Answer:
(2) The excess negative charge from the sphere spread out all over your body.
(7) After you touched it, the metal sphere was very nearly neutral.
Explanation:
Plastic pen repels magic tape so magic tape is also negatively charged . Further , magic tape repels small metal sphere that means small sphere also is negatively charged.
Now when small sphere is touched by a man insulated from ground , the charge is distributed between man and small sphere according to their capacitance .
Since human body will have greater capacitance ,it will acquire larger share of charge . Sphere being of very small size will retain very less charge and it will become almost neutral . Hence it will be attracted by charged tape .
<em>A</em> - <em>B</em> = (10<em>i</em> - 2<em>j</em> - 4<em>k</em>) - (<em>i</em> + 7<em>j</em> - <em>k</em>)
<em>A</em> - <em>B</em> = 9<em>i</em> - 9<em>j</em> - 3<em>k</em>
|<em>A</em> - <em>B</em>| = √(9² + (-9)² + (-3)²) = √189 = 3√19
Answer:
Sound intensity levels are quoted in decibels (dB) much more often than sound intensities in watts per meter squared. Decibels are the unit of choice in the scientific literature as well as in the popular media. The reasons for this choice of units are related to how we perceive sounds. How our ears perceive sound can be more accurately described by the logarithm of the intensity rather than directly to the intensity. The sound intensity level β in decibels of a sound having an intensity I in watts per meter squared is defined to be β(dB)=10log10(II0)β(dB)=10log10(II0), where I0 = 10−12 W/m2 is a reference intensity. In particular, I0 is the lowest or threshold intensity of sound a person with normal hearing can perceive at a frequency of 1000 Hz. Sound intensity level is not the same as intensity. Because β is defined in terms of a ratio, it is a unitless quantity telling you the level of the sound relative to a fixed standard (10−12 W/m2, in this case). The units of decibels (dB) are used to indicate this ratio is multiplied by 10 in its definition. The bel, upon which the decibel is based, is named for Alexander Graham Bell, the inventor of the telephone.
Table 1. Sound Intensity Levels and IntensitiesSound intensity level β (dB)Intensity I(W/m2)Example/effect01 × 10–12Threshold of hearing at 1000 Hz101 × 10–11Rustle of leaves201 × 10–10Whisper at 1 m distance301 × 10–9Quiet home401 × 10–8Average home501 × 10–7Average office, soft music601 × 10–6Normal conversation701 × 10–5Noisy office, busy traffic801 × 10–4Loud radio, classroom lecture901 × 10–3Inside a heavy truck; damage from prolonged exposure[1]1001 × 10–2Noisy factory, siren at 30 m; damage from 8 h per day exposure1101 × 10–1Damage from 30 min per day exposure1201Loud rock concert, pneumatic chipper at 2 m; threshold of pain1401 × 102Jet airplane at 30 m; severe pain, damage in seconds1601 × 104Bursting of eardrums
Answer:
a) When R is very small R << r, therefore the term R+ r will equal r and the current becomes
b) When R is very large, R >> r, therefore the term R+ r will equal R and the current becomes
Explanation:
<u>Solution :</u>
(a) We want to get the consumed power P when R is very small. The resistor in the circuit consumed the power from this battery. In this case, the current I is leaving the source at the higher-potential terminal and the energy is being delivered to the external circuit where the rate (power) of this transfer is given by equation in the next form
P=∈*I-I^2*r (1)
Where the term ∈*I is the rate at which work is done by the battery and the term I^2*r is the rate at which electrical energy is dissipated in the internal resistance of the battery. The current in the circuit depends on the internal resistance r and we can apply equation to get the current by
I=∈/R+r (2)
When R is very small R << r, therefore the term R+ r will equal r and the current becomes
I= ∈/r
Now let us plug this expression of I into equation (1) to get the consumed power
P=∈*I-I^2*r
=I(∈-I*r)
=0
The consumed power when R is very small is zero
(b) When R is very large, R >> r, therefore the term R+ r will equal R and the current becomes
I=∈/R
The dissipated power due toll could be calculated by using equation.
P=I^2*r (3)
Now let us plug the expression of I into equation (3) to get P
P=I^2*R=(∈/R)^2*R
=∈^2/R
