Oceanic crust that records negative magnetic anomalies formed when the earth's magnetic field was __________

Monochromatic light with wavelength 588 nm is incident on a slit with width 0.0351 mm. The distance from the slit to a screen is

Norma-Jean [14]

Answer:

Explanation:

A. Using

Sinစ= y/ L = 0.013/2.7= 0.00481

စ=0.28°

B.here we use

Alpha= πsinစa/lambda

= π x (0.0351)sin(0.28)/588E-9m

= 9.1*10^-2rad

C.we use

I(စ)/Im= (sin alpha/alpha) ²

So

{= (sin0.091/0.091)²

= 3*10^-4

6 0

3 years ago

0.5-lbm of a saturated vapor is converted to asaturated liquid by being cooled in a weighted piston-cylinder device maintained a

klio [65]

Answer:

The boiling point temperature of this substance when its pressure is 60 psia is 480.275 R

Explanation:

Given the data in the question;

Using the Clapeyron equation

$(\frac{dP}{dT} )_{sat } = \frac{h_{fg}}{Tv_{fg}}$

$(\frac{dP}{dT} )_{sat } = \frac{\frac{H_{fg}}{m} }{T\frac{V_{fg}}{m} }$

where $h_{fg$ is the change in enthalpy of saturated vapor to saturated liquid ( 250 Btu

T is the temperature ( 15 + 460 )R

m is the mass of water ( 0.5 Ibm )

$V_{fg$ is specific volume ( 1.5 ft³ )

we substitute

$(\frac{dP}{dT} )_{sat } =( \frac{250Btu\frac{778Ibf-ft}{Btu} }{0.5})$ / $( (15+460)\frac{1.5}{0.5})$

$(\frac{dP}{dT} )_{sat } =$ 272.98 Ibf-ft²/R

Now,

$(\frac{dP}{dT} )_{sat } =$ $(\frac{P_2 - P_1}{T_2 - T_1})_{sat$

where P₁ is the initial pressure ( 50 psia )

P₂ is the final pressure ( 60 psia )

T₁ is the initial temperature ( 15 + 460 )R

T₂ is the final temperature = ?

we substitute;

$T_2$ $= ( 15 + 460 ) + \frac{(60-50)psia(\frac{144in^2}{ft^2}) }{272.98}$

$T_2 = 475 + 5.2751\\$

$T_2 =$ 480.275 R

Therefore, boiling point temperature of this substance when its pressure is 60 psia is 480.275 R

3 0

2 years ago

What should be the speed of an artificial satellite moving on a circular orbit around the Earth at a distance of 400 km from the

Drupady [299]

Answer:

7.67001846 km/s or 17157.38529 mph

Explanation:

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

M = Mass of the Earth = 5.972 × 10²⁴ kg

m = Mass of satellite

v = Velocity of satellite

The distance between the Earth's center and the satellite is

r = 6371000+400000 = 6771000 m

As the centripetal force balances the force of gravity we have

$\frac{mv^2}{r}=\frac{GMm}{r^2}\\\Rightarrow v=\sqrt{\frac{GM}{r}}\\\Rightarrow v=\sqrt{\frac{6.67\times 10^{-11}\times 5.972\times 10^{24}}{6771000}}\\\Rightarrow v=7670.01846\ m/s=7.67001846\ km/s$

Converting to mph

$7670.01846\times \frac{3600}{1609.34}=17157.38529\ mph$

The velocity of the satellite is 7.67001846 km/s or 17157.38529 mph

5 0

3 years ago

Is this statement true or false? Most metals in the chart react with<br> oxygen.

n200080 [17]

Answer:

true

Hoped this helped(⁄ ⁄>⁄ ▽ ⁄<⁄ ⁄)

7 0

3 years ago

A fast Humvee drove from desert A to desert B. For the first 12

nalin [4]

Answer: 172

Explanation: Because yeah

4 0

3 years ago

Oceanic crust that records negative magnetic anomalies formed when the earth's magnetic field was ___________.