How do you comprehend the fourth dimension?

!!!!!<br> Is blood an incompressible fluid or a compressible fluid?

stiks02 [169]

Blood is an incompressible fluid. Hope this helps!

3 0

4 years ago

Object A has mass mA = 9 kg and initial momentum vector pA,i = < 20, -6, 0 > kg · m/s, just before it strikes object B, wh

love history [14]

Answer:

$p= kg m/s$

Explanation:

Momentum is a vector quantity that represents the "amount of motion" of an object.

Mathematically, the momentum of an object is given by

$p=mv$

where

m is the mass of the object

v is the velocity

Since momentum is a vector, it also has a direction, which is the same as the velocity.

Therefore, if we have two objects, the total momentum of the two objects will be obtained from the vector sum of the individual momenta of the two objects.

In this problem we have:

$p_A= kg m/s$ is the momentum of object A

$p_B = kg m/s$ is the momentum of object B

Therefore, the total momentum of objects A and B can be obtained by adding each components of A to the corresponding component of B, so:

$p_x = 20 +6 = 26 kg m/s\\p_y = -6 +6 = 0 kg m/s\\p_z = 0 + 0 = 0 kg m/s$

So the total initial momentum is

$p= kg m/s$

6 0

3 years ago

A rotating fan completes 1200 revolutions every minute. consider the tip of a blade, at a radius of 0.19 m. through what distanc

ser-zykov [4K]

The tip of the fan moves through the outer side of the circle.

So it moves a distance of perimeter of circle in one revolution.

Perimeter of circle = 2 $\pi$ r, where r is the radius of circle.

In this case radius of circular motion = 0.19 meter

So perimeter of circle = 2 $\pi$ *0.19 = 0.38 $\pi$ = 1.194 m

So distance does the tip move in one revolution = 1.194 meter

5 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

State how you agree or disagree with the following statement. A good circuit cannot have internal resistance.

Rashid [163]

Answer: I do

Explanation:

Resistance opposes current thereby reducing the amount of current that flows through a circuit. In other words, it leads to a loss of electrical energy.

Ideally speaking, a good circuit should have no internal resistance as this would lead to more energy having to be supplied to overcome that resistance. External resistance however, is not a bad thing. For instance, oxygen being removed from lightbulbs.

7 0

3 years ago