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3 years ago

5

1) The unit of power is

1 answer:

miss Akunina [59]3 years ago

4 0

Answer:

1) the unit of power is A.Watt

2)100J=50W

100=50s

divide both sides by 50s

S=2s(A)

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What is the highest energy photon that can be absorbed by a ground-state hydrogen atom without causing ionization?

lakkis [162]

Highest energy photon absorbed: $2.18\cdot 10^{-18}J$

Explanation:

An atom is said to be (positively) ionised when it absorbs a photon, and as a consequence, an electron becomes energetic enough to escape the atom, leaving an excess of positive charge behind.

In order for the electron to escape, the energy of the absorbed photon must be exactly equal to the (negative) energy of the level in which the electron lies.

For an hydrogen atom, the energy levels are given by

$E_n = -13.6 \frac{1}{n^2}$

where this energy is measured in electronvolts, and n is the number of the energy level.

Since the energy is negative, this means that the electron which requires most energy is the one lying in the ground state (n=1). Therefore, for an electron in the ground state, the most energy that can be absorbed from the incoming photon is

$E_1 = -13.6 eV$

Converting into Joules, this is equal to

$E_1 = 13.6 \cdot 1.6\cdot 10^{-19}=2.18\cdot 10^{-18}J$

Learn more about hydrogen atom:

brainly.com/question/2757829

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3 0

3 years ago

An air balloon is moving upward at a constant speed of 3 m/s. Suddenly a passenger realizes that she left her camera on the grou

frosja888 [35]

Answer:t=0.3253 s

Explanation:

Given

speed of balloon is $u=3\ m/s$

speed of camera $u_1=20\ m/s$

Initial separation between camera and balloon is $d_o=5\ m$

Suppose after t sec of throw camera reach balloon then,

distance travel by balloon is

$s=ut$

$s=3\times t$

and distance travel by camera to reach balloon is

$s_1=ut+\frac{1}{2}at^2$

$s_1=20\times t-\frac{1}{2}gt^2$

Now

$\Rightarrow s_1=5+s$

$\Rightarrow 20\times t-\frac{1}{2}gt^2 =5+3t$

$\Rightarrow 5t^2-17t+5=0$

$\Rightarrow t=\dfrac{17\pm \sqrt{17^2-4(5)(5)}}{2\times 5}$

$\Rightarrow t=\dfrac{17\pm 13.747}{10}$

$\Rightarrow t=0.3253\ s\ \text{and}\ t=3.07\ s$

There are two times when camera reaches the same level as balloon and the smaller time is associated with with the first one .

(b)When passenger catches the camera time is $t=0.3253\ s$

velocity is given by

$v=u+at$

$v=20-10\times 0.3253$

$v=16.747\ m/s$

and position of camera is same as of balloon so

Position is $=5+3\times 0.3253$

$=5.975\approx 6\ m$

8 0

3 years ago

If a specimen was being viewed using a 20x objective lens and 10x ocular lens, what would be the total magnification

Paraphin [41]

Answer:

As Per Given Information

20x objective lens was used by specimen

10x ocular lens was also used by him.

we have to find the total magnification.

For calculating the total magnification we 'll simply do multiplication

Total Magnification = 20x × 10x

Total Magnification = 200x

So , the total magnification will be 200x .

6 0

2 years ago

A bat hits a moving baseball. If the bat delivers a net eastward impulse of 1.3 N-s and the ball starts with an initial horizont

Georgia [21]

Answer:

$141.30grams$

Explanation:

First, denote our known values;

$J=1.3N-s(+,east)\\u=-3.8ms^-^1(-,west)\\v=5.4ms^-^1(+,east)\\m=?$

Mass is impulse divided by change in velocity:

$m=\frac{J}{v-u}\\=\frac{1.3}{5.4--3.8}\\\\=\frac{1.3}{9.2}\\=0.1413Kgs$

Hence, the mass of the ball is 141.30grams

3 0

3 years ago

4. A car accelerating at 60 meters/second is hit in an accident by a bus. The net force exerted on

Licemer1 [7]

Answer:

500kg

Explanation:

mass = newtons/force divided by the acceleration rate

m = 30,000/60

m = 500

3 0

3 years ago