All categories

3 years ago

Why can even a small injury to the cornea have a major effect on vision?

Physics

1 answer:

Rudik [331]3 years ago

8 0

Your eyes are the most delicate thing on your body. Little things can make you go blind

You might be interested in

A circus acrobat is shot out of a cannon with an initial upward speed of 34 ft/s. if the acrobat leaves the cannon 4 ft above th

jok3333 [9.3K]

<h2>It will take 0.125 seconds to reach the net.</h2>

Explanation:

Initial speed, u = 34 ft/s = 10.36 m/s

Acceleration, a = -9.81 m/s²

Displacement, s = Final height - Initial height = 8 - 4 = 4 ft = 1.22 m

We have equation of motion, s = ut + 0.5 at²

Substituting

s = ut + 0.5 at²

1.22 = 10.36 x t + 0.5 x -9.81 x t²

4.905t² - 10.36 t + 1.22 = 0

t = 1.99 s or t = 0.125 seconds

Minimum time is 0.125 seconds.

It will take 0.125 seconds to reach the net.

6 0

3 years ago

Which factors affect heat transfer between a warm and a cool substance?

damaskus [11]

Factors that affect heat transfer are:

1) Difference in temperature,
2) Mass of the object
3) Specific heat of the object

Hope this helps!

8 0

3 years ago

You're driving your new sports car at 80 mph over the top of a hill that has a radius of curvature of 540 m. What fraction of yo

luda_lava [24]

Answer:

75.84%

Explanation:

We were given Speed of the sports car, v as 80 mph , we can convert to m/s for unit consistency.

v=80mph= 35.76 m/s

The radius of curvature is given as , r = 540 m

✓ the normal weight can be denoted as Wn

✓ the apparent weight of the person can be denoted as Wa

Wn= normal weight= mg

Wa=apparent weight = (mg - mv^2/r)

g= acceleration due to gravity= 9.8m/s^2

The apparent weightand normal weight has a ratio of

Mn/Ma= [mg - mv^2/r]/mg ........eqn(1)

If we simplify eqn(1) we have

Mn/Ma=[g - vr^2/g].............eqn(2)

Then substitute the given values

Mn/Ma=9.8 - [(35.76^2)/540]/ 9.8

=0.758×100%

Mn/Ma=75.84%

Hence, the required fraction is 75.84%

4 0

3 years ago

Block A of mass M is on a horizontal surface of negligible friction. An identical block B is attached to block A by a light stri

miv72 [106K]

Answer:

T’= 4/3 T

The new tension is 4/3 = 1.33 of the previous tension the answer e

Explanation:

For this problem let's use Newton's second law applied to each body

Body A

X axis

T = m_A a

Axis y

N- W_A = 0

Body B

Vertical axis

W_B - T = m_B a

In the reference system we have selected the direction to the right as positive, therefore the downward movement is also positive. The acceleration of the two bodies must be the same so that the rope cannot tension

We write the equations

T = m_A a

W_B –T = M_B a

We solve this system of equations

m_B g = (m_A + m_B) a

a = m_B / (m_A + m_B) g

In this initial case

m_A = M

m_B = M

a = M / (1 + 1) M g

a = ½ g

Let's find the tension

T = m_A a

T = M ½ g

T = ½ M g

Now we change the mass of the second block

m_B = 2M

a = 2M / (1 + 2) M g

a = 2/3 g

We seek tension for this case

T’= m_A a

T’= M 2/3 g

Let's look for the relationship between the tensions of the two cases

T’/ T = 2/3 M g / (½ M g)

T’/ T = 4/3

T’= 4/3 T

The new tension is 4/3 = 1.33 of the previous tension the answer e

4 0

3 years ago

What is the speed of a wave in (m/s) with a 5 meter wavelength and a period of 20 seconds?

arlik [135]

Answer: 0.25 m/s

Explanation: Speed = wavelengt · frequency

v = λf and frequency is 1/period f = 1/T

Then v = λ/T = 5 m / 20 s = 0.25 m/s

6 0

3 years ago