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goldenfox [79]
3 years ago
9

State any two effects of gravitational force​

Physics
1 answer:
Dmitry_Shevchenko [17]3 years ago
3 0
Connection to Big Idea about energy: Gravity creates gravitational potential energy. Gravitational energy relies on the masses of two bodies and their distance.

Connection to Big Idea about the universe: Gravitational force is exerted by all objects with mass throughout the Universe. It is what keeps the Earth and the planets in orbit around the Sun, and our Solar System in orbit around the centre of the Milky Way. Gravity is one of the forces involved in the birth of stars, their evolution and finally their death.

Connection to Big Idea about Earth: The gravitational force is responsible for many physical properties of Earth and consequently it affects the existence and the properties of living creatures on it. For instance, the existence, the chemical composition and the structure of Earth’s atmosphere was determined by Earth’s gravitational force.

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A parallel-plate capacitor has circular plates and no dielectric between the plates. Each plate has a radius equal to 2.8 cm and
Snezhnost [94]

Answer:1.88\times 10^{14} V/m-s

Explanation:

Given

radius of capacitor Plate r=2.8 cm

Area A=\pi r^2=\pi \times 2.8^\times 10^{-4} m^2

A=24.63\times 10^{-4} m^2

current I=4.1 A

separation d=1.1 mm

Electric Field strength is given by

E=\frac{Q}{\epsilon _0A}

E=\frac{I\cdot t}{\epsilon _0A}

\frac{\mathrm{d} E}{\mathrm{d} t}=\frac{I}{\epsilon _0A}

\frac{\mathrm{d} E}{\mathrm{d} t}=\frac{4.1}{8.85\times 10^{-12}\times 24.63\times 10^{-4}}

\frac{\mathrm{d} E}{\mathrm{d} t}=1.88\times 10^{14} V/m-s

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A. Blocking light waves that vibrate in the plane. (?)
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My dad gifted me a calculator. I have observed that very small cells are used in a calculator. What are these cells called and w
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A disk between vertebrae in the spine is subjected to a shearing force of 600 N. Find its shear deformation, taking it to have a
Verizon [17]

Answer:

The shear deformation is \Delta x=3.34\times10^{-6}\ m.

Explanation:

Given that,

Shearing force F = 600 N

Shear modulus S = 1\times10^{9}\ N/m^2

length = 0.700 cm

diameter = 4.00 cm

We need to find the shear deformation

Using formula of shear modulus

S=\dfrac{Fl_{0}}{A\Delta x}

\Delta x=\dfrac{Fl_{0}}{(\dfrac{\pi d^2}{4})S}

\Delta x=\dfrac{4Fl_{0}}{\pi d^2 S}

Put the value into the formula

\Delta x=\dfrac{4\times600\times0.700\times10^{-2}}{3.14\times1\times10^{9}\times(4.00\times10^{-2})^2}

\Delta x=3.34\times10^{-6}\ m

Hence, The shear deformation is \Delta x=3.34\times10^{-6}\ m.

7 0
3 years ago
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