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3 years ago

An apple falls out of a tree from a height of 2.3m. what is the impact speed of the apple

Physics

1 answer:

Anuta_ua [19.1K]3 years ago

5 0

Answer:6.71 m/s

Explanation:

Given

Apple fall from a height of $h=2.3 m$

We need to find the impact speed of apple which can be given by using

$v^2-u^2=2gh$

where v=final velocity

u=initial velocity

h=Displacement

Assuming initial velocity to be zero

substituting the value we get

$v^2-0=2\times 9.8\times 2.3$

$v=\sqrt{2\times 9.8\times 2.3}$

$v=6.71\ m/s$

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Consider a Hydrogen atom with the electron in the n 8 shell. What is the energy of this system? (The magnitude of the ground sta

Shtirlitz [24]

Answer:

The energy of an electron in the 8th shell is given by: -0.2125 eV

The number of subshells is: 8

The number of orbitals is: 64

The number of electrons that fit on this shell is: 128

Explanation:

First, we find the energy of the electrons in the 8th shell. In order to do this, we recall that the energy of an electron (in the Hydrogen atom) whose principal number is n is given by:

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

Substituting n=8, we find that the energy is given by:

$E_{8} = -13.6\frac{1}{8^{2}}=-0.2125$

In order to find the number of subshells we recall that, for a given principal quantum number n, the possible values of the quantum number l, which corresponds to the number of subshells are:

0, 1, 2, ... , n-1

Since n = 8 in our problem, the possible values of l are: 0, 1, 2, 3, 4, 5, 6, 7. Therefore, the number of subshells are 8.

Now we continue with the number of orbitals. For every subshell l, we have 2l+1 possible values of m, which correspond to the orbitals. Since the possible values of l are: 0,1,2,3,4,5,6,7, therefore, we have to perform the sum:

$\sum_{l=0}^{7}(2l+1) = 8^2=64$

And we can conclude that the number of orbitals is equal to 64.

Finally, we know that we can fit two electrons per orbital, therefore we can have 64*2 = 128 electrons in the shell corresponding to n=8.

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3 years ago

Put the waves in order from highest frequency to lowest frequency

S_A_V [24]

BCA for sure, b the lines are showing more movement

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3 years ago

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A concave lens can only form a. A. real image. B. reversed image. C. virtual image. D. magnified image.

Umnica [9.8K]

A concave lens can only form a virtual image. The correct option among all the options that are given in the question is the third option or option "C". Concave lenses are mostly thinner in the middle compared to its edges. I hope that this answer has come to your help.

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3 years ago

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Your starship, the Aimless Wanderer,lands on the mysterious planet Mongo. As chief scientist-engineer,you make the following mea

melisa1 [442]

Answer:

m = 1.26*10²⁵ kg.

Explanation:

Assuming that the mass of the stone is much smaller than the mass of the planet, we can get the mass, applying the Universal Law of Gravitation to both masses, as follows:

Fg = G* ms* mp / rp²

Now, if we apply Newton's 2nd Law to the mass of the stone, we can get the gravitational acceleration, as follows:

Fg = ms*a = ms*g ⇒ g = G*mp / rp²

First of all, we need to get the value of g.

Assuming that this acceleration is constant, we can appy the kinematic equations to this situation.

We know that the stone is thrown upward with an initial velocity vo = 15 m/s.

At the highest point in the trajectory, just before of changing direction, the stone comes momentarily to a stop.

At this point, applying the definition of acceleration, we can write:

vf = vo -g*t ⇒ 0 = vo -gt ⇒ g = vo/t (1)

We have the total time since the stone was thrown upwards, not the one used for the upward trajectory.

It can be showed, using the expression for the displacement (which is the same in both directions) that the time used for going up, it's the same used to go down, so the time that we need to put in (1). is just the half of the total time.

So, replacing in (1) we get the value of g, as follows:

g = 15 m/s / 4.5 s = 3.33 m/s²

Now, we can replace this value in the equation that gives us g based in the Universal Law of Gravitation, as follows:

g=G*mp / rp² (2)

Before solving for mp, however, we need to get the value of the radius of the planet.

Assuming that it's a perfect sphere, we can get this value from the value of the circumference at the planet's equator:

rp = 2*π*rp / 2*π ⇒ rp = 1.0*10⁵ km / 2*π = 15,915 km.

With this value for rp, we can solve (2) for mp, as follows:

mp= g*rp² / G = 3.33 m/s² * (15,915 km)² / 6,67*10⁻¹¹ N.m²/kg²

mp = 1.26*10²⁵ kg.

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3 years ago

(a) In what direction would the ship in Exercise 3.57 have to travel in order to have a velocity straight north relative to the

shtirl [24]

Answer:

Direction of ship: 9.45° West of North

Ship's relative speed: 7.87m/s

Explanation:

A. Direction of ship: since horizontal of the velocity of boat relative to the ground is 0

Vx=0

Therefore, -VsSin∅+VcCos∅40°

Sin∅ = Vc/Vs × Cos 40°

Sin∅ = 1.5/7 ×Cos40°

Sin∅= 0.164

∅= Sin-¹ (0.164)

∅= 9.45° W of N

B. Ship's relative speed:

Vy= VsCos∅ + Vcsin40°

= 7Cos9.45° + 1.5sin40°

= 7×0.986 + 1.5×0.642

= 7.865

= 7.87m/s

4 0

3 years ago