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Aleksandr-060686 [28]
3 years ago
10

Ok, so I've answered this question but I really need someone to just check through it because I have this feeling that I've done

it wron

Physics
2 answers:
bonufazy [111]3 years ago
5 0
Your fill-ins for the first three lines of the table are correct.  Good work.

You have all the same answers for the last 2 lines of the table, like
it makes no difference to anything whether S2 is open or closed. 
That's not correct. It does make a difference.  If S2 is closed, then
the current would much rather go through S2 than through the bulbs
C and D because it's so much easier, and that's exactly what it does.
When S2 is closed, bulbs C and D are off. 

But I'll say again:  Except for those two middle boxes on the bottom line
of the table, everything else is correct.  That's 12 correct fill-ins for you,
and only 2 incorrect ones.  That's a score of 86% !  After posting so many
of these on Brainly this afternoon, doesn't it feel great to do your own work
for a change, and get it almost all correct ?
olga_2 [115]3 years ago
4 0
I believe you are correct
You might be interested in
A 3-m-high, 7-m-wide rectangular gate is hinged at the top edge and is restrained by a fixed ridge. Determine the hydrostatic fo
Shalnov [3]

Answer:

The Hydrostatic force is   F  =  137.2 kN

The location of pressure center is  Z  = 1.333 \ m  

Explanation:

From the question we are told that

   The height of the gate is  h =  3 \ m

     The weight of the gate is  w =  7 \  m

      The height of the water is  h_w  =  2 \ m

       The density of water is \rho_w  =  1000 \ kg/m^3

Note used h_w for height of water and height of gate immersed by water since both have the same value

The area of the gate immersed in water  is mathematically represented as

         A =  h_w  * w

substituting values

         A =  2*  7

         A =  14  \ m^2

The hydrostatic force is mathematically represented as

          F  =  \rho_w * g * h_f * A

Where

            h_f =h-  h_w

           h_f =3 -2

           h_f = 1\ m  

So  

              F  =  1000 * 9.8 * 1 * 14

            F  =  137.2 kN

The center of pressure is mathematically represented as

        Z  =  h_f + \frac{I_g}{h_f * A}

Where I_g is the moment of inertia of the gate which mathematically represented as

            I_g =  \frac{w * h_w^2}{12}

The h_w is the height of gate immersed in water

            I_g =  \frac{7  * 2^2 }{12}

             I_g = 4.667\ kg  m^2

Thus  

        Z  = 1  + \frac{4.66}{1 * 14}

        Z  = 1.333 \ m

3 0
2 years ago
Planet Nine is speculated to be on average 20 times farther away from the Sun than Neptune (on average distance from the Sun). H
saveliy_v [14]

Answer:

The distance is 55.636 billion miles, or 528.2 AU.

Explanation:

Since the distance from the Sun to Neptune is 2.7818 billion miles, the distance from the Sun to Planet Nine would be 20 times that, which is:

d=(20)(2781800000\ miles)=55636000000\ miles

or 55.636 billion miles.

Since 1 astronomical unit (AU) is 93 million miles, that distance is also:

d=(55636000000\ miles)(\frac{1AU}{93000000\ miles})=598.2\ AU

6 0
3 years ago
(a) Neil A. Armstrong was the first person to walk on the moon. The distance between the earth and the moon is . Find the time i
a_sh-v [17]

Answer:

a)<em> It took 1.28 seconds to Neil Armstrong's voice to reach the Earth via radio waves. </em>

b) <em>The minimum time that will be required for a message from Mars to reach the Earth via radio waves is 192 seconds. </em>

Explanation:

The electromagnetic spectrum is the distribution of radiation due to the different frequencies at which it radiates and its different intensitie. That radiation is formed by electromagnetic waves, which are transverse waves formed by an electric field and a magnetic field perpendicular to it.

The distribution of the radiation in the electromagnetic spectrum can also be given in wavelengths, but it is more frequent to work with it at frequencies:

  • Gamma rays
  • X-rays
  • Ultraviolet rays
  • Visible region
  • Infrared
  • Microwave
  • Radio waves.

Any radiation that belongs to electromagnetic spectrum has a speed in vacuum of 3x10^{8}m/s.  

<em>a) Find the time it took for his voice to reach the Earth via radio waves.</em>

To know the time that took for Neil Armstrong's voice to reach the Earth via radio waves, the following equation can be used:

c = \frac{d}{t}  (1)

Where v is the speed of light, d is the distance and t is the time.

Notice that t can be isolated from equation 1.

t = \frac{d}{c}  (2)

The distance from the Earth to the Moon is 3.85x10^{8} m, therefore.

t = \frac{3.85x10^{8} m}{3x10^{8}m/s}

t = 1.28s

Hence, it took 1.28 seconds to Neil Armstrong's voice to reach the Earth via radio waves.

<em>b) Determine the minimum time that will be required for a message from Mars to reach the Earth via radio waves.</em>

The distance from the Earth to the Mars at its closest approach is 5.76x10^{10}m, therefore.

t = \frac{5.76x10^{10}m}{3x10^{8}m/s}

t = 192s

Hence, the minimum time that will be required for a message from Mars to reach the Earth via radio waves is 192 seconds.

3 0
3 years ago
Review the Four Social Errors and Biases in the Highlights area on page 127 in Ch. 4 of THiNK: Critical Thinking and Logic Skill
AysviL [449]

Answer: Provided in the explanation

Explanation:

I have understood that I have been influenced by the 'affinity bias' for quite a while. It caused me to feel fascination and feel better and right about individuals who had comparable intrigue and thought designs. I sort of began to feel this is the one for me based on those likenesses yet later used to be miserable when I comprehended that those similitudes are not many and insufficient consistently. I have begun to beat this inclination by rehearsing self reflection. I have begun to think about what causes me to feel pulled in to somebody and on the off chance that I introspect that it's exclusively founded on likenesses, at that point I cause myself to get that and monitoring that helps in escaping the inclination.

I make an effort not to get into the snare of paradoxes in my own announcements yet I have been forced to bear deceptions during contentions. One deception which I have encountered most is the 'Foul play' error as there have been a great deal of occurrences when individuals began to tear down me and my family when they couldn't win on a contention regarding balanced and rationale.

Basic reasoning causes us in understanding our defects and those issues of our own which keeps us from taking better choices and furthermore forestalls us tackling issues in more successful manners. Through basic reasoning, one can reflect and recognize utilization of deceptions in the contentions and that will help him in deciphering the circumstance from an alternate perspective. He will assess the circumstance better and through appropriate induction, he will have the option to get over those issues in thinking and subsequently, have the option to take better and more powerful choices for taking care of an issue.

The most fascinating idea which I have learned is that of the utilization of reflection or explicitly self reflection to comprehend ourselves better and furthermore to display basic reasoning appropriately. Being said that, I am a still somewhat confused about the manners by which we can reflect appropriately at each circumstance in a successful manner. This disarray remains in light of the fact that most deceptions and predispositions are oblivious in nature while the basic considering aptitudes reflection is cognizant and I meander whether a cognizant ability will have the option to break down and distinguish each oblivious inclinations or whether a few guards will keep a few inclinations covered up.

4 0
3 years ago
a sphere of diameter 6•0cm is moulded into a thin uniform wire of diameter 0•2mm.calculate the length of the wire in metres​
Scilla [17]

The length of the wire is 36 m.

<u>Explanation:</u>

Given, Diameter of sphere = 6 cm

We know that, radius can be found by taking the half in the diameter value. So,

       \text { sphere radius, } R=\frac{D}{2}=\frac{6}{2}=3 \mathrm{cm}=3 \times 10^{-2} \mathrm{m}

Similarly,

      \text { wire radius, } r=\frac{0.2}{2}=0.1 \mathrm{mm}=1 \times 10^{-3} \mathrm{m}

We know the below formulas,

          \text {volume of sphere}=\frac{4}{3} \times \pi \times R^{3}

          \text {volume of wire}=\pi \times r^{2} \times l

When equating both the equations, we can find length of wire as below, where \pi=\frac{22}{7}

          \frac{4}{3} \times \pi \times R^{3}=\pi \times r^{2} \times l

         \frac{4}{3} \times \frac{22}{7} \times\left(3 \times 10^{-2}\right)^{3}=\frac{22}{7} \times\left(1 \times 10^{-3}\right)^{2} \times l

The \pi value gets cancelled as common on both sides, we get

           \frac{4}{3} \times 27 \times 10^{-6}=10^{-6} \times l

The 10^{-6} value gets cancelled as common on both sides, we get

           l=4 \times 9=36 m

7 0
3 years ago
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