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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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2 years ago

List at least 10 derived units of some physical quantities.

Artist 52 [7]

Answer:

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3 years ago

Read 2 more answers

A soccer player kicks a soccer ball initially at rest setting it in motion at a velocity of 30 m/s. If the ball has a mass of 0.

Alchen [17]

Answer:

F= 600 N

Explanation:

Given that

Initial velocity ,u= 0 m/s

Final velocity ,v= 30 m/s

mass ,m = 0.5 kg

time ,t= 0.025 s

The change in the linear momentum is given as

ΔP= m (v - u)

ΔP= 0.5 ( 30 - 0 ) kg.m/s

ΔP= 15 kg.m/s

We know that from second law of Newtons

$F=\dfrac{dP}{dt}$

$F=\dfrac{\Delta P}{t}$

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$F=\dfrac{15}{0.025}\ N$

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5 0

3 years ago

A ball is dropped from rest from a height h above the ground. another ball is thrown vertically upwards from the ground at the i

Darya [45]

The position of the first ball is

$y_1=h-\dfrac g2t^2$

while the position of the second ball, thrown with initial velocity $v$ , is

$y_2=vt-\dfrac g2t^2$

The time it takes for the first ball to reach the halfway point satisfies

$\dfrac h2=h-\dfrac g2t^2$

$\implies\dfrac h2=\dfrac g2t^2$

$\implies t=\sqrt{\dfrac hg}$

We want the second ball to reach the same height at the same time, so that

$\dfrac h2=v\sqrt{\dfrac hg}-\dfrac g2\left(\sqrt{\dfrac hg}\right)^2$

$\implies h=2v\sqrt{\dfrac hg}-g\left(\dfrac hg\right)$

$\implies h=v\sqrt{\dfrac hg}$

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3 years ago

A wad of clay of mass m1 = 0.49 kg with an initial horizontal velocity v1 = 1.89 m/s hits and adheres to the massless rigid bar

notka56 [123]

Answer:

<h2>The angular velocity just after collision is given as</h2><h2> $\omega = 0.23 rad/s$ </h2><h2>At the time of collision the hinge point will exert net external force on it so linear momentum is not conserved</h2>

Explanation:

As per given figure we know that there is no external torque about hinge point on the system of given mass

So here we will have

$L_i = L_f$

now we can say

$m_1v_1\frac{L}{2} = (m_2L^2 + m_1(\frac{L}{2})^2)\omega$

so we will have

$0.49(1.89)(0.45) = (2.13(0.90)^2 + 0.49(0.45)^2)\omega$

$\omega = 0.23 rad/s$

Linear momentum of the system is not conserved because at the time of collision the hinge point will exert net external force on the system of mass

So we can use angular momentum conservation about the hinge point

6 0

3 years ago