Ask question

Ask question

All categories

2 years ago

10

Suppose that the mirror described in Part A is initially at rest a distance R away from the sun. What is the critical value of a

rea density for the mirror at which the radiation pressure exactly cancels out the gravitational attraction from the sun?

1 answer:

Sati [7]2 years ago

5 0

Ans;

6.25m²/Kd

Explanation: see attached file

You might be interested in

A bookshelf is 2m long, with supports at its end (p and q). A book weighing 10N is placed in the middle of the shelf. what are t

mel-nik [20]

If the book is placed in the middle, the forces acting on <em>p</em> and <em>q</em> is 5N. If the book is moved 50 cm from <em>q</em>, the forces at <em>p</em> and <em>q</em> can be solved by doing a moment balance

<u>With </u><u><em>p</em></u><u> as the pivot;</u>

Fq (2 m) = 10 N (0.5 m)

Fq = 2.5 N

Fp = 10 N - 2.5 N = 7.5 N

5 0

1 year ago

On a warm summer day (31 ∘c), it takes 4.60 s for an echo to return from a cliff across a lake. on a winter day, it takes 5.00 s

xenn [34]

The question is missing, but I guess the problem is asking for the distance between the cliff and the source of the sound.

First of all, we need to calculate the speed of sound at temperature of $T=31^{\circ}C$ :
$v=(331+0.60 T) m/s = (331+0.6 \cdot 31) m/s = 349.6 m/s$

The sound wave travels from the original point to the cliff and then back again to the original point in a total time of t=4.60 s. If we call L the distance between the source of the sound wave and the cliff, we can write (since the wave moves by uniform motion):
$v= \frac{2L}{t}$
where v is the speed of the wave, 2L is the total distance covered by the wave and t is the time. Re-arranging the formula, we can calculate L, the distance between the source of the sound and the cliff:
$L= \frac{vt}{2}= \frac{(349.6 m/s)/4.60 s)}{2}= 804.1 m$

6 0

2 years ago

Stu wanted to calculate the resistance of a light bulb

Nonamiya [84]

Answer:

yes

Explanation:

He used the formula R = VI and obtained an

answer of 2

3 0

2 years ago

Read 2 more answers

Recent findings in astrophysics suggest that the observable universe can be modeled as a sphere of radius R = 13.7 × 109 light-y

-BARSIC- [3]

Answer:

$3.7\times 10^{51})$ kg

Explanation:

$R$ = radius of the sphere modeled as universe = $13\times 10^{25}$ m

Volume of sphere is given as

$V = \frac{4\pi R^{3}}{3}$

$V = \frac{4(3.14) (13\times 10^{25})^{3}}{3}$

$V = 9.2\times 10^{78}$ m³

$\rho$ = average total mass density of universe = $1\times 10^{-26}$ kg/m³

$m$ = Total mass of the universe = ?

We know that mass is the product of volume and density, hence

$m = \rho V$

$m = (1\times 10^{-26}) (9.2\times 10^{78})$

$m = 9.2\times 10^{52}$ kg

$M$ = mass of "ordinary" matter = ?

mass of "ordinary" matter is only about 4% of total mass, hence

$M = (0.04) m$

$M = (0.04)(9.2\times 10^{52})$

$M = 3.7\times 10^{51}$ kg

6 0

1 year ago

The different in size of each of the rope's pullers, correspond to a difference in the magnitude of the applied force, such that

olga55 [171]

Answer:

F = - 50 N

Hence, the magnitude of resultant force is 50 N and its direction is leftwards.

Explanation:

The magnitude of the resultant force is always equal to the sum of all forces. While, the direction of resultant force will be equal to the direction of the force with greater magnitude:

$Resultant\ Force = F = F_{1} - F_{2}$

considering right direction to be positive:

F₁ = Force applied on right rope = 150 N

F₂ = Force applied on left rope = 200 N

Therefore, the resultant force can be found by using these values in equation:

$F = 150\ N - 200\ N$

<u>F = - 50 N</u>

<u>Hence, the magnitude of resultant force is 50 N and its direction is leftwards.</u>

5 0

2 years ago