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2 years ago

a charged partocle produces an electric field with a magnitude of 2.0 N/C at a point that is 50cm away from the particle

Physics

1 answer:

zheka24 [161]2 years ago

4 0

The charge on the particle is 5.6 × 10⁻¹¹ C.

<h3>Calculation:</h3>

The magnitude of an electric field produced by a charge is given by:

E = q/ 4πε₀r²

where,

E = electric field

q = charge

r = distance

1/4πε₀ = 8.99 × 10⁹ Nm²/C²

Given,

E = 2.0 N/C

r = 50 cm = 0.5 m

To find,

q =?

Put the values in the above equation:

E = q/ 4πε₀r²

q = E (4πε₀r²)

q = 2.0 × (0.50²)/ 8.99 × 10⁹

q = 5.6 × 10⁻¹¹ C

Therefore, the particle has a charge of 5.6 × 10⁻¹¹ C.

<h3>What is an electric field?</h3>

The physical field that surrounds each electric charge and acts to either attract or repel all other charges in the field is known as an electric field. Electric charges or magnetic fields with different amplitudes are the sources of electric fields.

I understand the question you are looking for is this:

A charged particle produces an electric field with a magnitude of 2.0 N/C at a point that is 50 cm away from the particle. What is the magnitude of the particle's charge?

Learn more about electric field here:

brainly.com/question/14857134

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3 years ago

1 A 75-g ball is projected from a height of 1.6 m with a horizontal velocity of 2 m/s and bounces from a 400-g smooth plate supp

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Answer with explanation:

We are given that

Mass of ball, $m_1=$ 75 g= $\frac{75}{1000}=0075kg$

1 kg=1000 g

Height, $h_1=1.6 m$

$h_2=0.6 m$

Horizontal velocity, $v_x=2 m/s$

Mass of plate $m_2=400 g=\frac{400}{1000}=0.4 kg$

a.Initial velocity of plate, $u_2=0$

Velocity before impact= $u_1=\sqrt{2gh_1}=\sqrt{2\times 9.8\times 1.6}=5.6m/s$

Where $g=9.8 m/s^2$

Velocity after impact, $v_1=\sqrt{2gh_2}=\sqrt{2\times 9.8\times 0.6}=3.4m/s$

According to law of conservation of momentum

$m_1u_1+m_2u_1=-m_1v_1+m_2v_2$

Substitute the values

$0.075\times 5.6+0=-0.075\times 3.4+0.4v_2$

$0.4v_2=0.075\times 5.6+0.075\times 3.4$

$v_2=\frac{0.075\times 5.6+0.075\times 3.4}{0.4}=1.69 m/s$

Velocity of plate=1.69 m/s

b.Initial energy= $\frac{1}{2}m_1v^2_x+m_1gh_1=\frac{1}{2}(0.075)(2^2)+0.075\times 9.8\times 1.6=1.326 J$

Final energy= $\frac{1}{2}m_1v^2_x+m_1gh_2+\frac{1}{2}m_2v^2_2$

Final energy= $\frac{1}{2}(0.075)(2^2)+0.075\times 9.8\times 0.6+\frac{1}{2}(0.4)(1.69)^2=1.162 J$

Energy lost due to compact=Initial energy-final energy=1.326-1.162=0.164 J

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3 years ago

a specimen of oil having an initial volume of 5000cm³ is subjected to a pressure of 10⁴N/m² and the volume decreases by 0.20cm³.

inessss [21]

Answer:

B = 2.5 10⁸ Pa

Explanation:

The volume modulus is defined by

B = $- \frac{P}{ \frac{\Delta V}{V} }$

The negative fate is for the module to be positive since the volume change is negative

It is not necessary to reduce the volumes to the SI system, since they are both in the same units

B = $- \frac{10^4}{ \frac{-0.20}{5000} }$ = $\frac{10^4}{4 \ 10^{-5} }$

B = 2.5 10⁸ Pa

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3 years ago

Two astronauts (each with mass 100 kg) are drifting together through space. They are connected to each other by a rope 5 m in le

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Answer:

1000 kgm²/s, 400 J

1000 kgm²/s, 1000 J

600 J

Explanation:

m = Mass of astronauts = 100 kg

d = Diameter

r = Radius = $\frac{d}{2}$

v = Velocity of astronauts = 2 m/s

Angular momentum of the system is given by

$L=mvr+mvr\\\Rightarrow L=2mvr\\\Rightarrow L=2\times 100\times 2\times 2.5\\\Rightarrow L=1000\ kgm^2/s$

The angular momentum of the system is 1000 kgm²/s

Rotational energy is given by

$K=I\omega^2\\\Rightarrow K=\frac{1}{2}(mr^2)\left(\frac{v}{r}\right)^2\\\Rightarrow K=mv^2\\\Rightarrow K=100\times 2^2\\\Rightarrow K=400\ J$

The rotational energy of the system is 400 J

There no external toque present so the initial and final angular momentum will be equal to the initial angular momentum 1000 kgm²/s

$L_i=L_f\\\Rightarrow 2mv_ir_i=2mv_fr_f\\\Rightarrow v_f=\frac{v_ir_i}{r_f}\\\Rightarrow v_f=\frac{2\times 2.5}{0.5}\\\Rightarrow v_f=10\ m/s$

Energy

$E_2=mv_f^2\\\Rightarrow E_2=100\times 10\\\Rightarrow E_2=1000\ J$

The new energy will be 1000 J

Work done will be the change in the kinetic energy

$W=E_2-E\\\Rightarrow W=1000-400\\\Rightarrow W=600\ J$

The work done is 600 J

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3 years ago