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2 years ago

What is Diabetic Retinopathy? Describe it in at least 1 paragraph.

Physics

1 answer:

Irina18 [472]2 years ago

4 0

Answer:

Diabetic Retinopathy is a form of diabetes that affects the eyes. It can be caused by damage to the retinas, and can cause permanent damage to the eyes, and even blindness. Initially the patient is asymptomatic and become more visibly affected in later stages. It can be treated if caught early, or in mild cases.

Explanation:

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You are designing a delivery ramp for crates containing exercise equipment. The 1470-N crates will move at 1.8 m/s at the top of

alina1380 [7]

Answer:

The force constant ,I = 2394N/m

Explanation:

Given:

Weight of crate,Wg = 1470N

Theta = 22.0°

Kinetic friction,Fk= Fs(max) = 550N

Total length of ramp =8.0m

If y =0 at the bottom of the ramp

y1 = d Sin theta

y1 = 8 × Sin 22°

y1 = 3.0m

y2 = 0

V1=1.8m/s

V2 = 0

The equation combining the gravitational and elastic potential energy, and the work energy theorem is given by:

K1 + Ugrav1 + Uel1 + W = K2 + Uel1 ..eq1

Where KE is given by: 1/2mv^2

Gravitational potential energy is given by: Ugrav = mgy ...eq2

Elastic potential energy = Uel = 1/2Kx^2 ..eq3

The restoring force constant of a spring compressed by a distance x is given by:

Fx = Kx ..eq4

Workdone by a constant force is given by W = Fscostheta ...eq5

Where s = displacement

Theta = the angle between the force and the displacement

Work,W = Wf = 550 ×8 × cos 180°

W = -4400J

From the weight of the crate.mass is:

Wg/g = m

1470/9.8 = 150kg

K1 = 1/2 ×150×1.8^2 = 243m/s

The crate comes to rest at K2=0

Ugrav1 = 150 × 9.8 × 3 = 4410J

Ugrav2 = 150 × 9.8 x 0 = 0J

Uel1= 0 Spring at equilibrium

Substituting the values of the energies and work

253 + 4410 + 0 - 4400 = 0 + 0 + Uel1

Uel1 = 253J

Substituting into eq3

253 = 1/2 Kx^2 = 1/2 Kx(x)

Kx = 506/x

Since crate remains at rest,we use Newton's 2nd law

Fx = fs + Wsin theta

Fx = 550 + 1470 sin 22

Fx = 1100.7N

Substituting into eq4

Kx = 1100.7

X = 506/1100.7 = 0.46m

Kx = 1100.7

K = 1100.7/0.47

K = 2394N/m

5 0

3 years ago

A 62 kg skier is moving at 6.5 m/s on frictionless horizontal snow-covered plateau when she encounters a rough patch 3.50 m long

Degger [83]

Answer:

(A). The work done by friction in crossing the patch is -637.98 J.

(B). The speed of skier is 10.57 m/s.

Explanation:

Given that,

Mass of skier = 62 kg

Speed = 6.5 m/s

Length = 3.50 m

Coefficient kinetic friction = 0.30

Height = 2.5 m

(A) we need to calculate the work done by friction in crossing the patch

Using formula of work done

$W=-\mu mg\times l$

Put the value into the formula

$W=-0.30\times62\times9.8\times3.50$

$W=-637.98\ J$

The work done by friction in crossing the patch is -637.98 J.

(B) we need to calculate the speed of skier

Using conservation of energy

$K.E_{i}+U_{i}-W_{friction}=K.E_{f}+U_{f}$

$\dfrac{1}{2}mv_{1}^2+mgh-\mu mgl=\dfrac{1}{2}mv_{2}^2+U_{f}$

Final potential energy is zero

So, $\dfrac{1}{2}mv_{1}^2+mgh-\mu mgl=\dfrac{1}{2}mv_{2}^2$

$\dfrac{1}{2}v_{2}^2=\dfrac{1}{2}v_{1}^2+gh-\mu gl$

Put the value into the formula

$\dfrac{1}{2}v_{2}^2=\dfrac{1}{2}\times6.5^2+9.8\times2.5+0.30\times9.8\times3.50$

$v_{2}=\sqrt{2\times55.915}$

$v_{2}=10.57\ m/s$

The speed of skier is 10.57 m/s.

Hence, (A).The work done by friction in crossing the patch is -637.98 J.

(B).The speed of skier is 10.57 m/s.

6 0

3 years ago

Is electrical conductivity?

expeople1 [14]

Answer:

can't tell if this is question, it is not written correctly

Explanation:

Electrical conductivity is the measure of a material's ability to allow the transport of an electric charge. Its SI is the siemens per meter, (A2s3m−3kg−1) (named after Werner von Siemens) or, more simply, Sm−1. It is the ratio of the current density to the electric field strength.

8 0

3 years ago

It is measured that 3/4 of a body's volume is submerged in oil of density 800kg/m³

Evgesh-ka [11]

Complete question:

It is measured that 3/4 of a body's volume is submerged in oil of density 800kg/m³. What is the specific gravity of oil?

Answer:

The specific gravity of the oil is 0.8.

Explanation:

Given;

density of the oil, $\rho_o$ = 800 kg/m³

density of water, $\rho_w$ = 1000 kg/m³

The specific gravity of any substance is the ratio of the substance density to the density of water.

Specific gravity of the oil = density of the oil / density of water

Specific gravity of the oil = 800/1000

Specific gravity of the oil = 0.8

Therefore, the specific gravity of the oil is 0.8.

8 0

3 years ago

Hi i need help. thNKS

Ede4ka [16]

Answer:

Explanation:

Groups go down, periods go across :)

3 0

2 years ago