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3 years ago

Bone has a Young’s modulus of about

Physics

1 answer:

MArishka [77]3 years ago

6 0

Answer:

Explanation:

Bone has a Young’s modulus of about

1.8 × 1010 Pa . Under compression, it can

withstand a stress of about 1.66 × 108 Pa before breaking.

Assume that a femur (thigh bone) is 0.56 m

long, and calculate the amount of compression

this bone can withstand before breakin :).

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3 years ago

Read 2 more answers

HELPPPPPPPPP Which option is a series of movement exercises for physical and mental health and is based on Hindu philosophy? O p

nika2105 [10]

Answer:

The correct answer is option ( D ) Yoga

Explanation:

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3 years ago

A motorcycle travels 90.0 km/h. How many seconds will it take the motorcycle to cover 2.10 x 103 m?

mina [271]

Answer:

The time taken is 84 seconds, please give the brainliest award would be much appreciated.

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3 years ago

Four charges 7 × 10−9 C at (0 m, 0 m), −9 × 10−9 C at (3 m, 3 m), 7 × 10−9 C at (1 m, 3 m), and −8 × 10−9 C at (−3 m, 2 m), are

Ivanshal [37]

Answer:

Magnitude of the resulting force on the 7 nC charge at the origin:

Fn₁= 23.95*10⁻⁹ N

Explanation:

Look at the attached graphic:

Charges of positive signs exert repulsive forces on q₁ + and charges of negative signs exert attractive forces on q₁ +.

q₁ experiences three forces (F₂₁,F₃₁,F₄₁) and we calculate them with Coulomb's law:

F = (k*q₁*q)/(d)²

$d_{12} = \sqrt{3^{2}+3^{2} } = \sqrt{18}$ m : distance from q₁ to q₂

(d₁₂)² = 18 m²

$d_{13} =\sqrt{1^{2}+3^{2} } = \sqrt{10}$ m : distance from q₁ to q₃

(d₁₃)² = 10 m²

$d_{14} =\sqrt{3^{2}+2^{2} } = \sqrt{13}$ m : distance from q₁ to q₄

(d₁₄)² = 13 m²

K= 8.98755 × 10⁹ N *m²/C²

q₁= 7*10⁻⁹C

k*q₁=8.98755*10⁹ *7*10⁻⁹= 62.9

F₂₁= (62.9)*(9* 10⁻⁹) /(18) = 31.45*10⁻⁹ C

F₃₁= (62.9)*(7* 10⁻⁹) /(10) = 44*10⁻⁹ C

F₄₁= (62.9)*(8* 10⁻⁹) /(13) = 38.7*10⁻⁹ C

x-y components of the net force on q₁ (Fn₁):

α= tan⁻¹(3/3)= 45° , β= tan⁻¹(3/1)= 71.56° , θ= tan⁻¹(2/3)= 33.69°

Fn₁x = F₂₁x+ F₃₁x+F₄₁x

F₂₁x =+ F₂₁*cosα =+ (31.45*10⁻⁹)* (cos 45°) = +22.24 *10⁻⁹ N

F₃₁x= -F₃₁*cosβ = - ( 44*10⁻⁹)* (cos 71.56°) = -13.91 *10⁻⁹ N

F₄₁x= -F₄₁*cosθ = -(38.7*10⁻⁹)* (cos 33.69°) = -32.2*10⁻⁹ N

Fn₁x = (+22.24 - 13.91 - 32.2)*10⁻⁹ N

Fn₁x = -23.87 *10⁻⁹ N

Fn₁y = F₂₁y+ F₃₁y+F₄₁y

F₂₁x =+ F₂₁*sinα =+ (31.45*10⁻⁹)* (sin 45°) = +22.24 *10⁻⁹ N

F₃₁x= -F₃₁*sinβ = - ( 44*10⁻⁹)* (sin 71.56°) = -41.74 *10⁻⁹ N

F₄₁x= +F₄₁*sinθ = +(38.7*10⁻⁹)* (sin 33.69°) =+21.47*10⁻⁹ N

Fn₁y = (22.24 -41.74+21.47)*10⁻⁹ N

Fn₁y = 1.97*10⁻⁹ N

Magnitude of the resulting force on the 7 nC charge at the origin (q₁):

$F_{n1} =\sqrt{(Fn_{1x} )^{2}+(Fn_{1y} )^{2} }$

$F_{n1} =\sqrt{(23.87 )^{2}+(1.97 )^{2} }$

Fn₁= 23.95*10⁻⁹ N

8 0

3 years ago

(b)

m_a_m_a [10]

Answer:

Driving force increases, friction forces increase, the driving force is bigger than friction 12.

Explanation:

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3 years ago