Ask question

Ask question

All categories

3 years ago

10

Space satellites in the inner solar system (such as those in orbit around Earth) are often powered by solar panels. Satellites t

hat travel into the outer solar system, however, are usually powered by the decay of nuclear isotopes – a more expensive and dangerous technology. Why can’t they use solar panels as well?
a. At greater distances, sunlight is more likely to destructively interfere.
b. At greater distances, sunlight is more likely to constructively interfere.
c. At greater distances from the sun, the frequency of sunlight will be too low.
d. At greater distances from the sun, the intensity of sunlight will be too low.

1 answer:

erastova [34]3 years ago

7 0

Answer:

D

Explanation:

See attached file

You might be interested in

Richard is driving home to visit his parents. 135 mi of the trip are on the interstate highway where the speed limit is 65 mph .

Elis [28]

<h2>Answer:</h2>

He saves 13.2 minutes

<h2>Explanation:</h2>

Hey! The question is incomplete, but it can be found on the internet. The question is:

How many minutes did he save?

Let's call:

$t_{1}:Time \ at \ speed \ 65mph \\ \\ t_{2}:Time \ at \ speed \ 73mph \\ \\ v_{1}=65mph \\ \\ v_{2}=73mph$

We know that the 135 miles are on the interstate highway where the speed limit is 65 mph. From this, we can calculate the time it takes to drive on this highway. Assuming Richard maintains constant the speed:

$v=\frac{d}{t} \\ \\ d:distance \\ \\ t:time \\ \\ v:velocity \\ \\ t_{1}=\frac{d}{v_{1}} \\ \\ t=\frac{135}{65} \\ \\ t_{1}=2.07 \ hours$

Today he is running late and decides to take his chances by driving at 73 mph, so the new time it takes to take the trip is:

$t_{2}=\frac{135}{73} \\ \\ t_{2}=1.85 \ hours$

So he saves the time $t_{s}$ :

$t_{s}=t_{1}-t_{2}=2.07-1.85=0.22 \ hours$

In minutes:

$t_{s}=0.22h\left(\frac{60min}{1h}\right) \\ \\ \boxed{t_{s}=13.2min}$

5 0

2 years ago

I’m giving points and an easy answer for y’all hope your having a great day ( and I think some people birthdays are in the 31 so

WARRIOR [948]

Difference between an inherited trait and an acquired trait is that an inherited trait is a trait inherited from a family, such as your eye color or hair color, while a acquired trait is something that you learned yourself such as learning how to read.

8 0

2 years ago

Read 2 more answers

When you enter a toll road, your ticket is stamped 1:00 p.m. When you leave, after traveling 55 miles, your ticket is stamped 2:

Viefleur [7K]

27.5 because of you divide the 55miles with the time you get your velocity which is the speed.

4 0

2 years ago

Read 2 more answers

A large, cylindrical water tank with diameter 3.60 m is on a platform 2.00 m above the ground. The vertical tank is open to the

zysi [14]

To solve this problem it is necessary to apply the concepts related to the geometry of a cylindrical tank and its respective definition.

The volume of a tank is given by

$V = \frac{\pi d^2}{4}h$

Where

d = Diameter

h = Height

Considering that there are two stages, let's define the initial and final volume as,

$V_0 = \frac{\pi d^2}{4}H$

$V_f = \frac{\pi d^2}{4}h$

We know as well by definition that

$1gal = 3.785*10^{-3}m^3$

Then we have for the statement that

$V_f = V_0 -1gal$

$V_f = V_0 - 3.785*10^{-3}$

Replacing the previous data

$\frac{\pi d^2}{4}h = \frac{\pi d^2}{4}H- 3.785*10^{-3}$

$\frac{\pi (3.6)^2}{4}h = \frac{\pi (3.6)^2}{4}(2)- 3.785*10^{-3}$

Solving to get h,

$h = 1.99963m$

Therefore the change is

$\Delta h = H-h$

$\Delta h = 2- 1.99963$

$\Delta h = 3.7*10^{-4}m=0.37mm$

Therefore te change in the height of the water in the tank is 0.37mm

4 0

2 years ago

~~~NEED HELP ASAP~~~<br>Please solve each section and show all work for each section.

anastassius [24]

Explanation:

<u>Forces</u><u> </u><u>on</u><u> </u><u>Block</u><u> </u><u>A</u><u>:</u>

Let the x-axis be (+) towards the right and y-axis be (+) in the upward direction. We can write the net forces on mass $m_A$ as

$x:\:\:(F_{net})_x = f_N - T = -m_Aa\:\:\:\:\:\:\:(1)$

$y:\:\:(F_{net})_y = N - m_Ag = 0 \:\:\:\:\:\:\:\:\:(2)$

Substituting (2) into (1), we get

$\mu_km_Ag - T = -m_Aa \:\:\:\:\:\:\:\:\:(3)$

where $f_N= \mu_kN$ , the frictional force on $m_A.$ Set this aside for now and let's look at the forces on $m_B$

<u>Forces</u><u> </u><u>on</u><u> </u><u>Block</u><u> </u><u>B</u><u>:</u>

Let the x-axis be (+) up along the inclined plane. We can write the forces on $m_B$ as

$x:\:\:(F_{net})_x = T - m_B\sin30= -m_Ba\:\:\:\:\:\:\:(4)$

$y:\:\:(F_{net})_y = N - m_Bg\cos30 = 0 \:\:\:\:\:\:\:\:\:(5)$

From (5), we can solve for <em>N</em> as

$N = m_B\cos30 \:\:\:\:\:\:\:\:\:(6)$

Set (6) aside for now. We will use this expression later. From (3), we can see that the tension<em> </em><em>T</em><em> </em> is given by

$T = m_A( \mu_kg + a)\:\:\:\:\:\:\:\:\:(7)$

Substituting (7) into (4) we get

$m_A(\mu_kg + a) - m_Bg\sin 30 = -m_Ba$

Collecting similar terms together, we get

$(m_A + m_B)a = m_Bg\sin30 - \mu_km_Ag$

or

$a = \left[ \dfrac{m_B\sin30 - \mu_km_A}{(m_A + m_B)} \right]g\:\:\:\:\:\:\:\:\:(8)$

Putting in the numbers, we find that $a = 1.4\:\text{m/s}$ . To find the tension <em>T</em>, put the value for the acceleration into (7) and we'll get $T = 21.3\:\text{N}$ . To find the force exerted by the inclined plane on block B, put the numbers into (6) and you'll get $N = 50.9\:\text{N}$

8 0

2 years ago