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jonny [76]
3 years ago
14

How many calories would you take in if you ate the whole box of crackers in one sitting ?

Physics
1 answer:
sweet [91]3 years ago
7 0

Answer:

i'm pretty sure it's 81 calories

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When the tube is filled with mercury vapor, as in this case, a sharp drop in the collected current is observed when the accelera
Nata [24]

Answer:

The energy absorbed by the atomic electrons in the mercury atom is 7.84 \times 10^{-19} J

Explanation:

Given:

Potential V = 4.9 V

According to the conservation law,

   Loss in kinetic energy = Gain in potential energy

Here, energy absorbed by the atomic electrons is given by,

    E = eV

Where e = 1.6 \times 10^{-19} C       ( charge of electron )

    E = 1.6 \times 10^{-19 } \times 4.9

    E = 7.84 \times 10^{-19} J

Therefore, the energy absorbed by the atomic electrons in the mercury atom is 7.84 \times 10^{-19} J

5 0
3 years ago
The energy of the electron in Hydrogen atom can be shown to be given by En = -13.6/n^2 eV, where n represents the principal quan
Tpy6a [65]

Answer:

This represents radiation in ultra-violet region .

Explanation:

Energy of the orbit where n = 3 is given as follows

E_3 =- \frac{13.6 }{3^2}

E_3 =- \frac{13.6 }{9} = -1.511 eV

Energy of the orbit where n = 1 is given as follows

E_1 =- \frac{13.6 }{1^2}

E_1 =- \frac{13.6 }{1} = 13.6 eV

Difference of [tex]E_3 and [tex]E_1 = - 1.511+ 13.6

= 12.089 eV.

The wavelength of light having this energy in nm is given by the expression as follows

Wavelength in nm = 1244 / energy in eV

= 1244 / 12.089

= 102.90 nm

This represents radiation in ultra-violet region .

8 0
3 years ago
During a rainy day, as a result of colliding clouds an observer saw lighting and a heard thunder sound. The time between seeing
hodyreva [135]

Answer:

d = 1700 meters

Explanation:

During a rainy day, as a result of colliding clouds an observer saw lighting and a heard thunder sound.  The time between seeing the lighting and hearing the sound was 5 second, t = 5 seconds

Speed of sound, v = 340 m/s (say)

Let d is the distance of the colliding cloud from the observer. The distance covered by the object. It is given by :

d=v\times t

d=340\ m/s\times 5\ s  

d = 1700 meters

So, the distance of the colliding cloud from the observer is 1700 meters. Hence, this is the required solution.    

8 0
3 years ago
Please help me with a speech.The topic of my speech is Passion.​
Vikentia [17]

Answer:

Explanation:

Passion

For me, standing on the summit of Mt Everest was the result of following a process. The process of mountaineering. I love mountaineering. I am passionate about it. I love the months of planning for an expedition, the months of sweating and training to prepare my body physically. The meticulous preparation of my equipment. Most of all I love the huge mental challenge I have to overcome before each climb to confront my own fear. All these reasons are why I climb, they are why I climbed Mt Everest and that is why I continue to climb.

Passion is an enormously powerful force. It gives us the strength to get through hard times and setbacks. It gives us strength to overcome our fears, to ignore what other people think of us, to be disciplined and make sacrifices in pursuit of our dreams. Passionate people do not want to take shortcuts – they consider that ‘learning the process’ is an important part of the journey.

In mountaineering it’s easy to spot those who are not passionate about the process. They want to stand on top of the mountain but they are not really interested in the process of climbing the mountain. I feel for these people. Success without hard work is a hollow, empty feeling. They never last long in the sport.

Just as in life, successful mountaineers are the ones who are passionate. They are not there just to stand on the summit. Their passion gives them the energy to work the hardest, fight the longest, and in the words of Winston Churchill “never, never. never give-up”.

3 0
3 years ago
The origin of an x axis is placed at the center of a nonconducting solid sphere of radius R that carries a charge +qsphere distr
MA_775_DIABLO [31]

Answer:

q=49Q/64

and

x =16R/15

Explanation:

See  attached figure.

E_{Q}= E due to sphere

E_{q}= E due to particule

E_{total}=E_{Q}-E_{q}=0  (1)

according to the law of gauss and superposition Law:

E_{Q}=E_{1}+E_{2}=E_{2} ; electric field due to the small sphere with r1=R/4

E_{Q}=kq_{2}/(r_{1}^{2})=

q_{2}=density*4/3*pi*r_{1}^{3}=Q/(4/3*pi*R^{3})*4/3*pi*r_{1}^{3}=Q*r_{1}^{3}/R^{3}

then: E_{Q}=kq_{2}/(r_{1}^{2})=k*Q*r_{1}^{3}/(R^{3}*r_{1}^{2}) = kQ/(4*R^{2})  (2)

on the other hand, for the particule:

E_{q}=kq/(r_{p}^{2})

r_{p}=2R-R/4=7R/4   ⇒    E_{q}=16kq/(49R^{2})   (3)

We replace (2) y (3) in (1):

E_{total}=E_{Q}-E_{q}=0=kQ/(4*R^{2}) - 49kq/(16R^{2})

q=49Q/64

--------------------

if R<x<2R   AND E_{total}=E_{Q}-E_{q}=0

E_{total}=E_{Q}-E_{q}=0=kQ/(x^{2}) - kq/(2R-x^{2})

remember that  q=49Q/64

then:

Q(2R-x^{2})=49/64*x^{2}

solving:

x_{1} =16R/15

x_{2} =16R

but: R<x<2R  

so : x =16R/15

7 0
3 years ago
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