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2 years ago

5

Which term names a large collection of stars, often billions, grouped together in the universe? A. supernova B. nebula C. solar

system D. galaxy

2 answers:

larisa [96]2 years ago

8 0

It would be D. Galaxy

garik1379 [7]2 years ago

5 0

D. Galaxy hope this helped=)

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The ratio of the lift force L to the drag force D for the simple airfoil is L/D = 10. If the lift force on a short section of th

Yuri [45]

Answer:

$R=64.32\ lb\\\\\theta=84.3\°$

Explanation:

Given:

Ratio of lift force to drag force is, $\frac{L}{D}=10$

Lift force on a short section is, $L=64\ lb$

Magnitude of resultant, $R= ?$

The angle of 'R' with the horizontal is, $\theta=?$

We know that, lift force and drag are at right angles to each other. So, the resultant can be computed using Pythagoras theorem.

For calculating 'R', we first compute drag force 'D'.

As per question:

$\frac{L}{D}=10\\\\D=\frac{L}{10}=\frac{64\ lb}{10}=6.4\ lb$

Now, the magnitude of resultant 'R' is given as:

$R=\sqrt{L^2+D^2}$

Plug in the given values and solve for 'R'. This gives,

$R=\sqrt{64^2+6.4^2}\\\\R=\sqrt{4096+40.96}\\\\R=\sqrt{4136.96}=64.32\ lb$

Therefore, the magnitude of the resultant force 'R' is 64.32 lb.

Now, the angle $\theta$ is given as the arctan of the ratio of the lift and drag force.

Therefore,

$\theta=\tan^{-1}(L/D)\\\\\theta=\tan^{-1}(10)\\\\\theta=84.3\°$

Therefore, the angle made with the horizontal is 84.3°.

6 0

3 years ago

The closest stars are 4 light years away from us. How far away must you be from a 854 kHz radio station with power 50.0 kW for t

Lubov Fominskaja [6]

Answer:

The distance from the radio station is 0.28 light years away.

Solution:

As per the question:

Distance, d = 4 ly

Frequency of the radio station, f = 854 kHz = $854\times 10^{3}\ Hz$

Power, P = 50 kW = $50\times 10^{3}\ W$

$I_{p} = 1\ photon/s/m^{2}$

Now,

From the relation:

P = nhf

where

n = no. of photons/second

h = Planck's constant

f = frequency

Now,

$n = \frac{P}{hf} = \frac{50\times 10^{3}}{6.626\times 10^{- 34}\times 854\times 10^{3}} = 8.836\times 10^{31}\ photons/s$

Area of the sphere, A = $4\pi r^{2}$

Now,

Suppose the distance from the radio station be 'r' from where the intensity of the photon is $1\ photon/s/m^{2}$

$I_{p} = \frac{n}{A} = \frac{n}{4\pi r^{2}}$

$1 = \frac{8.836\times 10^{31}}{4\pi r^{2}}$

$r = \sqrt{\frac{8.836\times 10^{31}}{4\pi}} = 2.65\times 10^{15}\ m$

Now,

We know that:

1 ly = $9.4607\times 10^{15}\ m$

Thus

$r = \frac{2.65\times 10^{15}}{9.4607\times 10^{15}} = 0.28\ ly$

5 0

2 years ago

paula finds four unlabeled containers of clear, odorless liquid in her laboratory storage. one of the containers holds water. sh

aliya0001 [1]

The answer is c ............

5 0

2 years ago

Read 2 more answers

The escape speed from an object is v2 = 2GM/R, where M is the mass of the object, R is the object's starting radius, and G is th

Rom4ik [11]

Answer:

Approximate escape speed = 45.3 km/s

Explanation:

Escape speed

$v=\sqrt{\frac{2GM}{R}}$

Here we have

Gravitational constant = G = 6.67 × 10⁻¹¹ m³ kg⁻¹ s⁻²

R = 1 AU = 1.496 × 10¹¹ m

M = 2.3 × 10³⁰ kg

Substituting

$v=\sqrt{\frac{2\times 6.67\times 10^{-11}\times 2.3\times 10^{30}}{1.496\times 10^{11}}}=4.53\times 10^4m/s=45.3km/s$

Approximate escape speed = 45.3 km/s

6 0

3 years ago

How much charge is on each plate of a 3.00-Î¼F capacitor when it is connected toa 15.0-V battery? b) If this same capacitor is c

Sauron [17]

Answer:

(a) 45 micro coulomb

(b) 6 micro Coulomb

Explanation:

C = 3 micro Farad = 3 x 10^-6 Farad

V = 15 V

(a) q = C x V

where, q be the charge.

q = 3 x 10^-6 x 15 = 45 x 10^-6 C = 45 micro coulomb

(b)

V = 2 V, C = 3 micro Farad = 3 x 10^-6 Farad

q = C x V

where, q be the charge.

q = 3 x 10^-6 x 2 = 6 x 10^-6 C = 6 micro coulomb

6 0

3 years ago