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notka56 [123]
3 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.
Physics
1 answer:
Lynna [10]3 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

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sukhopar [10]

Answer:

v=2.58\times10^8m/s

Explanation:

The index of refraction is equal to the speed of light c in vacuum divided by its speed v in a substance, or n=\frac{c}{v}. For our case we want to use v=\frac{c}{n}, which for our values is equal to:

v=\frac{c}{n}=\frac{299792458m/s}{1.16}=258441774.138m/s

Which we will express with 3 significant figures (since a product or quotient must contain the same number of significant figures as the measurement with the  <em>least</em> number of significant figures):

v=2.58\times10^8m/s

4 0
3 years ago
7. When will an object's displacement and distance traveled be different?
mariarad [96]

If an object changes direction while travelling will an object's displacement and distance travelled be different.

Some people believe that distance and displacement are simply different names for the same quantity. However, distance and displacement are not the same thing. If an object changes direction while travelling, the total distance travelled is greater than the displacement between those two points.

The magnitude of the displacement is always less than or equal to the distance because it is measured along the shortest path between two points.

When the direction of displacement does not change, the magnitude of the displacement and distance are the same. When a body travels in a straight line, for example, its displacement and distance are the same.

Learn more about displacement and distance brainly.com/question/3243551

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8 0
2 years ago
A spring is stretched to a displacement of 3.4 m from equilibrium. Then the spring is released and allowed to recoil to a displa
Stells [14]
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5 0
3 years ago
On a perfect fall day, you are hovering at low altitude in a hot-air balloon, accelerated neither upward nor downward. The total
Stella [2.4K]

Explanation:

Since the balloon is not accelerating means that the net force on the balloon is zero. This implies that the weight of balloon must be equal to the buoyant force on balloon.

Hence, the buoyant force equals the weight of air displaced by the balloon, also 20,000 N.

Weight of the air displaced = density of air × volume

The density of air at 1 atm pressure and 20º C is 1.2 kg/m³  

the volume V = 20,000/(1.2×9.8) =  1700 m³

3 0
3 years ago
A space probe is fired as a projectile from the Earth's surface with an initial speed of 2.05 104 m/s. What will its speed be wh
Elanso [62]

Answer:

The value is  v  =  2.3359 *10^{4} \ m/s

Explanation:

From the question we are told that

  The  initial speed is u =  2.05 *10^{4} \  m/s

 Generally the total energy possessed by the space probe when on earth is mathematically represented as

             T__{E}} =  KE__{i}} +  KE__{e}}

Here  KE_i is the kinetic energy of the space probe due to its initial speed which is mathematically represented as

          KE_i =   \frac{1}{2}  *  m  *  u^2

=>       KE_i =   \frac{1}{2}  *  m  *  (2.05 *10^{4})^2

=>       KE_i =  2.101 *10^{8} \ \ m \ \ J

And  KE_e is the kinetic energy that the space probe requires to escape the Earth's gravitational pull , this is mathematically represented as

       KE_e =  \frac{1}{2}  *  m *  v_e^2

Here v_e is the escape velocity from earth which has a value v_e =  11.2 *10^{3} \  m/s

=>    KE_e =  \frac{1}{2}  *  m *  (11.3 *10^{3})^2

=>    KE_e =  6.272 *10^{7} \  \  m  \ \   J

Generally given that at a position that is very far from the earth that the is Zero, the kinetic energy at that position is mathematically represented as

        KE_p =  \frac{1}{2}  *  m *  v^2

Generally from the law energy conservation we have that

        T__{E}} =  KE_p

So

       2.101 *10^{8}  m  +  6.272 *10^{7}  m  =   \frac{1}{2}  *  m *  v^2

=>     5.4564 *10^{8} =   v^2

=>     v =  \sqrt{5.4564 *10^{8}}

=>     v  =  2.3359 *10^{4} \ m/s

4 0
3 years ago
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