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2 years ago

An object possessing an excess of 6.0x10^6 electrons has a net charge of

1 answer:

7 0

A single electron has a charge of
$e=-1.6 \cdot 10^{19}C$
Therefore, if we have an excess of $N=6.0\cdot10^6$ electrons, the total net charge will be the product between the charge of a single electron and the total number of electrons in excess:
$Q=Ne=(6.0\cdot10^6)(-1.6 \cdot 10^{-19}C)=9.6 \cdot 10^{-13}C$

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2 years ago

A constant amount of charge passes through a conductor. How is current affected if the same amount of charge passes in less time

baherus [9]

Answer:

B. The current increases.

Explanation:

As we know that rate of flow of charge through the conductor is known as electric current

So we have

$i = \frac{q}{t}$

here we know that charge Q flowing through the conductor is constant while the time in which it passes through it is decreased

so we can say that the ratio of charge and time will increase

so here we have

$i = increased$

So correct answer will be

B. The current increases.

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2 years ago

A uniform disk, a thin hoop, and a uniform sphere, all with the same mass and same outer radius, are each free to rotate about a

olga nikolaevna [1]

Answer:

4 hoop, disk, sphere

Explanation:

Because

We are given data that

Hoop, disk, sphere have Same mass and radius

So let

And Initial angular velocity, = 0

The Force on each be F

And Time = t

Also let

Radius of each = r

So let's find the inertia shall we!!

I1 = m r² /2

= 0.5 mr² the his is for dis

I2 = m r² for hoop

And

Moment of inertia of sphere wiil be

I3 = (2/5) mr²

= 0.4 mr²

ωf = ωi + α t

= 0 + ( τ / I ) t

= ( F r / I ) t

So we can see that

ωf is inversely proportional to moment of inertia.

And so we take the

Order of I ( least to greatest ) :

I3 (sphere) , I1 (disk) , I2 (hoop) , ,

Order of ωf: ( least to greatest)

That of omega xf is the reverse of inertial so

hoop, disk, sphere

Option - 4

5 0

2 years ago

An object has the acceleration graph shown in (Figure 1). Its velocity at t=0s is vx=2.0m/s. Draw the object's velocity graph fo

timama [110]

Answer:

Explanation:

We may notice that change in velocity can be obtained by calculating areas between acceleration lines and horizontal axis ("Time"). Mathematically, we know that:

$v_{b}-v_{a} = \int\limits^{t_{b}}_{t_{a}} {a(t)} \, dt$

$v_{b} = v_{a}+ \int\limits^{t_{b}}_{t_{a}} {a(t)} \, dt$

Where:

$v_{a}$ , $v_{b}$ - Initial and final velocities, measured in meters per second.

$t_{a}$ , $t_{b}$ - Initial and final times, measured in seconds.

$a(t)$ - Acceleration, measured in meters per square second.

Acceleration is the slope of velocity, as we know that each line is an horizontal one, then, velocity curves are lines with slopes different of zero. There are three region where velocities should be found:

Region I (t = 0 s to t = 4 s)

$v_{4} = 2\,\frac{m}{s} +\int\limits^{4\,s}_{0\,s} {\left(-2\,\frac{m}{s^{2}} \right)} \, dt$

$v_{4} = 2\,\frac{m}{s}+\left(-2\,\frac{m}{s^{2}} \right) \cdot (4\,s-0\,s)$

$v_{4} = -6\,\frac{m}{s}$

Region II (t = 4 s to t = 6 s)

$v_{6} = -6\,\frac{m}{s} +\int\limits^{6\,s}_{4\,s} {\left(1\,\frac{m}{s^{2}} \right)} \, dt$

$v_{6} = -6\,\frac{m}{s}+\left(1\,\frac{m}{s^{2}} \right) \cdot (6\,s-4\,s)$

$v_{6} = -4\,\frac{m}{s}$

Region III (t = 6 s to t = 10 s)

$v_{10} = -4\,\frac{m}{s} +\int\limits^{10\,s}_{6\,s} {\left(2\,\frac{m}{s^{2}} \right)} \, dt$

$v_{10} = -4\,\frac{m}{s}+\left(2\,\frac{m}{s^{2}} \right) \cdot (10\,s-6\,s)$

$v_{10} = 4\,\frac{m}{s}$

Finally, we draw the object's velocity graph as follows. Graphic is attached below.

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3 years ago

Is the 3 in the molecule a coefficient, subscript, or element? 3H₂O₂

JulijaS [17]

<span>coefficient

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2 years ago