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DochEvi [55]
4 years ago
10

The student uses a spring as a spring balance.when the student hangs a stone from this spring its extension is 72mm the spring d

oes not go past the limit of proportionality calculate the force exerted by the stone on the spring the spring constant = n/m
Physics
1 answer:
Maru [420]4 years ago
5 0

Force Of Spring Equation


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Stolb23 [73]

Answer:

a is the answer to the question

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An electric motor spins at a constant 2695.0 rpm. If the rotor radius is 7.165 cm, what is the linear acceleration of the edge o
Vikki [24]

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5707 m/s^2

Explanation:

5 0
3 years ago
Find the length of a microscope having magnifying power 100, if the focal lengths of the objective and eye piece are 2 cm and 5
sp2606 [1]

Answer:

   L = 40 cm

Explanation:

A microscope is an optical instrument built by two lenses in such a way that the image of the first is formed within the distance of the other (eyepiece), so that the latter creates an enlarged virtual image of the object, for which the magnification of the microscope is the same to the multiplication of the magnification of each lens

    M = - L / f₀ (25 cm /f_{e})

where fo and fe are the focal lengths of the objective and eyepiece, 25 cm is the near vision distance and L is the length of the microscope

     L = - M f_{o} f_{e} / 25

let's calculate

     L = - (-100) 2 5/25

      L = 40 cm

7 0
3 years ago
Whats the measure of how much energy a sound wave carries, the loudness of a sound?
musickatia [10]
Decibels I believe? I’m not 100% sure
6 0
3 years ago
(7)Figure 4 shows three charges: Q₁, Q₂ and Q3 . Determine the net force (Fnet) acting on Q3. (Hint: Draw a free body diagram of
NISA [10]

Remember Coulomb's law: the magnitude of the electric force F between two stationary charges q₁ and q₂ over a distance r is

F = \dfrac{kq_1q_2}{r^2}

where k ≈ 8,98 × 10⁹ kg•m³/(s²•C²) is Coulomb's constant.

8.1. The diagram is simple, since only two forces are involved. The particle at Q₂ feels a force to the left due to the particle at Q₁ and a downward force due to the particle at Q₃.

8.2. First convert everything to base SI units:

0,02 µC = 0,02 × 10⁻⁶ C = 2 × 10⁻⁸ C

0,03 µC = 3 × 10⁻⁸ C

0,04 µC = 4 × 10⁻⁸ C

300 mm = 300 × 10⁻³ m = 0,3 m

600 mm = 0,6 m

Force due to Q₁ :

F_{Q_2/Q_1} = \dfrac{k (6 \times 10^{-16} \,\mathrm C)}{(0,3 \, \mathrm m)^2} \approx \boxed{6,0 \times 10^{-5} \,\mathrm N} = 0,06 \,\mathrm{mN}

Force due to Q₃ :

F_{Q_2/Q_3} = \dfrac{k (12 \times 10^{-16} \,\mathrm C)}{(0,6 \, \mathrm m)^2} \approx \boxed{3,0 \times 10^{-5} \,\mathrm N} = 0,03 \,\mathrm{mN}

8.3. The net force on the particle at Q₂ is the vector

\vec F = F_{Q_2/Q_1} \, \vec\imath + F_{Q_2/Q_3} \,\vec\jmath = \left(-0,06\,\vec\imath - 0,03\,\vec\jmath\right) \,\mathrm{mN}

Its magnitude is

\|\vec F\| = \sqrt{\left(-0,06\,\mathrm{mN}\right)^2 + \left(-0,03\,\mathrm{mN}\right)^2} \approx 0,07 \,\mathrm{mN} = \boxed{7,0 \times 10^{-5} \,\mathrm N}

and makes an angle θ with the positive horizontal axis (pointing to the right) such that

\tan(\theta) = \dfrac{-0,03}{-0,06} \implies \theta = \tan^{-1}\left(\dfrac12\right) - 180^\circ \approx \boxed{-153^\circ}

where we subtract 180° because \vec F terminates in the third quadrant, but the inverse tangent function can only return angles between -90° and 90°. We use the fact that tan(x) has a period of 180° to get the angle that ends in the right quadrant.

8 0
2 years ago
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