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mrs_skeptik [129]

3 years ago

6

A vertebra is subjected to a shearing force of 535 N. Find the shear deformation, taking the vertebra to be a cylinder 3.00 cm h

igh and 4.00 cm in diameter. The shear modulus for vertebrae is 8.00 × 1010 N/m2.

1 answer:

Ganezh [65]3 years ago

5 0

4.00 is the answer i think im not sure either 4.00 or 8.00

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What is the current that reverses direction in a regular patter called

Oxana [17]

Answer:

Current that reverses direction in the regular pattern is called an alternating current, abbreviated as 'AC'.

Explanation:

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4 0

3 years ago

Develop an equation (with a proportionality constant) that describes the relationship between the gravitational force (fgrav), t

ivolga24 [154]

According to newton's law of gravitation, the gravitational force(F) is directly proportional to the product mass of the moon(Mm) and the mass of the planet (Mp) and it is inversely proportional to the square of the separation between them.

Fg ∝ (Mp)(Mm) →(1)

Fg ∝ 1/d²→(2)

Combining equation (1) and (2),

Fg ∝ (Mp)(Mm)/d²

Fg = G(Mp)(Mm)/d²

This is an equation that describes the relation between mass of moon (Mm) and mass of planet (Mp) and separation(d) between them.

To support the claim in favuor of this equation we use this equation to obtain the value of acceleration due to gravity on earth.

Let m be the mass of an object on earth then Fg between earth (Mp) and mass of an object is obtained by:

Fg = G(Mp)(m)/R², where R= Radius of earth

This force is equal to the weight of an object i.e.,

g= G(Mp)/R²

Putting the values of G, Mp and R , we get, g=9.81 m/s²

which is the value we obtained on earth for acceleration due to gravity.

To know more about gravitational constant, visit:

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5 0

1 year ago

Please Help!!!!! Quickly!!!! brainiest to correct answer !!!!!!

iragen [17]

Just show a picture so I can help ur information is misleading...

5 0

3 years ago

Suppose that a comet that was seen in 550 A.D. by Chinese astronomers was spotted again in year 1941. Assume the time between ob

pickupchik [31]

To solve the problem it is necessary to apply the concepts related to Kepler's third law as well as the calculation of distances in orbits with eccentricities.

Kepler's third law tells us that

$T^2 = \frac{4\pi^2}{GM}a^3$

Where

T= Period

G= Gravitational constant

M = Mass of the sun

a= The semimajor axis of the comet's orbit

The period in years would be given by

$T= 1941-550\\T= 1391y(\frac{31536000s}{1y})\\T=4.3866*10^{10}s$

PART A) Replacing the values to find a, we have

$a^3= \frac{T^2 GM}{4\pi^2}$

$a^3 = \frac{(4.3866*10^{10})^2(6.67*10^{-11})(1.989*10^{30})}{4\pi^2}$

$a^3 = 6.46632*10^{39}$

$a = 1.86303*10^{13}m$

Therefore the semimajor axis is $1.86303*10^{13}m$

PART B) If the semi-major axis a and the eccentricity e of an orbit are known, then the periapsis and apoapsis distances can be calculated by

$R = a(1-e)$

$R = 1.86303*10^{13}(1-0.997)$

$R= 5.58*10^{10}m$

7 0

3 years ago

An electron at Earth's surface experiences a gravitational force of magnitude F=(9.11×10−31 kg)⋅(9.8 m/s2). Part A How far away

katrin [286]

Answer:

r = 5,085 m

Explanation:

The force exerted by on the surface of the Earth on an electron is its weight

W = F = 9.11 10⁻³¹ 9.8

W = 8.9 10⁻³⁰ N

The electric force between an electron and a proton is given by Coulomb's Law

Fe = k q₁ q₂ / r²

Fe = - k q² / r²

They ask us that W = Fe

W = k q² / r²

r = √ k q² / W

Let's calculate

r = √ 8.99 10⁹ (1.6 10⁻¹⁹)² /8.9 10⁻³⁰

r = √ 25.86

r = 5,085 m

Let's look for the relationship of this distance with the harmonic distance

R / R_atomic = 5,085 / 10⁻¹⁰

R / R_Atomic = 5 10¹⁰

We see that this distance is 10¹⁰ times the interatomic distance, so the gravitational attraction force is very small at atomic scale

8 0

3 years ago