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3 years ago

A ball is thrown upward at a velocity of 19.6 m/s. What is its velocity after 3.0, assuming negligible air resistance?

2 answers:

5 0

If the ball travels 19.6 meters every second and it travels for 3 seconds, obviously you have to multiply the velocity by 3:

19.6 x 3= 58.8m/s

MrRa [10]3 years ago

5 0

Answer:

After 3 seconds the velocity becomes 49 m/s

Explanation:

Applying upward displacement equation:

$Y = V_yt +\frac{1}{2}gt^2$

differentiate this equation,

$\frac{dy}{dt} = V_y + gt$

where; dy/dt is the velocity function;

After 3 seconds, the velocity becomes;

V = 19.6 + (9.8)(3)

V = 19.6 + 29.4

V = 49 m/s

Therefore, after 3 seconds the velocity becomes 49 m/s

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2 years ago

Consider the hydrogen atom. How does the energy difference between adjacent orbit radii change as the principal quantum number i

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Answer:

the energy difference between adjacent levels decreases as the quantum number increases

Explanation:

The energy levels of the hydrogen atom are given by the following formula:

$E=-E_0 \frac{1}{n^2}$

where

$E_0 = 13.6 eV$ is a constant

n is the level number

We can write therefore the energy difference between adjacent levels as

$\Delta E=-13.6 eV (\frac{1}{n^2}-\frac{1}{(n+1)^2})$

We see that this difference decreases as the level number (n) increases. For example, the difference between the levels n=1 and n=2 is

$\Delta E=-13.6 eV(\frac{1}{1^2}-\frac{1}{2^2})=-13.6 eV(1-\frac{1}{4})=-13.6 eV(\frac{3}{4})=-10.2 eV$

While the difference between the levels n=2 and n=3 is

$\Delta E=-13.6 eV(\frac{1}{2^2}-\frac{1}{3^2})=-13.6 eV(\frac{1}{4}-\frac{1}{9})=-13.6 eV(\frac{5}{36})=-1.9 eV$

And so on.

So, the energy difference between adjacent levels decreases as the quantum number increases.

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3 years ago

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2 years ago

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2 years ago

An object has a height of 0.066 m and is held 0.210 m in front of a converging lens with a focal length of 0.140 m. (Include the

Anvisha [2.4K]

Explanation:

Given that,

Size of object, h = 0.066 m

Object distance from the lens, u = 0.210 m (negative)

Focal length of the converging lens, f = 0.14 m

If v is the image distance from the lens, we can find it using lens formula as follows :

$\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}\\\\\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{0.14 }+\dfrac{1}{(-0.21)}\\\\v=0.42\ m$

(a) Magnification,

$m=\dfrac{v}{u}\\\\m=\dfrac{0.42}{(-0.21)}\\\\m=-2$

(b) Magnification, $m=\dfrac{h'}{h}$

h' is image height

$-2=\dfrac{h'}{0.066}\\\\h'=-2\times 0.066\\\\h'=-0.132\ m$

Hence, this is the required solution.

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2 years ago