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3 years ago

What are the connective cells

1 answer:

eduard3 years ago

8 0

Special connective tissue consists of reticular connective tissue, adipose tissue, cartilage, bone, and blood. Other kinds of connective tissues include fibrous, elastic, and lymphoid connective tissues. New vascularised connective tissue that forms in the process of wound healing is termed granulation tissue.

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An object is formed by attaching a uniform, thin rod with a mass of mr = 6.85 kg and length L = 5.76 m to a uniform sphere with

Ket [755]

Answer:

Part a)

$I = 1879.7 kg m^2$

Part b)

$\alpha = 0.70 rad/s^2$

Part c)

$I = 153.8 kg m^2$

Part 4)

angular acceleration will be ZERO

Part 5)

$I = 345.6 kg m^2$

Explanation:

Part a)

Moment of inertia of the system about left end of the rod is given as

$I = \frac{m_r L^2}{3} + (\frac{2}{5} m_s R^2 + m_s(R + L)^2)$

So we have

$I = \frac{m_r(4R)^2}{3} + (\frac{2}{5}(5m_r) R^2 + (5m_r)(R + 4R)^2)$

$I = \frac{16}{3}m_r R^2 + (2m_r R^2 + 125 m_rR^2)$

$I = (\frac{16}{3} + 127)m_r R^2$

$I = (\frac{16}{3} + 127)(6.85)(1.44)^2$

$I = 1879.7 kg m^2$

Part b)

If force is applied to the mid point of the rod

so the torque on the rod is given as

$\tau = F\frac{L}{2}$

$\tau = 460(2R)$

$\tau = 460 \times 2 \times 1.44$

$\tau = 1324.8 Nm$

now angular acceleration is given as

$\alpha = \frac{\tau}{I}$

$\alpha = \frac{1324.8}{1879.7}$

$\alpha = 0.70 rad/s^2$

Part c)

position of center of mass of rod and sphere is given from the center of the sphere as

$x = \frac{m_r}{m_r + m_s}(\frac{L}{2} + R)$

$x = \frac{m_r}{6 m_r}(3R) = \frac{R}{2}$

so moment of inertia about this position is given as

$I = \frac{m_r L^2}{12} + m_r(\frac{L}{2} + \frac{R}{2})^2 + (\frac{2}{5} m_s R^2 + m_s(\frac{R}{2})^2)$

so we have

$I = \frac{m_r (16R^2)}{12} + m_r(\frac{5R}{2})^2 + \frac{2}{5}(5m_r)R^2 + (5m_r)(\frac{R^2}{4})$

$I = m_r R^2(\frac{16}{12} + \frac{25}{4} + 2 + \frac{5}{4})$

$I = 6.85(1.44)^2\times 10.83$

$I = 153.8 kg m^2$

Part 4)

If force is applied parallel to the length of rod

then we have

$\tau = \vec r \times \vec F$

$\tau = 0$

so angular acceleration will be ZERO

Part 5)

moment of inertia about right edge of the sphere is given as

$I = \frac{m_r L^2}{12} + m_r(\frac{L}{2} + 2R)^2 + (\frac{2}{5} m_s R^2 + m_s(R)^2)$

so we have

$I = \frac{m_r (16R^2)}{12} + m_r(4R)^2 + \frac{2}{5}(5m_r)R^2 + (5m_r)(R^2)$

$I = m_r R^2(\frac{16}{12} + 16 + 2 + 5)$

$I = 6.85(1.44)^2\times 24.33$

$I = 345.6 kg m^2$

6 0

3 years ago

a steam engine work on its vicinity 285 k heat is released with the help of 225 c energy absorbed to the system what is the effi

zvonat [6]

The efficiency of steam engine is 26.66 %.

<h3>What is Efficiency?</h3>

The efficiency is defined as the work done by the engine divided by the heat supplied.

Work done is the difference between the heat supplied and heat rejected.

So, efficiency η = (Qs -Qr) / Qs

Given is the heat supplied Qs = 225 Joules and heat rejected Qr =285 . then the work done is

W = 225 -225 = - 60 Joules

Efficiency η = 60/225 x 100%

η = 26.66 %

Thus, the efficiency of the steam engine is 26.66 %.

Learn more about efficiency.

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8 0

2 years ago

Read 2 more answers

According to Newton's law of cooling, the rate at which an object's temperature changes is directly proportional to the differen

Kryger [21]

Answer:

dT(t)/dt = k[T5 - T(t)]

Explanation:

Since T(t) represents the temperature of the object and T5 represents the temperature of the surroundings, according to Newton's law of cooling, the rate at which an object's temperature changes is directly proportional to the difference in temperature between the object and the surrounding medium, that is dT(t)/dt ∝ T5 - T(t)

Introducing the constant of proportionality

dT(t)/dt = k[T5 - T(t)]

which is the desired differential equation

4 0

3 years ago

Read 2 more answers

8. The force constant of a spring is 150. N/m. (a) How much force is required to stretch the spring 0.25 m? (b) How much work is

Ne4ueva [31]

Answer:

<em>b) 9.375Joules</em>

Explanation:

a) According to Hooke's law

F = ke

k is the spring constant

e is the extension;

F = 150 * 0.25

F = 37.5N

b) Work done on the spring = 1/2ke^2

Work done on the spring = 1/2 * 150 * 0.25^2

Work done on the spring = 75 * 0.0625

Work done on the spring = 9.375Joules

4 0

3 years ago

At what width does a totality of a Solar Eclipse span?

kari74 [83]

115km (71 miles) , the eclipse will also have a magnitude of 1.0306

6 0

4 years ago