1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
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.
You might be interested in
We can model a pine tree in the forest as having a compact canopy at the top of a relatively bare trunk. Wind blowing on the top
frez [133]

We need to consider for this exercise the concept Drag Force and Torque. The equation of Drag force is

F_D = c_D A \frac{\rho V^2}{2}

Where,

F_D = Drag Force

c_D = Drag coefficient

A = Area

\rho= Density

V = Velocity

Our values are given by,

c_D = 0.5 (That is proper of a cone-shape)

A = 9m^2

\rho = 1.2Kg/m^3

V = 6.5m/s

Part A ) Replacing our values,

F_D = 0.5*9*\frac{1.2*6.5^2}{2}

F_D = 114.075N

Part B ) To find the torque we apply the equation as follow,

\tau = F*d

\tau = (114.075N)(7)

\tau = 798.525N.m

3 0
3 years ago
How long will it take to travel 200 m traveling at 10 m/s? Follow example below.
polet [3.4K]

Answer:

variables  - d = 200m \: v = 10 m {s}^{ - 1}  \\ equation \:   - v =  \frac{d}{t}  \\ 10 = \frac{200}{t}  \\ cross \: multiply \\ 10t = 200 \\  \frac{10t}{10}  =   \frac{200}{10}  \\ t = 20s

It will take 10 seconds to travel 200m at a speed of 10m/s

Explanation:

HOPE THAT THIS IS HELPFUL.

HAVE A GREAT DAY.

4 0
2 years ago
Read 2 more answers
A pump is used to transport water to a higher reservoir. if the water temperature is 15ºc, determine the lowest pressure that ca
Reika [66]
Normally, the water pressure inside a pump is higher than the vapor pressure: in this case, at the interface between the liquid and the vapor, molecules from the liquid escapes into vapour form. Instead, when the pressure of the water becomes lower than the vapour pressure, molecules of vapour can go inside the water forming bubbles: this phenomenon is called cavitation. 

So, cavitation occurs when the pressure of the water becomes lower than the vapour pressure. In our problem, vapour pressure at 15^{\circ} is 1.706 kPa. Therefore, the lowest pressure that can exist in the pump without cavitation, at this temperature, is exactly this value: 1.706 kPa.
7 0
3 years ago
A pendulum has a length of 2 m and a 30 kg mass hanging on the end. What is the period of the
anastassius [24]

Answer:

T = 2.83701481512 seconds

Explanation:

Hi!

The formula that you will want to use to solve this question is:

T = 2\pi *\sqrt{\frac{L}{g}  }

T--> period

L --> length of the pendulum

g --> acceleration due to gravity (9.8m/s^2)

since we know that the mass of the bob at the end of the pendulum does not affect the period of the pendulum, we can go ahead and ignore that bit of information (unless, of course, the weight causes the pendulum to stretch)

so now we can plug in our given info into the formula above and solve!

T = 2*pi * sqrt(2/9.8)

T = 2.83701481512 seconds

*Note*

- I used 3.14 to pi, if you need to use a different value for pi (a longer version, etc) your answer will be slightly different

I hope this helped!

7 0
3 years ago
Two identical small metal spheres with q1 > 0 and |q1| > |q2| attract each other with a force of magnitude 72.1 mN when se
Brrunno [24]

1) +2.19\mu C

The electrostatic force between two charges is given by

F=k\frac{q_1 q_2}{r^2} (1)

where

k is the Coulomb's constant

q1, q2 are the two charges

r is the separation between the charges

When the two spheres are brought in contact with each other, the charge equally redistribute among the two spheres, such that each sphere will have a charge of

\frac{Q}{2}

where Q is the total charge between the two spheres.

So we can actually rewrite the force as

F=k\frac{(\frac{Q}{2})^2}{r^2}

And since we know that

r = 1.41 m (distance between the spheres)

F= 21.63 mN = 0.02163 N

(the sign is positive since the charges repel each other)

We can solve the equation for Q:

Q=2\sqrt{\frac{Fr^2}{k}}=2\sqrt{\frac{(0.02163)(1.41)^2}{8.98755\cdot 10^9}}}=4.37\cdot 10^{-6} C

So, the final charge on the sphere on the right is

\frac{Q}{2}=\frac{4.37\cdot 10^{-6} C}{2}=2.19\cdot 10^{-6}C=+2.19\mu C

2) q_1 = +6.70 \mu C

Now we know the total charge initially on the two spheres. Moreover, at the beginning we know that

F = -72.1 mN = -0.0721 N (we put a negative sign since the force is attractive, which means that the charges have opposite signs)

r = 1.41 m is the separation between the charges

And also,

q_2 = Q-q_1

So we can rewrite eq.(1) as

F=k \frac{q_1 (Q-q_1)}{r^2}

Solving for q1,

Fr^2=k (q_1 Q-q_1^2})\\kq_1^2 -kQ q_1 +Fr^2 = 0

Since Q=4.37\cdot 10^{-6} C, we can substituting all numbers into the equation:

8.98755\cdot 10^9 q_1^2 -3.93\cdot 10^4 q_1 -0.141 = 0

which gives two solutions:

q_1 = 6.70\cdot 10^{-6} C\\q_2 = -2.34\cdot 10^{-6} C

Which correspond to the values of the two charges. Therefore, the initial charge q1 on the first sphere is

q_1 = +6.70 \mu C

8 0
3 years ago
Other questions:
  • Most comets come from an area called the oort cloud true or false
    6·2 answers
  • A 2.0 kg mass weighs 10 Newtons on planet X. what is the acceleration due to gravity on planet X? Show the work.
    8·1 answer
  • Which listed property of alternating current is the MOST LIKELY reason it was chosen over direct current to provide electricity
    8·2 answers
  • In an undeveloped country, the rural areas will often be off the grid, with no access to consistent electricity. For a household
    12·1 answer
  • Chapter 16, Problem 63. A person is standing in a room at 18 ◦C. The exposed surface area and skin temperature of the person are
    7·1 answer
  • The distance between the fixed point of a Celsius thermometer is20cm. what is the temperature when the mercury level is 4.5cm ab
    12·1 answer
  • What is the fault-current circuit breaker? Describe its function.
    10·2 answers
  • Assuming things about someone based on your experiences with similar people you have encountered is called
    6·1 answer
  • The transfer of energy when particles of a fluid move from one place to another is called
    14·1 answer
  • What do active solar heat systems have that passive systems do not have?.
    11·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!