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katovenus [111]
3 years ago
5

In solar thermal electric generation, the sun’s energy is concentrated by ___ to heat the fluid in a pipe that is used to genera

te electricity.
Physics
1 answer:
pishuonlain [190]3 years ago
7 0

Answer:

<u>Mirrors</u>

Explanation:

In Solar thermal power generation system, mirrors are used to collect and focus sunlight  onto a fluid pipe to produce high temperature required to generate electricity.

There are two main components: mirrors capture and focus light onto a receiver. The heat transfer fluid is circulated in the receiver. Steam is produced which is converted to mechanical energy in a turbine. It powers a generator and electricity is produced.

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Hydraulics uses _______ to change the size and/or direction of a force
cupoosta [38]
I know this the answer is <span>pressurized liquids if you go on quizlet they will always give you the answer just so you know</span>
5 0
3 years ago
The North American Plate is moving west at a rate of approximately 20 mm/yr. How long will it take for New York to move 10° long
scZoUnD [109]

Answer:

55000000 years

Explanation:

Rate of North American Plate moving (Velocity) = 20 mm/yr

Degrees New York has to move = 10° west

New York's longitude = 110 km/°

Distance of New York = 110×10 = 1100 km

= 1100×10⁶ mm

\text {Time taken by the continental plate}=\frac {\text {Distance of New York}}{\text {Rate of North American Plate moving (Velocity)}}\\\Rightarrow \text {Time taken by the continental plate}=\frac{1100\times 10^6}{20}\\\Rightarrow \text {Time taken by the continental plate}=55\times 10^6\ years

∴ Time taken by New York to move 10° west is 55000000 years

6 0
3 years ago
*A car is going through a dip in the road whose curvature approximates a circle of radius 150m. At what velocity will the occupa
Valentin [98]

Answer:

v= 14.85 m/s

Explanation:

  • When at the bottom of the dip, the only force that keeps the car in the circular trajectory, is the centripetal force.
  • This force is not a new force, is just the net force aiming to the center of the circle.
  • In this case, is just the difference between the normal force (always perpendicular to the surface, pointing upward) and the force that gravity exerts on the car (which is known as the weight), pointing downward.
  • So, we can write the following expression:

       F_{cent} = F_{n} - F_{g}  (1)

  • It can be showed that the centripetal force is related to the speed by the following expression:
  • F_{cent} = m*\frac{v^{2}}{r} (2)
  • The normal force, it is called the apparent weight, because it would be the weight as measured by a scale.
  • Replacing (2) in (1), and solving for Fn, we get:

       F_{n} = m*\frac{v^{2} }{r} + m*g (3)

  • Now, we need to find the value of v that makes Fn, exactly 15% more than the weight m*g, so we can write the following equation:

      F_{n} = 1.15*F_{g} = m*\frac{v^{2}}{r} +F_{g}  (4)

  • Replacing Fg by its value, simplifying, and solving for v, we get:

       v = \sqrt{0.15*g*r} = \sqrt{9.8 m/s2*0.15*150m} = 14.85 m/s (5)

3 0
3 years ago
The solid rod shown has a radius of 0.75 in. if it is subjected to the force the 500lb determine the max normal stress developed
Kipish [7]
The normal stress follows the formula written below:

σ = F/A

There are two types of stress, axial and tangential. Since we are only given with the dimension of the radius (and not the length), the possible stress is axial. So, the area is,

A = πr² = π(0.75 in)² = 1.767 in²

So,
σ = F/A = 500 lb/1.767 in² = <em>282.94 psi</em>
7 0
3 years ago
An air-track glider with a mass of 239 g is moving at 0.81 m/s on a 2.4 m long air track. It collides elastically with a 513 g g
HACTEHA [7]

Answer:

Glider it stops just when it reaches the end of the runway

Explanation:

This is a shock between two bodies, so we must use the equations of conservation of the amount of movement, in the instant before the crash and the subsequent instant, with this we calculate the second glider speed, as the shock that elastic is also keep it kinetic energy

        Po = pf

        Ko = Kf

 Before crash

       Po = m1 Vo1 + 0

       Ko = ½ m1 Vo1²

 

After the crash

       Pf = m1 Vif + Vvf

       Kf = ½ m1 V1f² + ½ m2 V2f²

 

      m1 V1o = m1 V1f + m2 V2f           (1)

      m1 V1o² = m1 V1f² + m2 V2f²      (2)

We see that we have two equations with two unknowns, so the system is solvable,  we substitute in 1 and 2

   

     m1 (V1o -V1f) = m2 V2f      (3)

      m1 (V1o² - V1f²) = m2 V2f²

Let's use the relationship      (a + b) (a-b) = a² -b²

     m1 (V1o + V1f) (V1o -V1f) = m2 V2f²

We divide  with 3 and simplify

      (V1o + V1f) = V2f      (4)

Substitute in 3, group and clear

         m1 (V1o - V1f) = m2 (V1o + V1f)

         m1 V1o - m2 V1o = m2 V1f + m1 V1f

         V1f (m1 -m2) = V1o (m1 + m2)

         V1f = V1o  (m1-m2 / m1+m2)

We substitute in (4) and group

         V2f = V1o + (m1-m2 / m1 + m2) V1o

         V2f = V1o [1+ + (m1-m2 / m1 + m2)]

         V2f = V1o (2m1 / (m1+m2)

We calculate with the given values

         V1f = 0.81 (239-513 / 239 + 513)

         V1f = 0.81 (-274/752)

         V1f = - 0.295 m/s

The negative sign indicates that the planned one moves in the opposite direction to the initial one

         V2f = 0.81 [2 239 / (239 + 513)]

        V2f = 0.81 [0.636]

        V2f = 0.515 m / s

Now we analyze in the second glider movement only, we calculate the energy and since there is no friction,

         Eo = Ef

Where Eo is the mechanical energy at the lowest point and Ef is the mechanical energy at the highest point

         Eo = K = ½ m2 vf2²

         Ef = U = m2 g Y

   

         ½ m2 v2f² = m2 g Y

         Y = V2f² / 2g

         Y = 0.515²/2 9.8

         Y = 0.0147 m

At this height the planned stops, let's use trigonometry to find the height at the end of the track of the track

         tan θ = Y / x

         Y = x tan θ

The crash occurs in the middle of the track whereby x = 1.2 m

        Y = 1.2 tan 0.7

        Y = 0.147 m

As the two quantities are equal in glider it stops just when it reaches the end of the runway

7 0
3 years ago
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