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vlabodo [156]
2 years ago
8

A hydrogen atom contains a single electron that moves in a circular orbit about a single proton. Assume the proton is stationary

, and the electron has a speed of 7.5 105 m/s. Find the radius between the stationary proton and the electron orbit within the hydrogen atom.
Physics
1 answer:
padilas [110]2 years ago
4 0

Answer:

450 pm

Explanation:

The electron is held in orbit by an electric force, this works as the centripetal force. The equation for the centripetal acceleration is:

a = v^2 / r

The equation for the electric force is:

F = q1 * q2 / (4 * π * e0 * r^2)

Where

q1, q2: the electric charges, the charge of the electron is -1.6*10^-19 C

e0: electric constant (8.85*10^-12 F/m)

If we divide this force by the mass of the electron we get the acceleration

me = 9.1*10^-31 kg

a = q1 * q2 / (4 * π * e0 * me * r^2)

v^2 / r = q1 * q2 / (4 * π * e0 * me * r^2)

We can simplify r

v^2 = q1 * q2 / (4 * π * e0 * me * r)

Rearranging:

r = q1 * q2 / (4 * π * e0 * me * v^2)

r = 1.6*10^-19 * 1.6*10^-19 / (4 * π * 8.85*10^-12 * 9.1*10^-31 * (7.5*10^5)^2) = 4.5*10^-10 m = 450 pm

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DENIUS [597]

Answer:

1)

v_{oy}=11.29\ m/s

2)

y=7.39\ m

Explanation:

<u>Projectile Motion</u>

When an object is launched near the Earth's surface forming an angle \theta with the horizontal plane, it describes a well-known path called a parabola. The only force acting (neglecting the effects of the wind) is the gravity, which acts on the vertical axis.

The heigh of an object can be computed as

\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}

Where y_o is the initial height above the ground level, v_{oy} is the vertical component of the initial velocity and t is the time

The y-component of the speed is

v_y=v_{oy}-gt

1) We'll find the vertical component of the initial speed since we have not enough data to compute the magnitude of v_o

The object will reach the maximum height when v_y=0. It allows us to compute the time to reach that point

v_{oy}-gt_m=0

Solving for t_m

\displaystyle t_m=\frac{v_{oy}}{g}

Thus, the maximum heigh is

\displaystyle y_m=y_o+\frac{v_{oy}^2}{2g}

We know this value is 8 meters

\displaystyle y_o+\frac{v_{oy}^2}{2g}=8

Solving for v_{oy}

\displaystyle v_{oy}=\sqrt{2g(8-y_o)}

Replacing the known values

\displaystyle v_{oy}=\sqrt{2(9.8)(8-1.5)}

\displaystyle v_{oy}=11.29\ m/s

2) We know at t=1.505 sec the ball is above Julie's head, we can compute

\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}

\displaystyle y=1.5+(11.29)(1.505)-\frac{9.8(1.505)^2}{2}

\displaystyle y=1.5\ m+16,991\ m-11.098\ m

y=7.39\ m

5 0
2 years ago
A student, along with her backpack on the floor next to her, are in an elevator that is accelerating upward with acceleration a.
Anna007 [38]

Answer:

\mu_k = \frac{2(vt - L)}{(g + a) t^2}

Explanation:

As we know that backpack is kicked on the rough floor with speed "v"

So here as per force equation in vertical direction we know that

N - mg = ma

so normal force on the block is given as

N = mg + ma

now the magnitude of kinetic friction on the block is given as

F_f = \mu N

F_f = \mu (mg + ma)

now when bag is sliding on the floor then net deceleration of the block due to friction is given as

a = - \frac{F_f}{m}

a = -\mu_k(g + a)

now we know that bag hits the opposite wall at L distance away in time t

so we have

d = v t + \frac{1}{2}at^2

L = vt - \frac{1}{2}(\mu_k)(g + a) t^2

\mu_k = \frac{2(vt - L)}{(g + a) t^2}

8 0
1 year ago
You are given two rectangular blocks of shiny metal, Block A and Block B, and are asked to determine which one will float in a b
vladimir2022 [97]

Answer:

Explanation:

Volume of block A = 10 x 6 x 1 = 60 cm³

Mass of block A = 630 g

density of mass A = mass / density

= 630 / 60 = 10.5g / cm³

Volume of block B = 5 x 5 x 3 = 75 cm³

Mass of block A = 604 g

density of mass A = mass / density

= 604 / 75 = 8.05 g / cm³

Since density of both A and B are less than that of mercury , both will float in mercury.

7 0
2 years ago
a 75 kg man is standing at rest on ice while holding a 4kg ball. if the man throws the ball at a velocity of 3.50 m/s forward, w
AysviL [449]

Answer:

His resulting velocity will be 0.187 m/s backwards.

Explanation:

Given:

Mass of the man is, M=75\ kg

Mass of the ball is, m=4\ kg

Initial velocity of the man is, u_m=0\ m/s(rest)

Initial velocity of the ball is, u_b=0\ m/s(rest)

Final velocity of the ball is, v_b=3.50\ m/s

Final velocity of the man is, v_m=?\ m/s

In order to solve this problem, we apply law of conservation of momentum.

It states that sum of initial momentum is equal to the sum of final momentum.

Momentum is the product of mass and velocity.

Initial momentum = Initial momentum of man and ball

Initial momentum = Mu_m+mu_b=75\times 0+4\times 0 =0\ Nm

Final momentum = Final momentum of man and ball

Final momentum = Mv_m+mv_b=75\times v_m+4\times 3.50 =75v_m+14

Now, initial momentum = final momentum

0=75v_m+14\\\\75v_m=-14\\\\v_m=\frac{-14}{75}\\\\v_m=-0.187\ m/s

The negative sign implies backward motion of the man.

Therefore, his resulting velocity is 0.187 m/s backwards.

3 0
2 years ago
Consider a vibrating system described by the initial value problem. (A computer algebra system is recommended.) u'' + 1 4 u' + 2
GarryVolchara [31]

Answer:

Therefore the required solution is

U(t)=\frac{2(2-\omega^2)^2}{(2-\omega^2)^2+\frac{1}{16}\omega} cos\omega t +\frac{\frac{1}{2}\omega}{(2-\omega^2)^2+\frac{1}{16}\omega} sin \omega t

Explanation:

Given vibrating system is

u''+\frac{1}{4}u'+2u= 2cos \omega t

Consider U(t) = A cosωt + B sinωt

Differentiating with respect to t

U'(t)= - A ω sinωt +B ω cos ωt

Again differentiating with respect to t

U''(t) =  - A ω² cosωt -B ω² sin ωt

Putting this in given equation

-A\omega^2cos\omega t-B\omega^2sin \omega t+ \frac{1}{4}(-A\omega sin \omega t+B\omega cos \omega t)+2Acos\omega t+2Bsin\omega t = 2cos\omega t

\Rightarrow (-A\omega^2+\frac{1}{4}B\omega +2A)cos \omega t+(-B\omega^2-\frac{1}{4}A\omega+2B)sin \omega t= 2cos \omega t

Equating the coefficient of sinωt and cos ωt

\Rightarrow (-A\omega^2+\frac{1}{4}B\omega +2A)= 2

\Rightarrow (2-\omega^2)A+\frac{1}{4}B\omega -2=0.........(1)

and

\Rightarrow -B\omega^2-\frac{1}{4}A\omega+2B= 0

\Rightarrow -\frac{1}{4}A\omega+(2-\omega^2)B= 0........(2)

Solving equation (1) and (2) by cross multiplication method

\frac{A}{\frac{1}{4}\omega.0 -(-2)(2-\omega^2)}=\frac{B}{-\frac{1}{4}\omega.(-2)-0.(2-\omega^2)}=\frac{1}{(2-\omega^2)^2-(-\frac{1}{4}\omega)(\frac{1}{4}\omega)}

\Rightarrow \frac{A}{2(2-\omega^2)}=\frac{B}{\frac{1}{2}\omega}=\frac{1}{(2-\omega^2)^2+\frac{1}{16}\omega}

\therefore A=\frac{2(2-\omega^2)^2}{(2-\omega^2)^2+\frac{1}{16}\omega}   and        B=\frac{\frac{1}{2}\omega}{(2-\omega^2)^2+\frac{1}{16}\omega}

Therefore the required solution is

U(t)=\frac{2(2-\omega^2)^2}{(2-\omega^2)^2+\frac{1}{16}\omega} cos\omega t +\frac{\frac{1}{2}\omega}{(2-\omega^2)^2+\frac{1}{16}\omega} sin \omega t

5 0
2 years ago
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