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Bogdan [553]
3 years ago
14

Calculate the net force and the acceleration on the block

Physics
1 answer:
Darina [25.2K]3 years ago
7 0

Answer:

Net Force = 10N

Acceleration = 2m/s^2

Explanation:

calculate the net force and the acceleration on the block

Net force on the block F = mass * acceleration

Net force acting in the positive direction = 4N + 6N = 10N

Mass = 5kg

According to newton's second law;

a = F/m

a = 10N/5

a = 2m/s^2

hence the acceleration on the block is 2m/s^2

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The diameter of a copper wire is thought to be approximately 0.3mm. Which instrument should be used to obtain a more accurate me
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Answer:

C

Explanation:

The correct answer should be the micrometer.

<u>The micrometer is an instrument that is capable of measuring the component of an object to very high precision. It is used to obtain accurate measurements in engineering.</u>

<em>In the case of the copper wire, the diameter of 0.3 mm was an approximation and a more accurate measure of the diameter would be obtained by using the micrometer.</em>

The correct option is C.

4 0
3 years ago
How much work is done on the electron by the electric field of the sheet as the electron moves from its initial position to a po
miss Akunina [59]

Incomplete question as the charge density is missing so I assume charge density of 3.90×10^−12 C/m².The complete one is here.

An electron is released from rest at a distance of 0 m  from a large insulating sheet of charge that has uniform surface charge density 3.90×10^−12 C/m² .  How much work is done on the electron by the electric field of the sheet as the electron moves from its initial position to a point 3.00×10−2 m from the sheet?

Answer:

Work=1.06×10⁻²¹J

Explanation:

Given Data

Permittivity of free space ε₀=8.85×10⁻¹²c²/N.m²

Charge density σ=3.90×10⁻¹² C/m²

The electron moves a distance d=3.00×10⁻²m

Electron charge e=-1.6×10⁻¹⁹C

To find

Work done

Solution

The electric field due is sheet is given as

E=σ/2ε₀

E=\frac{3.90*10^{-12}C/m^{2}  }{2(8.85*10^{-12}C^{2} /N.m^{2} )}\\ E=0.22V/m

Now we need to find force on electron

F=eE\\F=(1.6*10^{-19}C )(0.22V/m)\\F=3.525*10^{-20}N

Now for Work done on the electron

W=F*d\\W=(3.525*10^{-20} N)(3.00*10^{-2}m)\\W=1.06*10^{-21}J

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