Ask question

Ask question

All categories

3 years ago

9

A common design for a spotlight uses a lens and a mirror, combined with a gas-discharge lamp, to project a powerful parallel bea

m of light. Where are the lens and mirror positioned?

1 answer:

wel3 years ago

7 0

Answer:

Explanation:

For answer please see the image attached .

You might be interested in

What is the measurement 111.009 mm rounded off to four significant digits?

dedylja [7]

111.0 because 111.009 rounds off to 111.01, thus rounding again off to 111.0

4 0

3 years ago

A mercury barometer reads 745.0 mm on the roof of a building and 760.0 mm on the ground. Assuming a constant value of 1.29 kg/m3

Ipatiy [6.2K]

Answer:

The height of the building is 158.140 meters.

Explanation:

A barometer is system that helps measuring atmospheric pressure. Manometric pressure is the difference between total and atmospheric pressures. Manometric pressure difference is directly proportional to fluid density and height difference. That is:

$\Delta P \propto \rho \cdot \Delta h$

$\Delta P = k \cdot \rho \cdot \Delta h$

Where:

$\Delta P$ - Manometric pressure difference, measured in kilopascals.

$\rho$ - Fluid density, measured in kilograms per cubic meter.

$\Delta h$ - Height difference, measured in meters.

Now, an equivalent height difference with a different fluid can be found by eliminating manometric pressure and proportionality constant:

$\rho_{air} \cdot \Delta h_{air} = \rho_{Hg} \cdot \Delta h_{Hg}$

$\Delta h_{air} = \frac{\rho_{Hg}}{\rho_{air}} \cdot \Delta h_{Hg}$

Where:

$\Delta h_{air}$ - Height difference of the air column, measured in meters.

$\Delta h_{Hg}$ - Height difference of the mercury column, measured in meters.

$\rho_{air}$ - Density of air, measured in kilograms per cubic meter.

$\rho_{Hg}$ - Density of mercury, measured in kilograms per cubic meter.

If $\Delta h_{Hg} = 0.015\,m$ , $\rho_{air} = 1.29\,\frac{kg}{m^{3}}$ and $\rho_{Hg} = 13600\,\frac{kg}{m^{3}}$ , the height difference of the air column is:

$\Delta h_{air} = \frac{13600\,\frac{kg}{m^{3}} }{1.29\,\frac{kg}{m^{3}} }\times (0.015\,m)$

$\Delta h_{air} = 158.140\,m$

The height of the building is 158.140 meters.

5 0

3 years ago

Read 2 more answers

If A.B=A.C<br>does this imply that B must be equal to c ? <br>I need answer with example please

Leno4ka [110]

Answer:

Yes

Explanation:

$Consider\ three\ variables\ a,b\ and\ c.\\If,\\ab=ac\\a*b=a*c\\We\ may\ now\ divide\ the\ LHS\ and\ RHS\ by\ a,\\Hence,\\\frac{a*b}{a} =\frac{a*c}{a}\\Hence,\\b=c$

$Ex: \\Consider\ a=5, b=6.\\Hence\ if,\\a*b=a*c\\5*6=5*c\\Hence,\\30=5c\\Hence, c=\frac{30}{5}\\c=6\\As\ b\ also\ equals\ to\ 6,\\b=c$

5 0

3 years ago

A record turntable is rotating at 33 rev/min. A watermelon seed is on the turntable 4.4 cm from the axis of rotation. (a) Calcul

romanna [79]

Answer:

a) a =0.53 m/s²

b) μ=0.054

c) μ = 0.068

Explanation:

a) If we assume that the turntable is rotating at a constant speed, the only force acting on the seed parallel to the surface, which keeps it from following a straight line trajectory, is the centripetal force.

So, we can apply Newton's 2nd Law to the seed in this way:

Fnet = m*a = m*ac = m*ω²*r

We have the value of the angular speed, ω, in rev/min, so it is advisable to convert it to rad/sec, as follows:

ω = 33 rev/min*(1 min/60 sec)*(2*π rad/ 1 rev) = 11/10*π rad/sec

So, replacing in (1), we can solve for ac, as follows:

ac = ω²*r = (11/10)²*π²*0.044 m = 0.53 m/s²

b) Now, the centripetal force that we found above, is not a new type of force, it must be a force that explains the behavior of the seed.

As the seed does not slip, the only force acting on it parallel to the surface, is the static friction force, which has a maximum value, as follows:

Ff = μ*N

As there is no movement in the vertical direction, this means that the normal force must be equal and opposite to Fg, so we can write the expression for Ff as follows:

Ff = μ*m*g

Now, this force is no other than centripetal force, so we can write this equation:

Ff = Fc ⇒ μ*m*g = m*ac

⇒ μ*g = ac

Solving for μ:

μ = ac/g = 0.53 m/s² / 9.8 m/s² = 0.054

c) During the acceleration period, added to the centripetal acceleration, as the angular speed is not constant, we will have also an angular acceleration, γ , which we can get as follows:

γ = Δω/Δt = (11/10)*π / 0.37 s = 9.34 rad/sec²

By definition of angular acceleration, there exists a fixed relationship between the angular acceleration and the tangential acceleration (same as the one between angular and tangential speed), as follows:

at = γ*r = 9.34 rad/sec²*0.044 m = 0.41 m/s

When the turntable has reached to its maximum angular velocity, it will have also the maximum value of the centripetal acceleration, which we have just found out.

So, the magnitude of the total acceleration (at the moment of maximum acceleration) as they are perpendicular each other) , is given by the following expression:

a = √(ac)²+(at)² = 0.67 m/s²

Now, as friction force opposes to the relative movement between surfaces (the seed and the turntable), it shall be larger than the product of the mass times the total acceleration, acting along the same action line, so we can say:

Ffmin = μ*m*g = m*a

⇒ μmin = a/g = 0.67 m/s²/9.8 m/s² = 0.068

5 0

4 years ago

Near the equator, rising air is associated with a pressure zone known as the _____.

umka2103 [35]

Equatorial low , i think,..

6 0

4 years ago

Read 2 more answers