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sleet_krkn [62]
3 years ago
10

Do machines create less work? Example: pulley system

Physics
1 answer:
kherson [118]3 years ago
6 0

Answer:

Technically, no the work of building machines is hard but in the long run, Yes they are easier after they're built.

Explanation:

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A type of star called a red dwarf gets its name because it is a relatively small star that appears redder than the Sun when view
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C) An identical spring is pulled with a force or 75 N The elastie limit of the spring is 72N
tiny-mole [99]

Answer:

Spring cannot return to its original, since a part of its deformation is <u>plastic</u>, not <u>elastic</u>.

Explanation:

Physically speaking, stress is equal to the axial force divided by effective transversal area of spring. In addition, springs have usually a linear relationship between stress and strain in <u>elastic region</u>, since they are made of ductile materials. Axial force is directly proportional to axial stress, which is also directly proportional to axial strain.

Then, if force is greater than force associated with elastic limit of the spring, then spring cannot return to its original, since a part of its deformation is <u>plastic</u>, not <u>elastic</u>.

8 0
3 years ago
A charge of 50 µC is placed on the y axis at y = 3.0 cm and a 77-µC charge is placed on the x axis at x = 4.0 cm. If both charge
zheka24 [161]

Answer:

The acceleration of an electron is 1.2\times10^{20}\ m/s^2

Explanation:

Given that,

One Charge = 50 μC

Distance on y axis = 3.0 cm

Second charge = 77 μC

Distance on x axis = 4.0 cm

We need to calculate the force on electron due to q₁

Using formula of force

F_{1}=\dfrac{kq_{1}q}{r^2}

Here, q = charge of electron

Put the value into the formula

F_{1}=\dfrac{9\times10^{9}\times50\times10^{-6}\times1.6\times10^{-19}}{(3\times10^{-2})^2}

F_{1}=8\times10^{-11}j\ \ N

We need to calculate the force on electron due to q₂

Using formula of force

F_{2}=\dfrac{kq_{2}q}{r^2}

Here, q = charge of electron

Put the value into the formula

F_{2}=\dfrac{9\times10^{9}\times77\times10^{-6}\times1.6\times10^{-19}}{(4\times10^{-2})^2}

F_{2}=6.93\times10^{-11}i\ \ N

We need to calculate the net force

Using formula of net force

F=F_{1}+F_{2}

Put the value into the formula

F=8\times10^{-11}j+6.93\times10^{-11}i

The magnitude of the net force

F=\sqrt{(8\times10^{-11})^2+(6.93\times10^{-11})^2}

F=1.058\times10^{-10}\ N

We need to calculate the acceleration of an electron

Using formula of force

F = ma

a=\dfrac{F}{m}

Put the value into the formula

a=\dfrac{1.058\times10^{-10}}{9.1\times10^{-31}}

a=1.2\times10^{20}\ m/s^2

Hence, The acceleration of an electron is 1.2\times10^{20}\ m/s^2

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4 years ago
The magnetic field of a magnet is strongest
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The answer is C !!!!
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