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2 years ago

a force 2.4E2 N exists between a positive charge of 8E-5 C and a positive charge of 3E-5 C. What distance separates the charges?

Physics

1 answer:

Verdich [7]2 years ago

6 0

The distance between the two charges is $0.3 m$

Explanation:

The electrostatic force between two charged objects is given by Coulomb's law:

$F=k\frac{q_1 q_2}{r^2}$

where:

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

$q_1, q_2$ are the charges of the two objects

r is the separation between the two charges

In this problem, we are given the following:

$q_1 = 8\cdot 10^{-5} C$

$q_2 = 3\cdot 10^{-5} C$

$F=2.4\cdot 10^2 N$

Therefore, we can rearrange the equation to solve for r, the distance between the two charges:

$r=\sqrt{\frac{kq_1 q_2}{F}}=\sqrt{\frac{(8.99\cdot 10^9)(8\cdot 10^{-5})(3\cdot 10^{-5})}{2.4\cdot 10^2}}=0.3 m$

Learn more about electrostatic force:

brainly.com/question/8960054

brainly.com/question/4273177

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A car of mass M = 1000 kg traveling at 50.0 km/hour enters a banked turn covered with ice. The road is banked at an angle θ , an

kupik [55]

The radius of the curved road at the given condition is 54.1 m.

The given parameters:

mass of the car, m = 1000 kg
speed of the car, v = 50 km/h = 13.89 m/s
banking angle, θ = 20⁰

The normal force on the car due to banking curve is calculated as follows;

$Fcos(\theta) = mg$

The horizontal force on the car due to the banking curve is calculated as follows;

$Fsin(\theta) = \frac{mv^2}{r}$

Divide the second equation by the first;

$\frac{Fsin(\theta)}{Fcos(\theta) } = \frac{mv^2}{rmg} \\\\tan(\theta) = \frac{v^2}{rg} \\\\r = \frac{v^2}{g \times tan(\theta)} \\\\r = \frac{13.89^2}{9.8 \times tan(20)} \\\\r = 54.1 \ m$

Thus, the radius of the curved road at the given condition is 54.1 m.

Learn more about banking angle here: brainly.com/question/8169892

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2 years ago

Infrared radiation from young stars can pass through the heavy dust clouds surrounding them, allowing astronomers here on Earth

Semmy [17]

Answer:

$\lambda=6.83\times 10^{-5}\ m$

Explanation:

Given that,

An infrared telescope is tuned to detect infrared radiation with a frequency of 4.39 THz.

We know that,

1 THz = 10¹² Hz

So,

f = 4.39 × 10¹² Hz

We need to find the wavelength of the infrared radiation.

We know that,

$\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{4.39\times 10^{12}}\\\\=6.83\times 10^{-5}\ m$

So, the wavelength of the infrared radiation is $6.83\times 10^{-5}\ m$ .

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2 years ago

What is the strength of the electric field inside the membrane just before the action potential?

gtnhenbr [62]

Answer:

Incomplete question, check attachment for the graph needed to solve problem.

A 8.1nm........

Explanation:

Electric Field is given as

E=V/d

Where V is voltage

And d is the distance apart

E is the electric field

The voltage V just before action of potential is -70mV,

The value d=8.1nm

d=8.1×10^-9m

E=V/d

E=-70×10^-3/8.1×10^-9

E=-8.6×10^6 N/C

Then the magnitude of the electric field is 8.6×10^6N/C

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3 years ago

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Ivahew [28]

Answer:

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