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ValentinkaMS [17]

3 years ago

15

C) An identical spring is pulled with a force or 75 N The elastie limit of the spring is 72N

1 answer:

tiny-mole [99]3 years ago

8 0

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>.

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A third point charge q3 is now positioned halfway between q1 and q2. The net force on q2 now has a magnitude of F2,net = 14.413

natima [27]

Answer:

The value of charge q₃ is 40.46 μC.

Explanation:

Given that.

Magnitude of net force $F=14.413\ N$

Suppose a point charge q₁ = -3 μC is located at the origin of a co-ordinate system. Another point charge q₂ = 7.7 μC is located along the x-axis at a distance x₂ = 8.2 cm from q₁. Charge q₂ is displaced a distance y₂ = 3.1 cm in the positive y-direction.

We need to calculate the distance

Using Pythagorean theorem

$r=\sqrt{x_{2}^2+y_{2}^2}$

Put the value into the formula

$r=\sqrt{(8.2\times10^{-2})^2+(3.1\times10^{-2})^2}$

$r=0.0876\ m$

We need to calculate the magnitude of the charge q₃

Using formula of net force

$F_{12}=kq_{2}(\dfrac{q_{3}}{r_{3}^2}+\dfrac{q_{1}}{r_{1}^2})$

Put the value into the formula

$14.413=9\times10^{9}\times7.7\times10^{-6}(\dfrac{q_{3}}{(0.0438)^2}+\dfrac{-3\times10^{-6}}{(0.0876)^2})$

$(\dfrac{q_{3}}{(4.38\times10^{-2})^2}+\dfrac{-3\times10^{-6}}{(0.0876)^2})=\dfrac{14.413}{9\times10^{9}\times7.7\times10^{-6}}$

$\dfrac{q_{3}}{(0.0438)^2}=207\times10^{-4}+3.909\times10^{-4}$

$q_{3}=0.0210909\times(0.0438)^2$

$q_{3}=40.46\times10^{-6}\ C$

$q_{3}=40.46\ \mu C$

Hence, The value of charge q₃ is 40.46 μC.

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3 years ago

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Answer:

A

Explanation:

6 0

2 years ago

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yarga [219]

Answer:

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3 years ago

A car stops in 130 m. If it has an acceleration of -5 m/s2 what was the cars starting velocity?

Tatiana [17]

Answer:

<u>We are given:</u>

displacement (s) = 130 m

acceleration (a) = -5 m/s²

final velocity (v) = 0 m/s [the cars 'stops' in 130 m]

initial velocity (u) = u m/s

<u>Solving for initial velocity:</u>

From the third equation of motion:

v² - u² = 2as

replacing the variables

(0)² - (u)² = 2(-5)(130)

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8 0

3 years ago

An object is projected from the ground with an upward speed of ų m/s has a speed of 23m/s when it is at a height of 5m above the

vovikov84 [41]

Answer:

25.08m/s

Explanation:

mgh1 + 0.5mv1² = mgh2 + 0.5mv2²

h1 = 0m

v1 = u

h2 = 5m

v2 = 23m/s

putting the values into the formula above;

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0.5mu² = 314.5m

dividing through by m

0.5u² = 314.5

u² = 629

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2 years ago