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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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The diagram shows organisms a diver observed during an ocean dive.
Nina [5.8K]

The diver most likely refers to the ocean's surface when describing the squid's location. Option A is correct.

<h3>What is the height?</h3>

The vertical distance between the object's top and the bottom is defined as height. It is measured in centimeters, inches, meters, and other units.

The organism is shown as;

Seaweed = - 20 meters

Clownfish  =  - 23 meters

Squid = - 44 meters.

The given data is a reference from the surface of the ocean. The negative sign in the data shows that the given height is below the ocean surface.

The diver most likely uses the ocean surface as a reference point to describe the position of the squid.

Hence, option A is correct.

To learn more about the height, refer to the link;

brainly.com/question/10726356

#SPJ1

5 0
2 years ago
How long does it take to raise the temperature of the air in a good-sized living room (3.00m×5.00m×8.00m) by 10.0∘C? Note that t
tekilochka [14]

Answer : The time required is, 16.1 minutes.

Explanation :

First we have to calculate the amount of heat required to increase the temperature is:

Q=mC\Delta T\\\\Q=\rho VC\Delta T

(m=\rho V)

where,

Q = amount of heat required = ?

m = mass

\rho = density of air = 1.20kg/m^3

V = volume of air

C = specific heat of air = 1006J/kg^oC

\Delta T = change in temperature = 10.0^oC

Now put all the given values in above formula, we get:

Q=\rho VC\Delta T

Q=(1.20kg/m^3)\times (3.00m\times 5.00m\times 8.00m)\times (1006J/kg^oC)\times (10.0^oC)

Q=1.449\times 10^6J

Now we have to calculate the time required.

Formula used :

t=\frac{Q}{P}

where,

t = time required = ?

Q = amount of heat required = 1.449\times 10^6J

P = power = 1500 W

Now put all the given values in above formula, we get:

t=\frac{1.449\times 10^6J}{1500W}

t=966s\times \frac{1min}{60s}=16.1min

Thus, the time required is, 16.1 minutes.

5 0
2 years ago
Please help with vectors (will give BRAINLIEST answer)
zhenek [66]

just analyze it in this way:

20cos30*=10( radical 3 )

20sin30*=10

7 0
3 years ago
How did ALS impact Stephen Hawking?
olga nikolaevna [1]

Answer:

Professor Hawking had just turned 21 when he was diagnosed with a very rare slow-progressing form of ALS, a form of motor neurone disease (MND). He was at the end of his time at Oxford when he started to notice early signs of his disease. He was getting more clumsy and fell over several times without knowing why.

Explanation:

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6 0
2 years ago
Not in book
umka2103 [35]

Answer:

x=2.4365\ m

and

x=-1.4365\ m

Explanation:

Given:

  • first charge, q_1=5\times 10^{-3}\ C
  • second charge, q_2=3\times 10^{-3}\ C
  • position of first charge, x_1=-2\ m
  • position of second charge, x_2=-1\ m

Now since there are only 2 charges and of the same sign so they repel each other. This repulsion will be zero at some point on the line joining the charges.

<u>Now, according to the condition, electric field will be zero where the effects of field due to both the charges is equal.</u>

E_1=E_2

  • since first charge is greater than the second charge so we may get a point to the right of the second charge and the distance between the two charges is 1 meter.

\frac{1}{4\pi.\epsilon_0} \frac{q_1}{(r+1)^2} =\frac{1}{4\pi.\epsilon_0} \frac{q_2}{(r)^2}

\frac{5\times 10^{-3}}{(r+1)^2} = \frac{3\times 10^{-3}}{(r)^2}

3(r^2+1+2r)=5r^2

2r^2-6r-3=0

r=3.4365 \&\ r=-0.4365

Since we have assumed that the we may get a point to the right of second charge so we calculate with respect to the origin.

x=-1+3.4365=2.4365\ m

and

x=-1-0.4365=-1.4365\ m

6 0
2 years ago
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