Power and it can also be written as J/s.

An hour has 3600s or 3.6 x 10³s, so:

E = 1.485 x 10³ $\frac{J}{s}$ (3.6 x 10³s)

E = 5.346 x 10⁶ J

Mega is equal to 10⁶, then:

E = 5.35 MJ

A 1.1 m by 1.5 m window collects, during 1 hour of daylight, 5.35 MJ

You might be interested in

1. A 25.35-g piece of iron absorbs 1562.75 Joules of heat energy, and its temperature

Lera25 [3.4K]

Answer:

Q = C M T where C is the specific, M the mass, T the temperature change

Note 1 cal = 4.19 Joules

1562.75 J / (4.19 J/cal) = 378 cal

C = Q / (M * T) = 378 cal / (25.35 g * 155 deg C)

C = .096 cal / g deg C

8 0

1 year ago

An electron and a proton are held on an x axis, with the electron at x = + 1.000 m and the proton at x = - 1.000 m. Part A How m

r-ruslan [8.4K]

PART A)

Electrostatic potential at the position of origin is given by

$V = \frac{kq_1}{r_1} + \frac{kq_2}{r_2}$

here we have

$q_1 = 1.6 \times 10^{-19} C$

$q_2 = -1.6 \times 10^{-19} C$

$r_1 = r_2 = 1 m$

now we have

$V = \frac{Ke}{r} - \frac{Ke}{r}$

$V = 0$

Now work done to move another charge from infinite to origin is given by

$W = q(V_f - V_i)$

here we will have

$W = e(0 - 0) = 0$

so there is no work required to move an electron from infinite to origin

PART B)

Initial potential energy of electron

$U = \frac{Kq_1e}{r_1} + \frac{kq_2e}{r_2}$

$U = \frac{9\times 10^9(-1.6\times 10^{-19}(-1.6 \times 10^{-19})}{19} + \frac{9\times 10^9(1.6\times 10^{-19}(-1.6 \times 10^{-19})}{21}$

$U = (2.3\times 10^{-28})(\frac{1}{19} - \frac{1}{21})$

$U = 1.15\times 10^{-30}$

Now we know

$KE = \frac{1}{2}mv^2$

$KE = \frac{1}{2}(9.1\times 10^{-31}(100)^2$

$KE = 4.55 \times 10^{-27} kg$

now by energy conservation we will have

So here initial total energy is sufficient high to reach the origin

PART C)

It will reach the origin

4 0

2 years ago

What is the acceleration of an object that goes from 45m/s to 10 m/s in 5 seconds?

kipiarov [429]

Answer:

$\boxed {\boxed {\sf a= -7 \ m/s^2}}$

Explanation:

Acceleration is the change in velocity over time.

$a= \frac {v_f-v_i}{t}$

The object accelerates <em>from</em> 45 meters per second <em>to </em>10 meters per second in 5 seconds. Therefore,

$v_f=10 \ m/s \\v_i= 45 \ m/s \\t= 5 \ s$

Substitute the values into the formula.

$a= \frac{ 10 \ m/s - 45 \ m/s}{5 \ s}$

Solve the numerator.

$a= \frac { -35 \ m/s}{5 \ s}$

Divide.

$a= -7 \ m/s/s$

$a= -7 \ m/s^2$

The acceleration of the object is -7 meters per square second. The acceleration is negative because the object's velocity decreases and the object slows down.

5 0

2 years ago

A ball is kicked at a speed of 16m/s at 33° and it eventually returns to ground level further down field.

saw5 [17]

Hi there!

We can begin by calculating the time taken to reach its highest point (when the vertical velocity = 0).

Remember to break the velocity into its vertical and horizontal components.

Thus:

0 = vi - at

0 = 16sin(33°) - 9.8(t)

9.8t = 16sin(33°)

t = .889 sec

Find the max height by plugging this time into the equation:

Δd = vit + 1/2at²

Δd = (16sin(33°))(.889) + 1/2(-9.8)(.889)²

Solve:

Δd = 7.747 - 3.873 = 3.8744 m

4 0

2 years ago

Read 2 more answers

A single strain gage has a nominal resistance of 120 ohm. For a quarter bridge with 120 ohm fixed resistors, strain gauge factor

natita [175]

Answer:

Output voltage is 1.507 mV

Solution:

As per the question:

Nominal resistance, R = $120\Omega$

Fixed resistance, R = $120\Omega$

Gauge Factor, G.F = 2.01

Supply Voltage, $V_{s} = 3\ V$

Strain, $\epsilon = 1000\times 10^{-6}\ strain$

Now,

To calculate the output voltage, $V_{o}$ :

WE know that strain is given by:

$\epsilon = \frac{(R + R')^{2}V_{o}}{RR'V_{s}\times G.F}$

Thus

$V_{o} = \frac{RR'V_{s}\epsilon \times G.F}{(R + R')^{2}}$

Now, substituting the suitable values in the above eqn:

$V_{o} = \frac{120\times 120\times 3\times 1000\times 10^{-6}\times 2.01}{(120 + 120)^{2}}$

$V_{o} = 1.507\ mV$

6 0

3 years ago