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BARSIC [14]
3 years ago
12

the electric current in a wire is 1.5 A. How many electrons flow past a given point in a time of 2 s?​

Physics
1 answer:
34kurt3 years ago
6 0

Answer:

The quantity of electrons that flows past a given point is 3.0 C.

Explanation:

An electric current (I) is the ratio of the quantity of charges (Q) that flows through a point to the time taken (t).

i.e            I = \frac{Q}{t}

It is measured in Ampere's by the use of an ammeter in the laboratory. The quantity of charge that flow through a given point is measured in Coulombs, while time is measured in seconds.

Given that;     I = 1.5A and t = 2s, find Q.

                          Q = It

                             = 1.5 × 2

                             = 3.0 C

The quantity of electrons that flows past a given point is 3.0 C.

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If you think about it, irregular galaxies dont really have a describable shape, as you can tell by the name. So B. 
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2 years ago
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What does the voltage-current graph above show about the relationship between voltage and current, and how do these properties a
zubka84 [21]

The answer is as voltage increases current increases and therefore resistance would remain constant

4 0
3 years ago
A huge tank of glycerine with a density of 1.260 g/cm3 is vertically stationed on a platform which is 15 m above the ground. The
EleoNora [17]

Answer:

The tank is losing 4.976*10^{-4}  m^3/s

v_g = 19.81 \ m/s

Explanation:

According to the Bernoulli’s equation:

P_1 + 1 \frac{1}{2} \rho v_1^2 + \rho gh_1 = P_2 +  \frac{1}{2}  \rho v_2^2 + \rho gh_2

We are being informed that both the tank and the hole is being exposed to air :

∴ P₁ = P₂

Also as the tank is voluminous ; we take the initial volume  v_1 ≅ 0 ;

then v_2 can be determined as:\sqrt{[2g (h_1- h_2)]

h₁ = 5 + 15 = 20 m;

h₂ = 15 m

v_2 = \sqrt{[2*9.81*(20 - 15)]

v_2 = \sqrt{[2*9.81*(5)]

v_2= 9.9 \ m/s  as it leaves the hole at the base.

radius r = d/2  = 4/2 = 2.0 mm

(a) From the law of continuity; its equation can be expressed as:

J = A_1v_2

J = πr²v_2    

J =\pi *(2*10^{-3})^{2}*9.9

J =1.244*10^{-4}  m^3/s

b)

How fast is the water from the hole moving just as it reaches the ground?

In order to determine that; we use the relation of the velocity from the equation of motion which says:

v² = u² + 2gh ₂

v² = 9.9² + 2×9.81×15

v² = 392.31

The velocity of how fast the water from the hole is moving just as it reaches the ground is : v_g = \sqrt{392.31}

v_g = 19.81 \ m/s

4 0
2 years ago
What is the period (in hours) of a satellite circling Mars 100 km above the planet's surface? The mass of Mars is 6.42 × 1023 kg
scZoUnD [109]

To solve this problem it is necessary to apply the concepts related to the Centrifugal Force and the Gravitational Force. Since there is balance on the body these two Forces will be equal, mathematically they can be expressed as

F_c = F_g

\frac{mv^2}{r} = \frac{GmM}{r^2}

Where,

m = Mass

G =Gravitational Universal Constant

M = Mass of the Planet

r = Distance/Radius

Re-arrange to find the velocity we have,

v^2 = \frac{GM}{r}

At the same time we know that the period is equivalent in terms of the linear velocity to,

T = \frac{2\pi}{\frac{v}{r}}

v = \frac{2\pi r}{T}

If our values are that the radius of mars is 3400 km and the distance above the planet is 100km more, i.e, 3500km we have,

v^2 = \frac{GM}{r}

( \frac{2\pi r}{T})^2 =  \frac{GM}{r}

T = \sqrt{\frac{4\pi^2 r^3}{GM}}

Replacing we have,

T = \sqrt{\frac{4\pi^2 (3500*10^3)^3}{(6.67430*10^{-11})(6.42*10^23)}}

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Therefore the correct answer is C.

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astra-53 [7]

Answer:

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matter is "anything that has mass and volume (occupies space)

     <em>Gases have mass. The space between gas particles is empty. Gases can be formed as products in chemical reactions. Gas particles can form bonds between them under certain conditions</em>

<em>       Gases have volume which isn't fixed </em>(no fixed volume)<em> and no fixed shape. Gases expand to fill the space available. They can also be compressed into a very small space.</em>

Explanation:

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