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makvit [3.9K]
3 years ago
5

4. Using the bone density of 2.0 kg/m3, calculate the mass of an adult femur bone that has a volume of 0.00027 m3.

Physics
1 answer:
kolbaska11 [484]3 years ago
6 0

Answer:

\boxed{\sf Mass \ of \ an \ adult \ femur \ bone = 0.00054 \ kg}

Given:

Bone density = 2.0 kg/m³

Volume of bone (V) = 0.00027 m³

To Find:

Mass of an adult femur bone (m).

Explanation:

\sf \implies Density = \frac{Mass (m)}{Volume (V)} \\ \\ \sf \implies \frac{Mass}{Volume} = Density \\ \\ \sf \implies Mass = Density \times Volume \\ \\ \sf \implies Mass = 2.0 \ kg/ \cancel{m^{3}} \times 0.00027 \ \cancel{m^{3}} \\ \\ \sf \implies Mass = 0.00054 \ kg

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3 years ago
An electron that has an instantaneous velocity of ???? = 2.0 × 106 m ???? ???? + 3.0 × 106 m ???? ???? is moving through the uni
Setler79 [48]

Explanation:

It is given that,

Velocity of the electron, v=(2\times 10^6i+3\times 10^6j)\ m/s

Magnetic field, B=(0.030i-0.15j)\ T

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(a) Let F_e is the force on the electron due to the magnetic field. The magnetic force acting on it is given by :

F_e=q_e(v\times B)

F_e=1.6\times 10^{-19}\times [(2\times 10^6i+3\times 10^6j)\times (0.030i-0.15j)]

F_e=-1.6\times 10^{-19}\times (-390000)(k)

F_e=6.24\times 10^{-14}k\ N

(b) The charge of electron, q_p=1.6\times 10^{-19}\ C

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3 years ago
Explain how do winds cause surface currents?
igor_vitrenko [27]

Answer:

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A 600kg car is moving at 5 m/s to the right and elastically collided with a stationary 900 kg car. What is the velocity of the 9
11111nata11111 [884]

Answer:

\mathrm{v}_{2} \text { velocity after the collision is } 3.3 \mathrm{m} / \mathrm{s}

Explanation:

It says “Momentum before the collision is equal to momentum after the collision.” Elastic Collision formula is applied to calculate the mass or velocity of the elastic bodies.

m_{1} v_{1}=m_{2} v_{2}

\mathrm{m}_{1} \text { and } \mathrm{m}_{2} \text { are masses of the object }

\mathrm{v}_{1} \text { velocity before the collision }

\mathrm{v}_{2} \text { velocity after the collision }

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\mathrm{m}_{2}=900 \mathrm{kg}

\text { Velocity before the collision } v_{1}=5 \mathrm{m} / \mathrm{s}

600 \times 5=900 \times v_{2}

3000=900 \times v_{2}

\mathrm{v}_{2}=\frac{3000}{900}

\mathrm{v}_{2}=3.3 \mathrm{m} / \mathrm{s}

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