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kap26 [50]
3 years ago
8

Connecting many devices in a single socket does not affect the flow of current in a

Physics
1 answer:
eimsori [14]3 years ago
7 0

<u>A</u><u>n</u><u>s</u><u>w</u><u>e</u><u>r</u><u>:</u><u>-</u><em> </em><em>F</em><em>a</em><em>l</em><em>s</em><em>e</em>

<u>E</u><u>x</u><u>p</u><u>l</u><u>a</u><u>n</u><u>a</u><u>t</u><u>i</u><u>o</u><u>n</u><u>:</u><u>-</u>

<em>False because it can leads to overloading and further to short circut.</em>

<h2><em><u>H</u></em><em><u>o</u></em><em><u>p</u></em><em><u>e</u></em><em><u> </u></em><em><u>I</u></em><em><u>t</u></em><em><u> </u></em><em><u>W</u></em><em><u>i</u></em><em><u>l</u></em><em><u>l</u></em><em><u> </u></em><em><u>H</u></em><em><u>e</u></em><em><u>l</u></em><em><u>p</u></em><em><u> </u></em><em><u>Y</u></em><em><u>o</u></em><em><u>u</u></em><em><u>!</u></em></h2>
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¿Qué resistencia debe ser conectada en paralelo con una de 20 Ω para hacer una
ValentinkaMS [17]

Answer:

60 Ω

Explanation:

R(com) = 15 Ω

1/R(com) = 1/R1 + 1/R2 + 1/R3 ..... + 1/Rn

1/15 = 1/20 + 1/R2

1/R2 = 1/15 - 1/20

1/R2 = (4 - 3) / 60

1/R2 = 1/60

R2 = 60 Ω

así, la combinada de resistencia necesaria es 60 Ω

5 0
3 years ago
A 1.45 kg falcon catches a 0.515 kg dove from behind in midair. What is their velocity after impact if the falcon's velocity is
True [87]

Answer:

Their velocity after the impact is 20.85 m/s.                            

Explanation:

Given that,

Mass of falcon, m_1=1.45\ kg

Mass of dove, m_2=0.515\ kg

Initial speed of the falcon, u_1=26.5\ m/s

Initial speed of the dove, u_2=4.95\ m/s

We need to find the final velocity after the impact. When the falcon catches the dove, it will becomes the case of inelastic collision. The conservation of momentum will be :

m_1u_1+m_2u_2=(m_1+m_2)V\\\\V=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}\\\\V=\dfrac{1.45\times 26.5+0.515\times 4.95}{(1.45+0.515)}\\\\V=20.85\ m/s

So, their velocity after the impact is 20.85 m/s.                          

8 0
3 years ago
The mohorovicic discontinuity marks the change in rock density between the core and the mantle. mantle and crust. asthenosphere
lions [1.4K]

The transition zone between the crust and mantle is called as mohorovicic discontinuity.

The mohorovicic discontinuity was discovered by Andrija Mohorovicic in the year of 1909. The Moho lies at the depth of 35km beneath the continents and 8km beneath the oceanic crust. The Moho separates both the continental crust and the oceanic crust from underlying mantle. The Moho lies almost entirely within the lithosphere, only beneath the Mid Oceanic Ridge does it define lithosphere and asthenosphere boundary. Immediate above the Moho velocity of the P wave is 6km/sec and just below the Moho it becomes 8km/sec. Moho is characterised by up to 500km thick.

The Mohorovicic Discontinuity marks the lower limit of Earth's crust. It occurs at an average depth of about 8 kilometers beneath the ocean basins and 32 kilometers beneath continental surfaces. Mohorovicic was able to use his discovery to study thickness variations of the crust.

Mohorovicic discontinuity is the layer which is between the earth's crust and mantle. It's density ranges from 3.3 to 3.7.

learn more about mohorovicic discontinuity from here: brainly.com/question/2887207

#SPJ4

3 0
1 year ago
If the speed of an object doubles its kinetic energy does what
Helga [31]
If<span> the </span>speed<span> of the </span>object<span> becomes </span>double<span>, </span>its kinetic energy<span> changes to four times the initial </span>kinetic energy<span>. Hope it help!</span>
6 0
3 years ago
Read 2 more answers
You deposit a thin film of magnesium difluoride on a glass lens (n &gt;1.60), reducing the reflection of yellow light, at normal
JulsSmile [24]

To solve this problem it is necessary to apply the concepts related to destructive interference.

The concept refers to an overlap of two or more waves of identical or similar frequency that, when interfering, creates a new pattern of waves of lower intensity

By definition destructive interference is given by

2Mt = (n+\frac{1}{2})\lambda

Where,

\lambda= wavelength

n=integer (1,2,3,4,5,6...etc)

t = thickness

M= Index of refrqaction

For minimum thickness to satisfy this condition n will be minimum there,

n=0

Therefore

2Mt = (0+\frac{1}{2})\lambda

Solving to find M,

M = \frac{\lambda}{4t}

M = \frac{585nm}{4*106nm}

M = 1.38

Therefore the correct answer is B. 1.38

6 0
4 years ago
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