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exis [7]
3 years ago
5

In addition to telling you which programs are currently running, what other information does the task bar display?

Physics
2 answers:
Papessa [141]3 years ago
8 0

Answer:

The answer is A. The current time

Explanation:

The task bar, in addition to the programs that are running, shows the current date, along with the time, also the language in which the computer is operated, the internet connection, in addition to hidden icons and if it is a laptop, also shows the battery charge of the computer.

uysha [10]3 years ago
4 0
The Task bar is the bar that(For the most part) is on the bottom of most computers.

The answer is A - The Current Time
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kolezko [41]
Electric current in a (circuit) is due to a (positive charge flow or just flow) produced by (sources such as batteries, solar cells, electric machines). It’s opposed by (resistance).. and I don’t understand the large and small.. but it’s a start. Hopefully I helped
4 0
3 years ago
The electric field on the surface of an irregularly shaped conductor varies from 74.0 kN/C to 14.0 kN/C.
mrs_skeptik [129]

Answer:

(a). The local surface charge density at the point on the surface where the radius of curvature of the surface is greatest is 123.9 nC/m².

(b). The local surface charge density at the point on the surface where the radius of curvature of the surface is greatest is 654.9 nC/m².

Explanation:

Given that,

Electric field E_{1}=74.0\ kN/C

Electric field E_{2}=14.0\ kN/C

When the radius of curvature is greatest, the electric field at the surface will be smaller.

Where the radius of curvature is greatest

(a). We need to calculate the local surface charge density at the point on the surface

Using formula of charge density

\sigma=\epsilon_{0}E_{2}

Put the value into the formula

\sigma=8.85\times10^{-12}\times14\times10^{3}

\sigma=1.239\times10^{-7}\ C/m^2

\sigma=123.9\times10^{-9}\ C/m^2

\sigma=123.9\ nC/m^2

The local surface charge density at the point on the surface where the radius of curvature of the surface is greatest is 123.9 nC/m².

(b). We need to calculate the local surface charge density at the point on the surface where the radius of curvature of the surface is smallest

Using formula of charge density

\sigma=\epsilon_{0}E_{1}

Put the value into the formula

\sigma=8.85\times10^{-12}\times74\times10^{3}

\sigma=6.549\times10^{-7}\ C/m^2

\sigma=654.9\times10^{-9}\ C/m^2

\sigma=654.9\ nC/m^2

The local surface charge density at the point on the surface where the radius of curvature of the surface is greatest is 654.9 nC/m².

Hence, (a). The local surface charge density at the point on the surface where the radius of curvature of the surface is greatest is 123.9 nC/m².

(b). The local surface charge density at the point on the surface where the radius of curvature of the surface is greatest is 654.9 nC/m².

7 0
3 years ago
A fellow student with a mathematical bent tells you that the wave function of a traveling wave on a thin rope is y(x,t)= 2.30mmc
MakcuM [25]
<span>y(x,t)= 2.30mmcos[(6.98rad/m)x + (742 rad/s)t]
</span>A) Amplitude is 2.30mm<span>
B) Frequency 1/</span>2.30mm<span>
C) Wavelength is </span>6.98rad/m<span>
D) Wave Speed is </span>742 rad/s<span>
E) Direction the wave is traveling

</span>
7 0
3 years ago
What fraction of the space within the atom is occupied by the nucleus?
ValentinkaMS [17]

Answer:

Approximately 6.8 x 10⁻¹⁵

Explanation:

To be able to get this fraction, there are some things we need to know.

1. The radius of nucleus = 1.0 x 10⁻¹³ cm

2. The radius of hydrogen atom = 52.9 pm

3. Volume of sphere = V1/V2 = (R1/R2)^3

4. 1 picometer (pm) = 10^-10 cm

CHECK ATTACHMENT FOR Step by step solution to the answer

3 0
3 years ago
An electron is released from rest at the negative plate of a parallel-plate capacitor. If the distance across the plate is 2.0 m
ddd [48]

Answer:

Velocity of electron will be 1.325\times 10^6m/sec

Explanation:

We have given distance across the plate d = 2 mm =2\times 10^{-3}m

Potential difference V = 6 volt

We know that potential difference at any distance is given by

V = Ed , here V is potential difference, E is electric field and d  is distance

So E=\frac{V}{d}=\frac{6}{2\times 10^{-3}}=3000N/C

Charge on electron e=1.6\times 10^{-19}C

We know that expression of velocity is given by v=\sqrt{\frac{2qEd}{m_e}}, here q is charge on electron, E is electric field and d is distance

So v=\sqrt{\frac{2\times 1.6\times 10^{-19}\times 3000\times 3000}{9.11\times 10^{-31}}}=1.325\times 10^6m/sec

8 0
4 years ago
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