answer.
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
Sedbober [7]
2 years ago
11

three point charges are positioned on the x-axis 64 uc at x=ocm , 80uc at x=25cm, and -160 uc at x=50 cm. what is the magnitude

of the electro static force acting on the 64-uc charge
Physics
1 answer:
LekaFEV [45]2 years ago
6 0

Answer:

The magnitude of the electrostatic force is F = 368.64 N

Explanation:

Given data,

The magnitude of charge, q₁ = 64 μ C

The magnitude of charge, q₂ = 80 μ C

The magnitude of charge, q₃ = -160 μ C

The charge q₁ is at the origin,

The position of charge q₁ in x-axis, x₁ = 0 cm

The position of charge q₂ in x-axis, x₂ = 25 cm

                                                               = 0.25 m

The position of charge q₃ in x-axis, x₃ = 50 cm

                                                               = 0.5 m

The magnitude of the electrostatic force experienced by the charge a is given by the formula,

The force acting on charge 'q₁' due to 'q₂' is,

                             F₁ = k  q₁ q₂ / x₂²

                                 = (9 x 10⁹ X 64 x 10⁻⁶ X 80 x 10⁻⁶) / 0.25²

                                =  + 737.28 N

The force acting on charge 'q₁' due to 'q₃' is,

                              F₂ = k q₁ q₃ / x₃²

                                   = [9 x 10⁹ X 64 x 10⁻⁶ X (-160 x 10⁻⁶)] / 0.5²

                                    = - 368.64 N

The net electrostatic force acting on the charge, q₁ is,

                              F = F₁ + F₂

                                  = + 737.28 N + (- 368.64 N)

                                   = 368.64 N

Hence, the magnitude of the electrostatic force is F = 368.64 N

You might be interested in
Water floats on a liquid called "carbon tetrachloride." The two liquids do not mix. A light ray passing from water into carbon t
Oduvanchick [21]

Answer:

1.26

Explanation:

index  \: of  \: refraction \:  =  \:  \ \frac{ \sin(angle \: of \: incidence) }{ \sin(angle \: of \: refraction) }

\frac{ \sin(45.0) }{ \sin(40.1) }

= 1.26

5 0
2 years ago
A skateboarder rides swiftly up the edge of a bowl-shaped surface and leaps into the air. While in the air, the skateboarder fli
Katen [24]

Answer:

Option(a) is the correct answer to the given question .

Explanation:

The main objective of the angular momentum is evaluating however much the rotational movement as well as the angular velocity in the entity does have.The angular momentum is measured in terms of kgm^{2 }\  / s.

  • In the given question the skateboarder rides quickly up the bottom of a bowl-shaped surface and climb into the air.it means it is rotational movement also it is not touching anything so it is angular momentum.
  • All the other option is incorrect because it is not follows the given scenario
3 0
2 years ago
A box of mass 3.1kg slides down a rough vertical wall. The gravitational force on the box is 30N . When the box reaches a speed
Nostrana [21]

Answer:

Explanation:

a) The net force on the box in vertical direction:

Fnet=Fg−f−Fp  *sin45  °

 here Fg is the  gravitational force .f is the force of friction and , Fp is the pushing force.

Fnet=ma

ma=Fg−f−Fp  *sin45  °

​a=\frac{30-13-23*sin(45)}{3.1}

 =0.24 m/s²  

Vf =Vi +at

     =0.48+0.24*2

Vf=2.98 m/s

b)

                                Fnet=Fg−f−Fp  *sin45  °

                                        =Fg−0.516Fp−Fp  *sin45  °

                                        =30-1.273Fp

                                Fnet=0               (Because speed is constant)

                                Fp=30/1.273

                                      =23.56 N

5 0
2 years ago
Read 2 more answers
A 2.70 kg cat is sitting on a windowsill. The cat is sleeping peacefully until a dog barks at him. Startled, the cat falls from
Alchen [17]

Answer:

The speed of the cat when it hits the ground is approximately 7.586 meters per second.

Explanation:

By Principle of Energy Conservation and Work-Energy Theorem, we have that initial potential gravitational energy of the cat (U_{g}), in joules, is equal to the sum of the final translational kinetic energy (K), in joules, and work losses due to air resistance (W_{l}), in joules:

U_{g} = K +W_{l} (1)

By definition of potential gravitational energy, translational kinetic energy and work, we expand the equation presented above:

m \cdot g\cdot h = \frac{1}{2}\cdot m \cdot v^{2}+W_{l} (2)

Where:

m - Mass of the cat, in kilograms.

g - Gravitational acceleration, in meters per square second.

h - Initial height of the cat, in meters.

v - Final speed of the cat, in meters per second.

If we know that m = 2.70\,kg, g = 9.807\,\frac{m}{s^{2}}, h = 5.20\,m and W_{l} = 120\,J, then the final speed of the cat is:

v = \sqrt{\frac{2\cdot (m\cdot g\cdot h-W_{l})}{m} }

v = \sqrt{2\cdot g\cdot h-\frac{W_{l}}{m} }

v \approx 7.586\,\frac{m}{s}

The speed of the cat when it hits the ground is approximately 7.586 meters per second.

4 0
2 years ago
A supersonic nozzle is also a convergent–divergent duct, which is fed by a large reservoir at the inlet to the nozzle. In the re
Lady_Fox [76]

Answer:

155.38424 K

2.2721 kg/m³

Explanation:

P_1 = Pressure at reservoir = 10 atm

T_1 = Temperature at reservoir = 300 K

P_2 = Pressure at exit = 1 atm

T_2 = Temperature at exit

R_s = Mass-specific gas constant = 287 J/kgK

\gamma = Specific heat ratio = 1.4 for air

For isentropic flow

\frac{T_2}{T_1}=\frac{P_2}{P_1}^{\frac{\gamma-1}{\gamma}}\\\Rightarrow T_2=T_1\times \frac{P_2}{P_1}^{\frac{\gamma-1}{\gamma}}\\\Rightarrow T_2=00\times \left(\frac{1}{10}\right)^{\frac{1.4-1}{1.4}}\\\Rightarrow T_2=155.38424\ K

The temperature of the flow at the exit is 155.38424 K

From the ideal equation density is given by

\rho_2=\frac{P_2}{R_sT_2}\\\Rightarrow \rho=\frac{1\times 101325}{287\times 155.38424}\\\Rightarrow \rho=2.2721\ kg/m^3

The density of the flow at the exit is 2.2721 kg/m³

4 0
2 years ago
Other questions:
  • A 12.0-kg shell is launched at an angle of 55.0 ∘ above the horizontal with an initial speed of 150 m/s. when it is at its highe
    14·1 answer
  • The average radius of Mars is 3,397 km. If Mars completes one rotation in 24.6 hours, what is the tangential speed of objects on
    14·2 answers
  • An electrical short cuts off all power to a submersible diving vehicle when it is a distance of 28 m below the surface of the oc
    8·1 answer
  • Lilli suggests that they explore the simulation starting with varying only a single parameter in order to understand the role of
    7·1 answer
  • A 0.500-kg ball traveling horizontally on a frictionless surface approaches a very massive stone at 20.0 m/s perpendicular to wa
    5·1 answer
  • In 1923, the United States Army (there was no U.S. Air Force at that time) set a record for in-flight refueling of airplanes. Us
    13·1 answer
  • A 6.0-cm-diameter, 11-cm-long cylinder contains 100 mg of oxygen (O2) at a pressure less than 1 atm. The cap on one end of the c
    6·1 answer
  • Astronomers have discovered a new planet called "Xandar" beyond the orbit of Pluto (No, not really but I need a fake planet for
    7·1 answer
  • 13. Calculate the total heat energy in Joules needed to convert 20 g of substance X from -10°C to 70°C?
    9·1 answer
  • An object of mass m moves horizontally, increasing in speed from 0 to v in a time t. The power necessary to accelerate the objec
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!