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3 years ago

How can plates move? O A, apart OB) together OC) sideways OD) in all these ways

Physics

2 answers:

noname [10]3 years ago

6 0

Answer:

The answer is D.

If I am right could you mark me as Brainliest.

nordsb [41]3 years ago

4 0

D is the correct answer

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What happens in the red blood cells in the lungs?

hichkok12 [17]

The correct answer is A hemoglobin

3 0

3 years ago

Consider a coin which is tossed straight up into the air. After it is released it moves upward, reaches its highest point and fa

avanturin [10]

Answer:

GRAVITATIONAL FORCE

Explanation:

We may have noticed that a body thrown upward in air falls back down again after attaining a particular height. The object was able to fall down back due to the effect of gravity acting on it. If there are no force of gravity acting on the body, the body will not fall back but rather disappears into the thin air.

A coin tossed upward in the air which falls back down when released is therefore under the influence of gravity i.e GRAVITATIONAL FORCE while it moves upward after it is released

3 0

3 years ago

You are working at a company that manufactures electri- cal wire. Gold is the most ductile of all metals: it can be stretched in

Alona [7]

Explanation:

We know that the relation between volume and density is as follows.

Volume = $\frac{\text{mass}}{\text{density}}$

So, V = $\frac{10^{-3}}{19.3 \times 10^{3} kg/m^{3}}$

= $5.181 \times 10^{-8} m^{3}$

Now, we will calculate the area as follows.

Area = $\frac{\text{volume}}{\text{length}}$

= $\frac{5.181 \times 10^{-8} m^{3}}{2.4 \times 10^{3}}$

= $2.15 \times 10^{-11} m^{2}$

Formula to calculate the resistance is as follows.

R = $\rho \frac{l}{A}$

= $\frac{2.44 \times 10^{-8} \times 2400}{}2.15 \times 10^{-11}}$

= $2.71 \times 10^{6} ohm$

Thus, we can conclude that the resistance of given wire is $2.71 \times 10^{6} ohm$ .

4 0

3 years ago

Three point charges are arranged along the x axis. Charge q1=-4.00nC is located at x= .250 m and q2= 2.40 nC is at the x= -.300m

Umnica [9.8K]

Answer:

q₃=5.3nC

Explanation:

First, we have to calculate the force exerted by the charges q₁ and q₂. To do this, we use the Coulomb's Law:

$F= k\frac{|q_aq_b|}{r^{2} } \\\\\\F_{13}=(9*10^{9} Nm^{2} /C^{2} )\frac{|(-4.00*10^{-9}C)q_3|}{(.250m)^{2} } =576q_3N/C\\\\F_{23}=(9*10^{9} Nm^{2} /C^{2} )\frac{|(2.40*10^{-9}C)q_3|}{(.300m)^{2} } =240q_3N/C\\$

Since we know the net force, we can use this to calculate q₃. As q₁ is at the right side of q₃ and q₁ and q₃ have opposite signs, the force F₁₃ points to the right. In a similar way, as q₂ is at the left side of q₃, and q₂ and q₃ have equal signs, the force F₂₃ points to the right. That means that the resultant net force is the sum of these two forces:

$F_{Net}=F_{13}+F_{23}\\\\4.40*10^{-9} N=576q_3N/C+240q_3N/C\\\\4.40*10^{-6} N=816q_3N/C\\\\\implies q_3=5.3*10^{-9}C=5.3nC$

In words, the value of q₃ must be 5.3nC.

7 0

3 years ago

Waves diffract the most when their wavelength is

djverab [1.8K]

<h2>Answer: about the same size of the gap or slit</h2>

Diffraction happens when a wave (mechanical or electromagnetic wave, in fact, any wave) meets an obstacle or a slit .When this occurs, the wave bends around the corners of the obstacle or passes through the opening of the slit that acts as an obstacle, forming multiple patterns with the shape of the aperture of the slit.

Note that the principal condition for the occurrence of this phenomena is that the obstacle must be comparable in size (similar size) to the size of the wavelength.

In other words, when the gap (or slit) size is larger than the wavelength, the wave passes through the gap and does not spread out much on the other side, but when the gap size is equal to the wavelength, maximum diffraction occurs.

Therefore:

<h2>Waves diffract the most when their wavelength is <u>about the same size of the gap </u></h2>

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5 0

3 years ago