A race car decelerating by 3 m /s to 10 m / s has an acceleration of 25

A horizontal pipeline is 660 mm in diameter and carries oil at a rate of 150 kg/s. this oil has a mass density of 850 kg/m3 and

timurjin [86]

A boiling pot of water (the water travels in a current throughout the pot), a hot air balloon (hot air rises, making the balloon rise) , and cup of a steaming, hot liquid (hot air rises, creating steam) are all situations where convection occurs.
Read more on Brainly.com - brainly.com/question/1581851#readmore

6 0

3 years ago

The inductance of a closely packed coil of 560 turns is 8.9 mH. Calculate the magnetic flux through the coil when the current is

Y_Kistochka [10]

Answer:

$0.11\times 10^{-6}weber$

Explanation:

We have given number of turns N = 560

Inductance L = 8.9 mH

Current through the coil = 7 mA

Inductance of the coil is given as $L=\frac{N\Phi }{I}$

Where N is number of turns I is current and $\Phi$ is flux

So $\Phi =\frac{LI}{N}=\frac{8.9\times 10^{-3}\times 7\times 10^{-3}}{560}=0.11\times 10^{-6}weber$

6 0

3 years ago

Read 2 more answers

What is the energy per photon absorbed during the transition from n = 2 to n = 3 in the hydrogen atom?

adelina 88 [10]

Answer : The energy of one photon of hydrogen atom is, $3.03\times 10^{-19}J$

Explanation :

First we have to calculate the wavelength of hydrogen atom.

Using Rydberg's Equation:

$\frac{1}{\lambda}=R_H\left(\frac{1}{n_i^2}-\frac{1}{n_f^2} \right )$

Where,

$\lambda$ = Wavelength of radiation

$R_H$ = Rydberg's Constant = 10973731.6 m⁻¹

$n_f$ = Higher energy level = 3

$n_i$ = Lower energy level = 2

Putting the values, in above equation, we get:

$\frac{1}{\lambda}=(10973731.6)\left(\frac{1}{2^2}-\frac{1}{3^2} \right )$

$\lambda=6.56\times 10^{-7}m$

Now we have to calculate the energy.

$E=\frac{hc}{\lambda}$

where,

h = Planck's constant = $6.626\times 10^{-34}Js$

c = speed of light = $3\times 10^8m/s$

$\lambda$ = wavelength = $6.56\times 10^{-7}m$

Putting the values, in this formula, we get:

$E=\frac{(6.626\times 10^{-34}Js)\times (3\times 10^8m/s)}{6.56\times 10^{-7}m}$

$E=3.03\times 10^{-19}J$

Therefore, the energy of one photon of hydrogen atom is, $3.03\times 10^{-19}J$

3 0

3 years ago

The speed that a tsunami can travel is modeled by the equation , where s is the speed in kilometers per hour and d is the averag

CaHeK987 [17]

The speed of tsunami is a.0.32 km.

Steps involved :

The equation s = 356d models the maximum speed that a tsunami can move at. It reads as follows: s = 200 km/h d =?

Let's now change s to s in the equation to determine d: s = 356√d 200 = 356√d √d = 200 ÷ 356 √d = 0.562 Let's square the equation now by squaring both sides: (√d)² = (0.562) ² d = (0.562)² = 0.316 ≈ 0.32

As a result, 0.32 km is roughly the depth (d) of water for a tsunami moving at 200 km/h.

To learn more about tsunami refer : brainly.com/question/11687903

#SPJ4

6 0

1 year ago

"Which of the following is true about the atom shown? Choose all that apply.

aalyn [17]

Option B & Option D is your correct answers.

'cause here, atom has eight electrons in it's valence shell, so it means it has stable structure which falls in group 18

Hope this helps!

8 0

2 years ago

Read 2 more answers