One of the earliest vertebrate animal groups that evolved in the early Paleozoic Era are

Because of the curvature of the earth, the maximum distance D that you can see from the top of a tall building of height h is es

mojhsa [17]

Answer:

D = 9.9 10⁶ mi

Explanation:

In the exercise they give the expression for maximum viewing distance

D = 2 r h + h²

Ask for this distance for a height of 1100 feet

Let's calculate

D = 2 3960 1100 + 1100²

D = 8.712 10⁶ + 1.21 10⁶

D = 9.92 10⁶ mi

D = 9.9 10⁶ mi

8 0

3 years ago

Parallel rays of monochromatic light with wavelength 571nm illuminate two identical slits and produce an interference pattern on

Natasha2012 [34]

Answer:

$8.8\times 10^{-6} W/m^2$

Explanation:

We are given that

Wavelength, $\lambda=571 nm=571\times 10^{-9} m$

$1 nm=10^{-9} m$

R=75 cm= $\frac{75}{100}=0.75 m$

1 m=100 cm

$d=0.640 mm=0.64\times 10^{-3} m$

$1 mm=10^{-3} m$

$a=0.434 mm=0.434\times 10^{-3} m$

$y=0.830 mm=0.830\times 10^{-3} m$

$I_0=5.00\times 10^{-4}W/m^2$

$tan\theta=\frac{y}{R}$

$\theta=\frac{0.830\times 10^{-3}}{0.75\times 10^{-2}}$

$\theta=1.1\times 10^{-3}rad$

$tan\theta\approx \theta$

Because $\theta$ is small.

$\phi=\frac{2\pi dsin\theta}{\lambda}$

$\sin\theta\approx \theta$ ,

Therefore

$\phi=\frac{2\times\pi\times 0.64\times 10^{-3}\times 1.1\times 10^{-3}}{571\times 10^{-9}}$

$\phi=7.74 rad$

$\beta=\frac{2\pi a\theta}{\lambda}$

$\beta=5.3 rad$

$I=I_0cos^2(\frac{\phi}{2})(\frac{sin\frac{\beta}{2}}{\frac{\beta}{2}})^2$

$I=5\times 10^{-4}cos^2(\frac{7.74}{2})(\frac{sin\frac{5.3}{2}}{\frac{5.3}{2}})^2$

$I=8.8\times 10^{-6} W/m^2$

6 0

3 years ago

What formula allows you to calculate the x component of a projectile?

Kay [80]

X-component of a projectile in flight =

(initial x-component)

plus

(initial horizontal component of velocity) x (flight time so far)

8 0

3 years ago

Read 2 more answers

POR FAVOR AYUDA !!!!!!!!!

lara [203]

Lololololololololololol

8 0

3 years ago

Part A By what distance do two objects carrying 1.0 C of charge each have to be separated before the electric force exerted on e

aalyn [17]

Answer:

Part A: 4.74×10⁴ m

Part B: 0.000211 C

Explanation:

Part A:

From coulombs law,

F = kq²/r²........................ Equation 1

Note: The two object are carrying equal charges.

Where F = Electric force, q = charge, r = distance of separation, k = proportionality constant

Given: F = 4.0 N, q = 1.0 C,

Constant: k = 1/4πe₀ = 9×10⁹ Nm²/C²

Substituting these values into equation 1,

4 = 9×10⁹(1²)/r²

r² = (9×10⁹)/4

r² = 2.25×10⁹

r = √(2.25×10⁹)

r = 4.74×10⁴ m.

Part 2:

F = kq²/r²

Making q the subject of the equation

q = √(Fr²/k)........................... Equation 2

Where: F = 4.0 N, r = 1.0 m, k = 9×10⁹ Nm²/c²

Substituting these values into equation 2

q = √(4×1²/(9×10⁹ )

q = 2.11/10⁴

q = 0.000211 C.

6 0

3 years ago