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julia-pushkina [17]
2 years ago
9

Connective Tissue in a tendon is

Physics
1 answer:
mariarad [96]2 years ago
4 0

Overview. A tendon is a fibrous connective tissue which attaches muscle to bone. Tendons may also attach muscles to structures such as the eyeball. A tendon serves to move the bone or structure.
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How much heat is contained in 100 kg of water at 60.0 °C?
attashe74 [19]
First we need to write down heat capacity for water which is constant.
cp=4186 J/(kg*K)

The equation for Energy that we will be calculating is:

E=cp*m*T
where m is mass and T is absolute temperature (273,15 + 60 in this case). Replacing all the values in equation we get:

E = 4186*100*333,15 = 139 456 590 J
6 0
3 years ago
How many degrees equal π in radians? How many revolutions (turns) equal π radians? ​
kodGreya [7K]

Answer:

It follows that the magnitude in radians of one complete revolution (360 degrees) is the length of the entire circumference divided by the radius, or 2πr / r, or 2π. Thus 2π radians is equal to 360 degrees, meaning that one radian is equal to 180/π ≈ 57.295779513082320876 degrees.

Explanation:

6 0
2 years ago
What happens to steam as it releases thermal energy inside a radiator?
marshall27 [118]
It condenses into liquid water.
7 0
3 years ago
A small meteorite with mass of 1 g strikes the outer wall of a communication satellite with a speed of 2Okm/s (relative to the s
strojnjashka [21]

Answer:

The energy coverted to heat is 200 kilojoules.

Explanation:

GIven the absence of external forces exerted both on the small meteorite and on the communication satellite, the Principle of Linear Momentum is considered and let suppose that collision is completely inelastic and that satellite is initially at rest. Hence, the expression for the satellite-meteorite system:

m_{M}\cdot v_{M} + m_{S}\cdot v_{S} = (m_{M}+m_{S})\cdot v

Where:

m_{M}, m_{S} - Masses of the small meteorite and the communication satellite, measured in kilograms.

v_{M}, v_{S} - Speeds of the small meteorite and the communication satellite, measured in meters per second.

v - Final speed of the satellite-meteorite system, measured in meters per second.

The final speed of the satellite-meteorite system is cleared:

v = \frac{m_{M}\cdot v_{M}+m_{S}\cdot v_{S}}{m_{M}+m_{S}}

If m_{M} = 1\times 10^{-3}\,kg, m_{S} = 200\,kg, v_{M} = 20000\,\frac{m}{s} and v_{S} = 0\,\frac{m}{s}, the final speed is now calculated:

v = \frac{(1\times 10^{-3}\,kg)\cdot \left(20000\,\frac{m}{s} \right)+(200\,kg)\cdot \left(0\,\frac{m}{s} \right)}{1\times 10^{-3}\,kg+200\,kg}

v = 0.1\,\frac{m}{s}

Which means that the new system remains stationary and all mechanical energy from meteorite is dissipated in the form of heat. According to the Principle of Energy Conservation and the Work-Energy Theorem, the change in the kinetic energy is equal to the dissipated energy in the form of heat:

K_{S} + K_{M} - K - Q_{disp} = 0

Q_{disp} = K_{S}+K_{M}-K

Where:

K_{S}, K_{M} - Initial translational kinetic energies of the communication satellite and small meteorite, measured in joules.

K - Kinetic energy of the satellite-meteorite system, measured in joules.

Q_{disp} - Dissipated heat, measured in joules.

The previous expression is expanded by using the definition for the translational kinetic energy:

Q_{disp} = \frac{1}{2}\cdot [m_{M}\cdot v_{M}^{2}+m_{S}\cdot v_{S}^{2}-(m_{M}+m_{S})\cdot v^{2}]

Given that m_{M} = 1\times 10^{-3}\,kg, m_{S} = 200\,kg, v_{M} = 20000\,\frac{m}{s}, v_{S} = 0\,\frac{m}{s} and v = 0.1\,\frac{m}{s}, the dissipated heat is:

Q_{disp} = \frac{1}{2}\cdot \left[(1\times 10^{-3}\,kg)\cdot \left(20000\,\frac{m}{s} \right)^{2}+(200\,kg)\cdot \left(0\,\frac{m}{s} \right)^{2}-(200.001\,kg)\cdot \left(0.001\,\frac{m}{s} \right)^{2}\right]Q_{disp} = 200000\,J

Q_{disp} = 200\,kJ

The energy coverted to heat is 200 kilojoules.

4 0
3 years ago
A thin rod of length 1.4 m and mass 140 g is suspended freely from one end. It is pulled to one side and then allowed to swing l
m_a_m_a [10]

Answer:

a The kinetic energy is  KE = 0.0543 J

b The height of the center of mass above that position is  h = 1.372 \ m    

Explanation:

From the question we are told that

  The length of the rod is  L = 1.4m

   The mass of the rod m = 140 = \frac{140}{1000} = 0.140 \ kg  

   The angular speed at the lowest point is w = 1.09 \ rad/s

Generally moment of inertia of the rod about an axis that passes through its one end is

                   I = \frac{mL^2}{3}  

Substituting values

               I = \frac{(0.140) (1.4)^2}{3}

               I = 0.0915 \ kg \cdot m^2

Generally the  kinetic energy rod is mathematically represented as

             KE = \frac{1}{2} Iw^2

                    KE = \frac{1}{2} (0.0915) (1.09)^2

                           KE = 0.0543 J

From the law of conservation of energy

The kinetic energy of the rod during motion =  The potential energy of the rod at the highest point

   Therefore

                   KE = PE = mgh

                        0.0543 = mgh

                             h = \frac{0.0543}{9.8 * 0.140}

                                h = 1.372 \ m    

                 

5 0
2 years ago
Read 2 more answers
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