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3 years ago

11

In 1913 Niels Bohr formulated a method of calculating the different energy levels of the hydrogen atom.

1 answer:

Svetlanka [38]3 years ago

4 0

They are right the answer is A true

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In a "Rotor-ride" at a carnival, people rotate in a vertical cylindrically walled "room." If the room radius is 5.5 m, and the r

Crank

Answer:

0.181

Explanation:

We can convert the 0.5 rps into standard angular velocity unit rad/s knowing that each revolution is 2π:

ω = 0.5 rps = 0.5*2π = 3.14 rad/s

From here we can calculate the centripetal acceleration

$a_c = \omega^2r = 3.14^2*5.5 = 54.3 m/s^2$

Using Newton 2nd law we can calculate the centripetal force that pressing on the rider, as well as the reactive normal force:

$F = N = a_cm = 54.3 m$

Also the friction force and friction acceleration

$F_f = N\mu = 54.3 m \mu N$

$a_f = F_f / m = 54.3 \mu$

For the rider to not slide down, friction acceleration must win over gravitational acceleration g = 9.81 m/s2:

$g = a_f = 54.3 \mu$

$9.81 = 54.3 \mu$

$\mu = 9.81 / 54.3 = 0.181$

6 0

3 years ago

A wave is sent along the first rope transmitting a power of 57.3 W. It has a wavelength of 5.54 cm and velocity of 13.87 m/s The

Serhud [2]

Answer:

$A = 2.43*10^{-3} m$

Explanation:

power through string can be determined as shown in figure

$P = 2\pi ^2 HVA^2F^2$

Where

P = 57.3 W

V = 13.87 m/s

H = 567 g/m

we know that

$V = f *\lambda$ $\lambda = \frac{v}{f}$

therefore $P = 2\pi ^2 HVA^2(\frac{v}{f})^2$

$57.3 = \frac{2\pi ^2 * 0.567 *13.87^{3}* A^2}{(5.54*10^{-2})^2}$

$A = 2.43*10^{-3} m$

7 0

3 years ago

Someone please help on this? I just need the diagram to be drawn.

gregori [183]

Well the diagram would look like the water cycle I think

7 0

3 years ago

A vehicle accelerates from 0 to 30 m/s in 10 seconds on a straight road

Solnce55 [7]

What's the question?

6 0

3 years ago

Read 2 more answers

A proton is shot perpendicularly at an infinite plane of charge. The charge density of the plane is +7.65×10^−4 C/m^2. If the pr

Alexeev081 [22]

Answer:

The velocity is $2.94\times10^{6}\ m/s$ .

Explanation:

Given that,

Charge density of the plane $\sigma=7.65\times10^{−4}\ C/m^2$

Distance = 1.05 mm

We need to calculate the electric field due to plane of charge

Using formula of electric field

$E=\dfrac{\sigma}{2\epsilon}$

Put the value into the formula

$E=\dfrac{7.65\times10^{−4}}{2\times8.85\times10^{-12}}$

$E=4.322\times10^{7}\ N/C$

We need to calculate potential difference

Using formula of potential difference

$V=E\times r$

Put the value into the formula

$V=4.322\times10^{7}\times1.05\times10^{-3}$

$V=4.5381\times10^{4}\ Volt$

We need to calculate the work requires to be done to reach the surface of the plane

Using formula of work done

$W=qV$

Put the value into the formula

$W = 1.6\times10^{-19}\times4.5381\times10^{4}$

$W=7.26096\times10^{-15}\ J$

We need to calculate the velocity

Using work energy theorem

$W=\dfrac{1}{2}mv^2$

$v^2=\dfrac{2W}{m}$

$v=\sqrt{\dfrac{2\times7.26096\times10^{-15}}{1.67\times10^{-27}}}$

$v=2.94\times10^{6}\ m/s$

Hence, The velocity is $2.94\times10^{6}\ m/s$ .

8 0

3 years ago