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Lorico [155]
4 years ago
5

Diagram of the dermis

Physics
1 answer:
lions [1.4K]4 years ago
6 0
Unlike the epidermis, this layer is connected to the blood and lymph supply as well as the nerves. Running through the dermis are sweat and sebaceous glands, hair follicles and muscles cells. Fibroblasts: responsible for the production of areolar tissue, collagen and elastin.
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Why are units of measurement useful?
Zanzabum

Without the ability to measure, it would be difficult for scientists to conduct experiments or form theories. Not only is measurement important in science and the chemical industry, it is also essential in farming, engineering, construction, manufacturing, commerce, and numerous other occupations and activities.

5 0
4 years ago
Read 2 more answers
n the figure, a uniform ladder 12 meters long rests against a vertical frictionless wall. The ladder weighs 400 N and makes an a
Leto [7]

Since the ladder is standing, we know that the coefficient of friction is at least something. This [gotta be at least this] friction coefficient can be calculated. As the man begins to climb the ladder, the friction can even be less than the free-standing friction coefficient. However, as the man climbs the ladder, more and more friction is required. Since he eventually slips, we know that friction is less than what's required at the top of the ladder.

 

The only "answer" to this problem is putting lower and upper bounds on the coefficient. For the lower one, find how much friction the ladder needs to stand by itself. For the most that friction could be, find what friction is when the man reaches the top of the ladder.

Ff = uN1

Fx = 0 = Ff + N2

Fy = 0 = N1 – 400 – 864

N1 = 1264 N

Torque balance

T = 0 = N2(12)sin(60) – 400(6)cos(60) – 864(7.8)cos(60)

N2 = 439 N

Ff = 439= u N1

U = 440 / 1264 = 0.3481

3 0
3 years ago
Read 2 more answers
The light bulb transfers electrical energy into light. What is one type of energy that is also generated that is NOT a desired e
zimovet [89]

The old style (incandescent) light bulb converts more energy
into heat than it does into light.  If you're using it mainly as a
source of light, then it's a bummer, and its efficiency is very low. 
BUT ... if you're using an incandescent light bulb as a heater, then
its efficiency is much better.  It all depends on your point of view.

8 0
3 years ago
Read 2 more answers
A block of mass m = 2.5 kg is attached to a spring with spring constant k = 730 N/m. It is initially at rest on an inclined plan
Alona [7]

Answer:

A) Em = 4.41 J

B) L = 0.33m

Explanation:

A) The total mechanical energy of the block is the elastic potential energy due to the compressed spring. The gravitational energy is zero. Then you have:

E_m=\frac{1}{2}k(\Delta x)^2

k: constant's spring = 730 N/m

Δx: distance of the compression = 0.11m

You replace the values of k and Δx:

E_m=\frac{1}{2}(730N/m)(0.11m)^2=4.41\ J

B) To find the distance L traveled by the block you take into account that the total mechanical energy of the block is countered by the work done by the friction force, and also by the work done by the gravitational energy.

Then, you have:

E_m-W_f-W_g=0\\\\W_f=(\mu Mg cos\theta)L\\\\W_g=(Mgsin\theta)L

μ:  coefficient of kinetic friction = 0.19

g: gravitational acceleration = 9.8m/s^2

M: mass of the block = 2.5kg

θ: angle of the inclined plane = 21°

You replace the values of all parameters:

E_m-W_f-W_g=0\\\\4.41-(0.19)(2.5kg)(9.8m/s^2)(cos21\°)L-(2.5kg)(9.8m/s^2)(sin21\°)L=0\\\\4.41-4.34L-8.78L=0\\\\4.41-13.12L=0\\\\L=0.33m

hence, the distance L in which the block stops is 0.33m

5 0
3 years ago
1. Swordfish are capable of stunning output power for short bursts. A 650 kg swordfish has a cross-sectional area of 0.92 m2 and
Oksana_A [137]

Answer:

P_{sp}=178.4W/kg

Explanation:

From the question we are told that:

Mass of fish m_f=650kg

Cross-sectional area A=0.92 m^2

Drag coefficient of \mu= 0.0091

Seawater  density \rho= 1026 kg/m^3.

Speed of  Fish v=30 m/s  

Generally the equation for Drag force F_d is mathematically given by

F_d = \mu * \rho *A v^2 /2

F_d = 0.0091* 0.92* 1026* 30^2/2 \\F_d= 3865. 35 N  

Generally the equation for high speed  Power  P_{sp} is mathematically given by

P_{sp}=3865*35*\frac{v}{m_f}

P_{sp}=F_d*35*\frac{30}{650}

P_{sp}=178.4W/kg

5 0
3 years ago
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