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2 years ago

Explain the versatility of epithelial tissue

2 answers:

8 0

Epithelial cells are classified by the following three factors : 1. Shape . 2. Stratification. 3.Specialization . It's a main aspect to the mehe family , and plays a huge part in helping the movement of bobbing your head.

mrs_skeptik [129]2 years ago

6 0

Answer:

Explanation:

The epithelial tissues cover the body surfaces, forms a lining over the body cavities and hollow organs. These tissues also forms a protective barrier between the external environment and the body.

The epithelial tissues are the tissues which performs various functions. These tissues are composed of cells that are laid close to each other with the cells that lies in the close association with each other. The structure and function of the epithelial cells vary from both upper and lower surfaces. Due to variety in the functions of the epithelial tissues like protection, secretion, cellular transportation, filtration, selective absorption and sensing, these are called as the versatile tissues.

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How can controlling the way light bends and reflects be used to help people?

Vinil7 [7]

too much sun is dangerous for humans and can cause cancer so it's important that light is reflected for example a pool reflects water back to space that is why water sometimes is cold because it reflects light

4 0

2 years ago

Each shot of the laser gun most favored by Rosa the Closer, the intrepid vigilante of the lawless 22nd century, is powered by th

mote1985 [20]

Answer:

U = 1794.005 × 10⁶ J

Explanation:

Data provided;

Capacitance of the original capacitor, C = 1.27 F

Potential difference applied to the original capacitor, V = 59.9 kV

= 59.9 × 10³ V

Now,

The Potential energy (U) for the capacitor is calculated as:

Potential energy of the original capacitor, U = $\frac{\textup{1}}{\textup{2}}$ × C × V²

on substituting the respective values, we get

U = $\frac{\textup{1}}{\textup{2}}$ × 1.27 × ( 59.9 × 10³ )²

U = 1794.005 × 10⁶ J

7 0

2 years ago

A truck with a heavy load has a total mass of 7100 kg. It is climbing a 15∘ incline at a steady 15 m/s when, unfortunately, the

Andrej [43]

Answer:

The load has a mass of 2636.8 kg

Explanation:

Step 1 : Data given

Mass of the truck = 7100 kg

Angle = 15°

velocity = 15m/s

Acceleration = 1.5 m/s²

Mass of truck = m1 kg

Mass of load = m2 kg

Thrust from engine = T

Step 2:

⇒ Before the load falls off, thrust (T) balances the component of total weight downhill:

T = (m1+m2)*g*sinθ

⇒ After the load falls off, thrust (T) remains the same but downhill component of weight becomes m1*gsinθ .

Resultant force on truck is F = T – m1*gsinθ

F causes the acceleration of the truck: F= m*a

This gives the equation:

T – m1*gsinθ = m1*a

T = m1(a + gsinθ)

Combining both equations gives:

(m1+m2)*g*sinθ = m1*(a + gsinθ)

m1*g*sinθ + m2*g*sinθ =m1*a + m1*g*sinθ

m2*g*sinθ = m1*a

Since m1+m2 = 7100kg, m1= 7100 – m2. This we can plug into the previous equation:

m2*g*sinθ = (7100 – m2)*a

m2*g*sinθ = 7100a – m2a

m2*gsinθ + m2*a = 7100a

m2* (gsinθ + a) = 7100a

m2 = 7100a/(gsinθ + a)

m2 = (7100 * 1.5) / (9.8sin(15°) + 1.5)

m2 = 2636.8 kg

The load has a mass of 2636.8 kg

6 0

2 years ago

An elevator has a weight of 14,700 N and has an acceleration of –0.30 m/s2. The free-body diagram shows the forces acting on the

qwelly [4]

14250. I just took it

4 0

2 years ago

Read 2 more answers

An electron moving at right angles to a 0.1 T magnetic field experiences an acceleration of 6 × 1015 m.s-2. What is the speed of

GaryK [48]

Explanation:

It is given that,

Magnetic field, B = 0.1 T

Acceleration, $a=6\times 10^{15}\ m/s^2$

Charge on electron, $q=1.6\times 10^{-19}\ C$

Mass of electron, $m=9.1\times 10^{-31}\ kg$

(a) The force acting on the electron when it is accelerated is, F = ma

The force acting on the electron when it is in magnetic field, $F=qvB\ sin\theta$

Here, $\theta=90$

So, $ma=qvB$

Where

v is the velocity of the electron

B is the magnetic field

$v=\dfrac{ma}{qB}$

$v=\dfrac{9.1\times 10^{-31}\ kg\times 6\times 10^{15}\ m/s^2}{1.6\times 10^{-19}\ C\times 0.1\ T}$

v = 341250 m/s

$v=3.41\times 10^5\ m/s$

So, the speed of the electron is $3.41\times 10^5\ m/s$

(b) In 1 ns, the speed of the electron remains the same as the force is perpendicular to the cross product of velocity and the magnetic field.

7 0

2 years ago