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DiKsa [7]
3 years ago
13

If you have charged an electroscope by contact with a positively charged object, describe how you could use it to determine the

charge of other objects. Specifically, what would the leaves of the electroscope do if other charged objects were brought near its knob?
Physics
1 answer:
maxonik [38]3 years ago
7 0

Explanation:

  • When we have a pre-charged electroscope that is charged using a positive object then we observe that the leaves of the electroscope are spread wide in the diverging position due to the like charges of positive nature on the two leaves.
  • As we know that the charges can only travel through the conductors hence the assembly of the electroscope from the contact knob to the leaves is made of conductor. So the charges spread uniformly on this conductor.
  • Now we can use this as a tester of the other charges. When we bring a negatively charged object near the knob of the electroscope then the opposite charges i.e. positive charges accumulate near the knob due to attraction and the leaves of the electroscope go unstrained and get closer naturally.
  • Contrarily when a positive charge is brought near the knob of the electroscope then the the uniformly distributed charges move away from the knob and accumulate at the farthest point and more deviation is observed on the leaves.
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Answer:

(a) Time t = 16.46 sec

(b) Time t =13.466 sec

(c) Deceleration = 2.677m/sec^2

Explanation:

(a) As the train starts from rest its initial velocity u = 0 m/sec

Acceleration a=1.35m/sec^2

Final speed v = 80 km/hr

80km/hr=\frac{80\times 1000}{3600sec}=22.22m/sec

From first equation of motion v =u+at

So t=\frac{v-u}{a}=\frac{22.22-0}{1.35}=16.46 sec

(b) Now initial speed u = 22.22 m/sec

As finally train comes to rest so final speed v=0 m/sec

Deceleration a=1.65m/sec^2

So t=\frac{v-u}{a}=\frac{0-22.22}{-1.65}=13.466 sec

(c) We have given that initial velocity = 80 km/hr = 22.22 m/sec

Final velocity v = 0 m/sec

Time t = 8.30 sec

So acceleration is given by

a=\frac{v-u}{t}=\frac{0-22.22}{8.3}=-2.6771m/sec^2

As acceleration is negative so it is a deceleration

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Statistical time division multiplexing does not require the capacity of the circuit to be equal to the sum of the combined circu
aleksklad [387]

Answer:

The answer is True

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2 years ago
A glass flask whose volume is 1000 cm^3 at a temperature of 1.00°C is completely filled with mercury at the same temperature. W
Marat540 [252]

Answer:

the coefficient of volume expansion of the glass is \mathbf{  ( \beta_{glass} )= 1.333 *10^{-5} / K}

Explanation:

Given that:

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temperature of the glass flask and mercury= 1.00° C

After heat is applied ; the final temperature = 52.00° C

Temperature change ΔT = 52.00° C - 1.00° C = 51.00° C

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the coefficient of volume expansion of mercury is 1.80 × 10⁻⁴ / K

The increase in the volume of the mercury =  10⁻³ m³ ×  51.00 × 1.80 × 10⁻⁴

The increase in the volume of the mercury = 9.18*10^{-6} \ m^3

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Now; the mercury overflow = Increase in volume of the mercury - increase in the volume of the flask

the mercury overflow = (9.18*10^{-6}  -  51.00* \beta_{glass}*10^{-3})\ m^3

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8.50*10^{-6} - 9.18*10^{-6} = ( -51.00* \beta_{glass}* 10^{-3} )\ m^3

-6.8*10^{-7} =  ( -51.00* \beta_{glass}* 10^{-3} )\ m^3

6.8*10^{-7} =  ( 51.00* \beta_{glass}* 10^{-3} )\ m^3

\dfrac{6.8*10^{-7}}{51.00 * 10^{-3}}=  ( \beta_{glass} )

\mathbf{  ( \beta_{glass} )= 1.333 *10^{-5} / K}

Thus; the coefficient of volume expansion of the glass is \mathbf{  ( \beta_{glass} )= 1.333 *10^{-5} / K}

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