All categories

3 years ago

All the functional parts of the _______________ system are called the alimentary canal.

Physics

2 answers:

polet [3.4K]3 years ago

3 0

I’m pretty sure it’s digestive

Lapatulllka [165]3 years ago

3 0

Ans is digestive system hope that will help you

You might be interested in

A hot-water stream at 80 ℃ enters a mixing chamber with a mass flow rate of 0.5 kg/s where it is mixed with a stream of cold wat

rewona [7]

Answer:

$\dot{m_{2}}=0.865 kg/s$

Explanation:

$\dot{m_1}= 0.5kg/s$

from steam tables , at 250 kPa, and at

T₁ = 80⁰C ⇒ h₁ = 335.02 kJ/kg

T₂ = 20⁰C⇒ h₂ = 83.915 kJ/kg

T₃ = 42⁰C ⇒ h₃ = 175.90 kJ/kg

we know

$\dot{m_{in}}=\dot{m_{out}}$

$\dot{m_{1}}+\dot{m_{2}}=\dot{m_{3}}$

according to energy balance equation

$\dot{m_{in}}h_{in}=\dot{m_{out}}h_{out}$

$\dot{m_{1}}h_{1}+\dot{m_{2}}h_{2}=\dot{m_{3}}h_{3}$

$\dot{m_{1}}h_{1}+\dot{m_{2}}h_{2}=(\dot{m_{1}}+\dot{m_{2}})h_{3}\\(0.5\times 335.02)+(\dot{m_{2}}\times 83.915)=(0.5+\dot{m_{2}})175.90\\\dot{m_{2}}=0.865 kg/s$

4 0

4 years ago

Any 3 differences between telescope and microscope

Keith_Richards [23]

An instrument used to observe or imagine very small object using an optical mangifier
mirco cell.
Telescope is a magnifer of distance object

4 0

4 years ago

8. An unpowered flywheel is slowed by a constant frictional torque. At time t = 0 it has an angular velocity of 200 rad/s. Ten s

allsm [11]

Answer:

a) $\omega = 50\,\frac{rad}{s}$ , b) $\omega = 0\,\frac{rad}{s}$

Explanation:

The magnitude of torque is a form of moment, that is, a product of force and lever arm (distance), and force is the product of mass and acceleration for rotating systems with constant mass. That is:

$\tau = F \cdot r$

$\tau = m\cdot a \cdot r$

$\tau = m \cdot \alpha \cdot r^{2}$

Where $\alpha$ is the angular acceleration, which is constant as torque is constant. Angular deceleration experimented by the unpowered flywheel is:

$\alpha = \frac{170\,\frac{rad}{s} - 200\,\frac{rad}{s} }{10\,s}$

$\alpha = -3\,\frac{rad}{s^{2}}$

Now, angular velocities of the unpowered flywheel at 50 seconds and 100 seconds are, respectively:

a) t = 50 s.

$\omega = 200\,\frac{rad}{s} - \left(3\,\frac{rad}{s^{2}} \right) \cdot (50\,s)$

$\omega = 50\,\frac{rad}{s}$

b) t = 100 s.

Given that friction is of reactive nature. Frictional torque works on the unpowered flywheel until angular velocity is reduced to zero, whose instant is:

$t = \frac{0\,\frac{rad}{s}-200\,\frac{rad}{s} }{\left(-3\,\frac{rad}{s^{2}} \right)}$

$t = 66.667\,s$

Since $t > 66.667\,s$ , then the angular velocity is equal to zero. Therefore:

$\omega = 0\,\frac{rad}{s}$

7 0

4 years ago

To reduce broadcasts in a lan, the most suitable device is a

Nookie1986 [14]

Wow I have no idea kid............

5 0

3 years ago

a beam of electrons passes through an electric field where the magnitude of the electric field strength is 3.00 * 10^3 newtons p

nikdorinn [45]

Electric Field Strength E = Force F/ Charge q

E = F/q

Force , F = E*q

But charge on an electron q = 1.6 * 10⁻¹⁹ C

E = 3 * 10³ N/C from the question.

Force F = 3* 10³ * 1.6 * 10⁻¹⁹

F = 4.8 * 10⁻¹⁶ N.

Electrostatic force F = 4.8 * 10⁻¹⁶ N

5 0

4 years ago