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tankabanditka [31]
3 years ago
9

The pathologic changes that occur in the development of coronary atherosclerotic lesions include call damage resulting from whic

h of the following? (Select all that apply):
1. a decrease in smooth muscle cells2. a chronic calcium buildup3. the effects of oxidized lipids4. an inflammatory response5. the formation of plaques
Physics
1 answer:
ale4655 [162]3 years ago
4 0

Answer:

3 effect of oxidized lipids

4 an inflammatory response

5 the formation of plaques

Explanation:

Destruction of cells due to oxidation of lipids whereby free radicals steal electrons in cell membrane.

This occurs when tissues get injured by trauma, bacteria or toxins, thereby causing damages cells to release chemicals like histamine, brakykinn that cause vessels to leak fluid into the injured tissues causing swelling.

-plaques are regions of destroyed cells which are visible structures formed inside a cell culture.

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Practice with Density
bekas [8.4K]

Answer:

5.7 g/cm3

Explanation:

7 0
3 years ago
Read 2 more answers
A particle covers 10m distance in 2sec. if final velocity is 8m/s, find initial velocity​
klasskru [66]

Explanation:

It is given that a particle covers 10m in first 5s and 10m in next 3s. so using the equation of motion

Case I

s=ut+

2

1

at

2

10=5u+

2

1

a(5)

2

20=10u+25a

4=2u+5a..............(1)

Case 2

In next 3s the particle covers more 10m distance. So

20=8u+

2

1

a(8)

2

5=2u+8a.........(2)

On solving equation (1) and (2)

4=2u+5a

5=2u+8a

a=

3

1

m/s

2

Put the value of a in equation (1)

u=

6

7

m/s

Now to find distance in next 10 s. total time will be 10s

s=

6

7

×10+

2

1

×

3

1

×(10)

2

s=28.33m

Distance travelled in next 2 sec

s=28.33−20=8.33m

4 0
3 years ago
Read 2 more answers
Two objects that are not initially in thermal equilibrium are placed in close contact. After a while, the temperature of the cod
Dima020 [189]

Answer:

If the temperature of  the colder object rises by the same amount as the temperature of the hotter object drops, then <u>the specific heats of both objects will be equal.</u>

Explanation:

If the temperature of  the colder object rises by the same amount as the temperature of the hotter object drops when the two<u> objects of same mass</u> are brought into contact, then their specific heat capacity is equal.

<u>We can prove this by the equation of heat for the two bodies:</u>

<em>According to given condition,</em>

\Delta T_1=\Delta T_2

\frac{Q_1}{m_1.c_1} = \frac{Q_2}{m_2.c_2}

<em>when there is no heat loss from the system of two bodies then </em>Q_1=Q_2

\frac{1}{m.c_1} =\frac{1}{m.c_2}

\Rightarrow c_1=c_2

  • Thermal conductivity is ultimately affects the rate of heat transfer, however the bodies will attain their final temperature based upon their mass and their specific heat capacities.

The temperature of the colder object will rise twice as much as the temperature of the hotter object only in two cases:

  • when the specific heat of the colder object is half the specific heat of the hotter object while mass is equal for both.

OR

  • the mass of colder object is half the mass of the hotter object while their specific heat is same.
3 0
3 years ago
The specification limits for a product are 8 cm and 10 cm. A process that produces the product has a mean of 9.5 cm and a standa
My name is Ann [436]

Answer:

The value of Cpk is 0.83.

Explanation:

Given that,

Upper specification limits = 10 cm

lower specification limits = 8 cm

Mean = 9.5

Standard deviation = 0.2 cm

We need to calculate the process capability

Using formula of Cpk

Cpk=min(\dfrac{USL-mean}{3\times SD}, \dfrac{mean-LSL}{3\times SD})

Put the value into the formula

Cpk=min(\dfrac{10-9.5}{3\times0.2}, \dfrac{9.5-8}{3\times0.2})

Cpk=min(0.83,2.5)

Cpk=0.83

Hence, The value of Cpk is 0.83.

4 0
2 years ago
It takes a minimum distance of 57.46 m to stop a car moving at 13.0 m/s by applying the brakes (without locking the wheels). Ass
vivado [14]

Answer:

The minimum stopping distance when the car is moving at

29.0 m/sec = 285.94 m

Explanation:

We know by equation of motion that,

v^{2}=u^{2}+2\cdot a \cdot s

Where, v= final velocity m/sec

u=initial velocity m/sec

a=Acceleration m/Sec^{2}

s= Distance traveled before stop m

Case 1

u=  13 m/sec, v=0, s= 57.46 m, a=?

0^{2} = 13^{2}  + 2 \cdot a \cdot57.46

a = -1.47 m/Sec^{2} (a is negative since final velocity is less then initial velocity)

Case 2

u=29 m/sec, v=0, s= ?, a=-1.47 m/Sec^{2} (since same friction force is applied)

v^{2} = 29^{2}  - 2 \cdot 1.47 \cdot S

s = 285.94 m

Hence the minimum stopping distance when the car is moving at

29.0 m/sec = 285.94 m

4 0
2 years ago
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