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3 years ago

Nervous tissue is made up of neuroglia and what other type of cell?

1 answer:

5 0

Nervous tissue is composed of neurons and supporting cells called neuroglia, or ” glial cells.” . Neurons are the other the other type of cell that comprise nervous tissue. Neurons have cell bodies, dendrites, and axons.

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A ball is being rolled by a normal push of 180N. It is opposed by friction which has a force of 61N and air resistance which has

Nina [5.8K]

Resultant force is basically the force left after everything is added.

if a ball is being pushed one one side with 180N, and being pushed on teh opposite side with 84N (I added friction and air resistance since they're acting on the same side), then the resultant force would be:

180N - 84N =<u> 96N</u> (you can determine whether it's positive or negative based on the direction of the vector)

7 0

4 years ago

A car has the velocity of 2.35 after 4.67 it's velocity is 9.89 what is the average acceleration

kiruha [24]

Average acceleration = (change in speed) / (time for the change) .

Change in speed = (ending speed) - (beginning speed)

=  (9.89 miles/hour) - (2.35 yards/second) = 26,839.2 ft/hr

Acceleration = (26,839.2 ft/hr) / (4.67 days) = 2,873.58 inch/hour²

6 0

3 years ago

Write the first equation of motion. Under what condition(s) is this equation valid?

Zepler [3.9K]

Explanation:

The first equation of motion in kinematics is given by :

$v=u+at$ .....(1)

u is initial speed

a is acceleration

v is final speed

t is time

Equation (1) is valid when the object is moving with constant acceleration. This equation gives relation between velocity and time.

3 0

3 years ago

The atmosphere of Mars is almost all carbon dioxide and the average surface pressure is 610 Pa (as compared with 101,000 Pa on E

Karolina [17]

Answer:

z = 3,737 10⁵ m

Explanation:

a) As they indicate that the atmosphere behaves like an ideal gas, we can use the equation

P V = n R T

P = (n r / V) T

We replace

P = (n R / V) T₀ $e^{- C z}$

b) Let's apply this equation in the points

Lower

.z = 0

P₀ = 610 Pa

P₀ = (nR / V) T₀

Higher.

P = 10 Pa

P = (n R / V) T₀ e^{- C z}

We replace

P = P₀ e^{- C z}

e^{- C z} = P / P₀

C z = ln P₀ / P

z = 1 / C ln P₀ / P

Let's calculate

z = 1 / 1.1 10⁻⁵ ln (610/10)

z = 3,737 10⁵ m

4 0

3 years ago

One of the waste products of a nuclear reactor is plutonium-239 . This nucleus is radioactive and decays by splitting into a hel

Gekata [30.6K]

Answer:

a) $v_{U-235} = 2.68 \cdot 10^{5} m/s$

$v_{He-4} = -1.57 \cdot 10^{7} m/s$

b) $E_{He-4} = 8.23 \cdot 10^{-13} J$

$E_{U-235} = 1.41 \cdot 10^{-14} J$

Explanation:

Searching the missed information we have:

E: is the energy emitted in the plutonium decay = 8.40x10⁻¹³ J

m(⁴He): is the mass of the helium nucleus = 6.68x10⁻²⁷ kg

m(²³⁵U): is the mass of the helium U-235 nucleus = 3.92x10⁻²⁵ kg

a) We can find the velocities of the two nuclei by conservation of linear momentum and kinetic energy:

Linear momentum:

$p_{i} = p_{f}$

$m_{Pu-239}v_{Pu-239} = m_{He-4}v_{He-4} + m_{U-235}v_{U-235}$

Since the plutonium nucleus is originally at rest, $v_{Pu-239} = 0$ :

$0 = m_{He-4}v_{He-4} + m_{U-235}v_{U-235}$

$v_{He-4} = -\frac{m_{U-235}v_{U-235}}{m_{He-4}}$ (1)

Kinetic Energy:

$E_{Pu-239} = \frac{1}{2}m_{He-4}v_{He-4}^{2} + \frac{1}{2}m_{U-235}v_{U-235}^{2}$

$2*8.40 \cdot 10^{-13} J = m_{He-4}v_{He-4}^{2} + m_{U-235}v_{U-235}^{2}$

$1.68\cdot 10^{-12} J = m_{He-4}v_{He-4}^{2} + m_{U-235}v_{U-235}^{2}$ (2)

By entering equation (1) into (2) we have:

$1.68\cdot 10^{-12} J = m_{He-4}(-\frac{m_{U-235}v_{U-235}}{m_{He-4}})^{2} + m_{U-235}v_{U-235}^{2}$

$1.68\cdot 10^{-12} J = 6.68 \cdot 10^{-27} kg*(-\frac{3.92 \cdot 10^{-25} kg*v_{U-235}}{6.68 \cdot 10^{-27} kg})^{2} +3.92 \cdot 10^{-25} kg*v_{U-235}^{2}$

Solving the above equation for $v_{U-235}$ we have:

$v_{U-235} = 2.68 \cdot 10^{5} m/s$

And by entering that value into equation (1):

$v_{He-4} = -\frac{3.92 \cdot 10^{-25} kg*2.68 \cdot 10^{5} m/s}{6.68 \cdot 10^{-27} kg} = -1.57 \cdot 10^{7} m/s$

The minus sign means that the helium-4 nucleus is moving in the opposite direction to the uranium-235 nucleus.

b) Now, the kinetic energy of each nucleus is:

For He-4:

$E_{He-4} = \frac{1}{2}m_{He-4}v_{He-4}^{2} = \frac{1}{2} 6.68 \cdot 10^{-27} kg*(-1.57 \cdot 10^{7} m/s)^{2} = 8.23 \cdot 10^{-13} J$

For U-235:

$E_{U-235} = \frac{1}{2}m_{U-235}v_{U-235}^{2} = \frac{1}{2} 3.92 \cdot 10^{-25} kg*(2.68 \cdot 10^{5} m/s)^{2} = 1.41 \cdot 10^{-14} J$

I hope it helps you!

3 0

3 years ago