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2 years ago

14

A stone is thrown horizontally at 15 m/s from the top of a cliff 40 meters high. How far from the base does the stone hit the gr

ound?
PLS HELP

1 answer:

Reil [10]2 years ago

5 0

Answer:

She stone hit ground 42.86 m far from base of cliff.

Explanation:

Initial height from ground = 40 meter.

We have equation of motion , $s= ut+\frac{1}{2} at^2$ , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

In this the velocity of body in vertical direction = 0 m/s, acceleration = 9.8 $m/s^2$ , we need to calculate time when s = 40 meter.

Substituting

$40=0*t+\frac{1}{2} *9.8*t^2\\ \\ t = 2.86 seconds$

So it will take 2.86 seconds to reach ground.

So, stone travels horizontally at 15 m/s for 2.86 seconds.

Distance travelled = 15 x 2.86 = 42.86 m

So, the stone hit ground 42.86 m far from base of cliff.

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Find the potential inside and outside a uniformly charged solid sphere whose radius is R and whose total charge is q. Use infini

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Answer:

Recall that the electric field outside a uniformly charged solid sphere is exactly the same as if the charge were all at a point in the centre of the sphere:

$E_{outside} =\frac{1}{4\pi(e_{0})}\frac{Q}{r^{2} } r^{'}$

lnside the sphere, the electric field also acts like a point charge, but only for the proportion of the charge further inside than the point r:

$E_{inside} =\frac{1}{4\pi(e_{0})}\frac{Q}{R^{2} } \frac{r}{R} r^{'}$

To find the potential, we integrate the electric field on a path from infinity (where of course, we take the direct path so that we can write the it as a 1 D integral):

$V(r>R)=\int\limits^r_\infty {\frac{1}{4\pi(e_{0)} }\frac{Q}{r^2} } \, dr=\frac{q}{4\pi(e_{0)} } \frac{1}{r} \\V(r$

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3 years ago

Use the Bohr model to address this question. When a hydrogen atom makes a transition from the 5 th energy level to the 2nd, coun

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Answer:

A. 2.82 eV

B. 439nm

C. 59.5 angstroms

Explanation:

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$E_{n1,n2}=-13.4(\frac{1}{n_2^2}-\frac{1}{n_1^2})$ (1)

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n2: initial state = 2

Where the energy is electron volts. You replace the values of n1 and n2 in the equation (1):

$E_{5,2}=-13.6(\frac{1}{5^2}-\frac{1}{2^2})=2.82eV$

B. The energy of the emitted photon is given by the following formula:

$E=h\frac{c}{\lambda}$ (2)

h: Planck's constant = 6.62*10^{-34} kgm^2/s

c: speed of light = 3*10^8 m/s

λ: wavelength of the photon

You first convert the energy from eV to J:

$2.82eV*\frac{1J}{6.242*10^{18}eV}=4.517*10^{-19}J$

Next, you use the equation (2) and solve for λ:

$\lambda=\frac{hc}{E}=\frac{(6.62*10^{-34} kg m^2/s)(3*10^8m/s)}{4.517*10^{-19}J}=4.39*10^{-7}m=439*10^{-9}m=439nm$

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$r_n=n^2a_o$ (3)

where ao is the Bohr's radius = 2.380 Angstroms

You use the equation (3) with n=5:

$r_5=5^2(2.380)=59.5$

hence, the radius of the atom in its 5-th state is 59.5 anstrongs

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2 years ago

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