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
PtichkaEL [24]
3 years ago
8

Two metal spheres of different sizes are charged such that the electric potential is the same at the surface of each. Sphere A h

as a radius 2 times that of sphere B. Let QA and QB be the charges on the two spheres, and let EA and EB be the electric-field magnitudes at the surfaces of the two spheres.What is the ratio QB/QA?What is the ratio EB/EA?
Physics
1 answer:
vodka [1.7K]3 years ago
8 0

Answer:

Explanation:

Given

Radius of A is twice of B i.e.

R_A=2R_B

Also Potential of both sphere is same

V_A=V_B

V=\frac{kQ}{R}

thus

k\frac{Q_A}{R_A}=K\frac{Q_B}{R_B}

\frac{Q_A}{Q_B}=\frac{R_A}{R_B}

\frac{Q_A}{Q_B}=\frac{2}{1}=2

\frac{Q_B}{Q_A}=\frac{1}{2}

(b)Ratio of \frac{E_B}{E_A}

Electric Field is given by E=\frac{kQ}{R^2}

thus E_A=\frac{kQ_A}{R_A^2}----1

E_B=\frac{kQ_B}{R_B^2}----2

Divide 2 by 1

\frac{E_B}{E_A}=\frac{Q_B}{R_B^2}\times \frac{R_A^2}{Q_A}

\frac{E_B}{E_A}=\frac{1}{2}\times 4=2

You might be interested in
Carbon-14 has a half-life of 5,730 years. if the age of an object older than 50,000 years cannot be determined by radiocarbon da
Aleonysh [2.5K]
<h3><u>Answer;</u></h3>

Carbon-14 levels in a sample are undetectable after approximately 9 half lives

<h3><u>Explanation;</u></h3>
  • <em><u>The half life of Carbon-14 is 5,730 years . Half life is the time taken by a radioactive material to decay by half of its original mass.  Therefore, it  would take a time of 5730 years for a sample of 100 g of carbon-14 to decay to 50 grams</u></em>
  • <em><u>A period of 50,000 years, is equivalent to; </u></em>

<em><u>  50,000÷5,730 </u></em>

<em><u>= 8.73 half lives</u></em>

<em>Which is approximately equal to 9 half lives.</em>

  • Therefore, if the age of an object older than 50,000 years cannot be determined by radiocarbon dating, then <em><u>Carbon-14 levels in a sample are undetectable after approximately 9 half lives</u></em>.
6 0
3 years ago
Read 2 more answers
The Sun delivers an average power of 1.575 W/m2 to the top of Neptune's atmosphere. Find the magnitudes of max and max for the e
FrozenT [24]

Answer:

1.1486813808\times 10^{-7}\ T

34.46 V/m

Explanation:

\mu_0 = Vacuum permeability = 4\pi \times 10^{-7}\ H/m

c = Speed of light = 3\times 10^8\ m/s

I = Intensity = 1.575 W/m²

The maximum magnetic field intensity is given by

B_m=\sqrt{\dfrac{2\mu_0I}{c}}\\\Rightarrow B_m=\sqrt{\dfrac{2\times 4\pi \times 10^{-7}\times 1.575}{3\times 10^8}}\\\Rightarrow B_m=1.1486813808\times 10^{-7}\ T

The magnetic field intensity is 1.1486813808\times 10^{-7}\ T

The maximum electric field intensity is given by

E_m=B_m\times c\\\Rightarrow E_m=1.1486813808\times 10^{-7}\times 3\times 10^8\\\Rightarrow E_m=34.46\ V/m

The  electric field intensity is 34.46 V/m

8 0
3 years ago
Find the weight of an object with mass 80 kg on the moon ( g = 1.6 m/s^2)
SpyIntel [72]

Answer:

80kg = 133 Newtons I'm pretty sure this is right.

5 0
2 years ago
An object with a mass of 2.0 kg is accelerated at 5.0 m/s/s. the net force acting on the mass is
Vitek1552 [10]
First we need to know the equation:
F= Mass times acceleration
F = 2.0 kg times 5.0 m/s^2 
multiply them to get the net force!
F= 10 N 
N is newton 
Hope this heps
7 0
3 years ago
Read 2 more answers
Help with 5 and the question below. BRAINLIEST FOR THE CORRECT ANSWER! Urgent Physics help!
krok68 [10]

Sound level at distance of 15 m is given as 20 dB

so intensity at this distance is given as

L = 10 Log\frac{I}{I_0}

20 = 10 Log{I}{10^{-12}}

I_1 = 10^{-10}W/m^2

now if we move closer to some some distance the sound level is now 50 dB

now the intensity is given as

L = 10 Log\frac{I}{I_0}

50 = 10Log\frac{I}{10^{-12}}

I_2 = 10^{-7}W/m^2

now we know that

\frac{I_1}{I_2} = \frac{r_2^2}{r_1^2}

\frac{10^{-10}}{10^{-7}} = \frac{r_2^2}{15^2}

r_2 = 47 cm

so now the distance from friend must be 47 cm

8 0
3 years ago
Other questions:
  • A particle of mass 7.00 10-8 kg and charge 7.8-µc is traveling due east. it enters perpendicularly a magnetic field whose magnit
    12·1 answer
  • A bug zapper consists of two metal plates connected to a high-voltage power supply. The voltage between the plates is set to giv
    13·1 answer
  • A 4 kg body is traveling at 3 m/s with no external force acting on it. At a certain instant an internal explosion occurs, splitt
    8·1 answer
  • How many atoms are in something determines its?
    6·1 answer
  • A 8.0 n force acts on a 0.70-kg object for 0.50 seconds. by how much does the object's momentum change (in kg-m/s)? (never inclu
    6·1 answer
  • You hold a 0.125 kg glider A and a 0.500 kg glider B at rest on an air track with a compressed spring of negligible mass between
    13·1 answer
  • When light with a wavelength of 209 nm is incident on a certain metal surface, electrons are ejected with a maximum kinetic ener
    10·1 answer
  • Compare the practical uses and limitations of nuclear fission and fusion. Include in your answer a detailed description of the t
    14·1 answer
  • Suppose an ice hockey puck strikes a hockey stick that lies flat on the ice and is free to move in any direction. Which
    14·1 answer
  • A cylinder of volume 3 liter has Argon gas initially at 300 K, and 1.00 atm pressure. The piston compresses the gas to a new pre
    11·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!