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4 years ago

Which part of the telescope gathers light from a faraway object and reflects it to another mirror?

Physics

1 answer:

melamori03 [73]4 years ago

5 0

Answer:

2)primary mirror

Explanation:

A reflecting telescope contains mirrors and a refracting telescope contains lenses. There are two mirrors in basic reflecting telescope. The primary mirror gathers the light from the bodies at distance and reflects it towards the secondary mirror. The secondary mirror is placed at such an angle that it then reflects the light towards the eye piece and we can observe the faraway body.

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Mutations _____.

NeTakaya

Answer:

add new genes to a gene pool

8 0

3 years ago

1) A fan is to accelerate quiescent air to a velocity of 8 m/s at a rate of 9 m3/s. Determine the minimum power that must be sup

azamat

Answer:

$\dot{W} = 339.84 W$

Explanation:

given data:

flow Q = 9 m^{3}/s

velocity = 8 m/s

density of air = 1.18 kg/m^{3}

minimum power required to supplied to the fan is equal to the POWER POTENTIAL of the kinetic energy and it is given as

$\dot{W} =\dot{m}\frac{V^{2}}{2}$

here $\dot{m}$ is mass flow rate and given as

$\dot{m} = \rho*Q$

$\dot{W} =\rho*Q\frac{V^{2}}{2}$

Putting all value to get minimum power

$\dot{W} =1.18*9*\frac{8^{2}}{2}$

$\dot{W} = 339.84 W$

7 0

3 years ago

Can you solve it descriptively . thanks

Solnce55 [7]

Answer:

$|M_y| = 170.82 \ N.mm$

Explanation:

From the diagram affixed below completes the question

Now from the diagram; We need to resolve the force at point A into (3) components ; i.e x.y. & z directions which are equivalent to $F_x \ , F_y \ , F_z$

So;

$F_x$ = positive x axis

$F_y =$ Negative y axis

$F_z$ = positive z axis

Then;

$|M_x| = F_y *27-F_z*11 = 77 ----- equation(1) \\ \\ |M_z| = F_y*4 - F_x*11 = 81 ---- equation (2) \\ \\ |M_y| = F_x *27 - F_z *4 = ? ---- equation (3)$

From equation (1); Let's make $F_y$ the subject of the formula ; then :

$F_y = \frac{77+11F_z}{27}$

Substituting the value for $F_y$ into equation (2) ; we have:

$(\frac{77+11F_z}{27})4-F_x*11=81 \\ \\ 11(\frac{7+F_z}{27} ) 4- F_x -11 =81 \\ \\ 28+4 F_z - 27F_x = \frac{81*27}{11} \\ \\ 4F_z - 27F_x = 198.82 -28 \\ \\ 4F_z - 27F_x = 170.82 \\ \\ Since \ |M_y| = 4F_z-27F_x \\ \\ Then: \\ \\ \\ |M_y| = 170.82 \ N.mm$

4 0

3 years ago

Ashley is flying a plane that has to reach a gradient of 360m/km in order to take off and not crash. Her goal is to travel from

KengaRu [80]

Answer:

She is likely to crash because her flight gradient is lesser than the flight gradient required gradient to avoid crashing

Explanation:

The given parameters are;

The required gradient of the plane Ashley is flying needs to reach in order to take off and not crash = 360 m/km

The initial elevation of the plane Ashley is flying = Sea level = 0 m

The goal Ashley intends to make = Elevation of 1000 m at 2.8 km. distance

∴ Ashley's goal = Traveling from sea level to 1000 m at 2.8 km horizontal distance

We have;

The gradient = Rate of change of elevation/(Horizontal distance)

Therefore;

The gradient of Ashley's flight = (1000 - 0)/(2.8 - 0) = 357.143 m/km

The gradient of Ashley's flight ≈ 357.143 m/km which is lesser than the required 360 m/km in order to take off and not crash, therefore, she will crash.

6 0

3 years ago

A soccer ball is kicked from the top of one building with a height of H1 = 30.2 m to another building with a height of H2 = 12.0

viktelen [127]

Hi there!

Initially, we have gravitational potential energy and kinetic energy. If we set the zero-line at H2 (12.0m), then the ball at the second building only has kinetic energy.

We also know there was work done on the ball by air resistance that decreased the ball's total energy.

Let's do a summation using the equations:
$KE = \frac{1}{2}mv^2 \\\\PE = mgh$

Our initial energy consists of both kinetic and potential energy (relative to the final height of the ball)

$E_i = \frac{1}{2}mv_i^2 + mg(H_1 - H_2)$

Our final energy, since we set the zero-line to be at H2, is just kinetic energy.

$E_f = \frac{1}{2}mv_f^2$

And:
$W_A = E_i - E_f$

The work done by air resistance is equal to the difference between the initial energy and the final energy of the soccer ball.

Therefore:
$W_A = \frac{1}{2}mv_i^2 + mg(H_1 - H_2) - \frac{1}{2}mv_f^2$

Solving for the work done by air resistance:
$W_A = \frac{1}{2}(.450)(15.1^2)+ (.450)(9.8)(30.2 - 12) - \frac{1}{2}(.450)(19.89^2)$

$W_A = \boxed{42.552 J}$

8 0

2 years ago