Could anyone help me with this?

Should people eat animals?

Lubov Fominskaja [6]

Answer:

I hope it's helpful for you ☺️

4 0

2 years ago

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The energy required to ionize magnesium is 738 kj/mol. What minimum frequency of light is required to ionize magnesium

Cerrena [4.2K]

Answer:

The minimum frequency required to ionize the photon is 111.31 × $10^{37}$ Hertz

Given:

Energy = 378 $\frac{kJ}{mol}$

To find:

Minimum frequency of light required to ionize magnesium = ?

Formula used:

The energy of photon of light is given by,

E = h v

Where E = Energy of magnesium

h = planks constant

v = minimum frequency of photon

Solution:

The energy of photon of light is given by,

E = h v

Where E = Energy of magnesium

h = planks constant

v = minimum frequency of photon

738 × $10^{3}$ = 6.63 × $10^{-34}$ × v

v = 111.31 × $10^{37}$ Hertz

The minimum frequency required to ionize the photon is 111.31 × $10^{37}$ Hertz

7 0

3 years ago

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1 If you measured the distance travelled by a snail in

Fittoniya [83]

The answer is m/s hope it helps

3 0

2 years ago

Elect the correct answer. Two identical projectiles, A and B, are launched with the same initial velocity, but the angle of laun

allsm [11]

Answer:

A. The range of A and B are equal.

Explanation:

Let u be the initial speed of both the projectile.

The gravitational force acting in the projectile is in the downward direction, sp the speed of the projectile in the horizontal direction remains constant and equals to the initial horizontal speed.

For projectile, a projectile having initial velocity u at an angle \theta with the horizontal direction,

The speed in the horizontal direction $= u\cos\theta$

and the speed in the vertical direction is $= u\sin\theta$ upward.

For A:

The speed in the horizontal direction $= u\cos75^{\circ}$

and the speed in the vertical direction is $= u\sin75^{\circ}$ upward.

For B:

The speed in the horizontal direction $= u\cos15^{\circ}$

and the speed in the vertical direction is $= u\sin15^{\circ}$ upward.

Let $t_A$ and $t_B$ are the time of flight for projectile A and B respectively.

As the range is the horizontal distance traveled by the projectile, so

The range for the projectile A $= u\cos75^{\circ}\times t_A\cdots(i)$

The range for the projectile B = $u\cos15^{\circ}\times t_B\cdots(ii)$

At the highest point, the vertical velocity is 0.

Bu using the equation of motion $v^2=u^2 +2a s.$

Here, the final velocity v=0, the initial velocity $u = u \sin \theta$ , h= vertical distance up to the highest point, and $a= -g$ (as per sign convention).