The -is the process by which gases in the atmosphere absorb and reradiate heat?

gulaghasi [49]

Visceral epithelial cells

5 0

3 years ago

German physicist Werner Heisenberg related the uncertainty of an object's position (Δx) to the uncertainty in its velocity (Δ???

fredd [130]

Answer:

$\Delta x = 5.47 \times 10^{-9} m$

Explanation:

As we know by the principle of uncertainty that the product of uncertainty in position and uncertainty in momentum is given as

$\Delta x \times \Delta P = \frac{h}{4\pi}$

so here we know that

$\Delta v = 0.01 \times 10^6 m/s$

$m = 9.11 \times 10^{-31} kg$

so we have

$\Delta x \times (9.11 \times 10^{-31})(0.01 \times 10^6) = \frac{6.26 \times 10^{-34}}{4\pi}$

$\Delta x = 5.47 \times 10^{-9} m$

5 0

3 years ago

Would time travel mean that parallel universes would be created off our linear time sequence to ensure the universe doesn't coll

MatroZZZ [7]

In theroy, yes. But time travel has yet to happen and thus, we dont know. I would assume that that if an alternate universe were NOT created, it would cause a rip in the fabic of time creating a worm hole and allowing all entitys to pass through simultaneously,and that would cause great stress on it, possibly causing it to rip large enough to encase the whole universe and it would implode.

4 0

3 years ago

Read 2 more answers

A ball of moist clay falls 17.3 m to the ground. It is in contact with the ground for 24.0 ms before stopping. (a) What is the a

gizmo_the_mogwai [7]

Answer:

Acceleration, $767.08\ m/s^2$

Explanation:

Given that,

Height from a ball falls the ground, h = 17.3 m

It is in contact with the ground for 24.0 ms before stopping.

We need to find the average acceleration the ball during the time it is in contact with the ground.

Firstly, find the velocity when it reached the ground. So,

$v^2=u^2+2ah$

u = initial velocity=0 m/s

a = acceleration=g

$v=\sqrt{2gh} \\\\v=\sqrt{2\times 9.8\times 17.3} \\\\v=18.41\ m/s$

It is in negative direction, u = -18.41 m/s

Let a is average acceleration of the ball. Consider, v = and u = -18.41 m/s.

$a=\dfrac{v-u}{t}\\\\a=\dfrac{0-18.41}{24\times 10^{-3}}\\\\a=767.08\ m/s^2$

So, the average acceleration of the ball during the time it is in contact is $767.08\ m/s^2$ .

4 0

3 years ago

A 5-cm-high peg is placed in front of a concave mirror with a radius of curvature of 20 cm.

Andrei [34K]

Answer:

Explanation:

Using the magnification formula.

Magnification = Image distance(v)/object distance(u) = Image Height(H1)/Object Height(H2)

M = v/u = H1/H2

v/u = H1/H2...1

3) Given the radius of curvature of the concave lens R = 20cm

Focal length F = R/2

f = 20/2

f = 10cm

Object distance u = 5cm

Object height H2= 5cm

To get the image distance v, we will use the mirror formula

1/f = 1/u+1/v

1/v = 1/10-1/5

1/v = (1-2)/10

1/v =-1/10

v = -10cm

Using the magnification formula

(10)/5 = H1/5

10 = H1

H1 = 10cm

Image height of the peg is 10cm

4) If u = 15cm

1/v = 1/f-1/u

1/v = 1/10-1/15

1/v = 3-2/30

1/v = 1/30

v = 30cm

30/15 = H1/5

15H1 = 150

H1/= 10cm

5) if u = 20cm

1/v = 1/f-1/u

1/v = 1/10-1/20

1/v = 2-1/20

1/v = 1/20

v = 20cm

20/20 = H1/5

20H1 = 100

H1 = 5cm

6) If u = 30cm

1/v = 1/f-1/u

1/v = 1/10-1/30

1/v = 3-1/30

1/v = 2/30

v = 30/2 cm

v =>15cm

15/30 = Hi/5

30H1 = 75

H1 = 75/30

H1 = 2.5cm

4 0

3 years ago