Please help would it be 10?

A line in the paschen series of hydrogen has a wavelength of 1880 nm. From what state did the electron originate?

NARA [144]

Answer:

$n=4$

Explanation:

Paschen series of hydrogen, is the series of transitions resulting from the hydrogen atom when electron jumps from a state of $n\geq 4$ to $n_f=3$ .

Emission lines for hydrogen are given by:

$\frac{1}{\lambda}=R_H(\frac{1}{n_f^2}-\frac{1}{n^2})$

$\lambda$ is the wavelength of the line emitted, $R_H$ is the Rydberg constant for hydrogen, $n_f$ is the final energy state of the electron and $n$ is the energy state where the electron transition originated.

We have $\lambda=1880nm=1880*10^{-9}m$ and $n_f=3$ . Solving for n:

$\frac{1}{\lambda R_H}=\frac{1}{n_f^2}-\frac{1}{n^2}\\\frac{1}{n^2}=\frac{1}{n_f^2}-\frac{1}{\lambda R_H}\\n^2=\frac{1}{\frac{1}{n_f^2}-\frac{1}{\lambda R_H}}\\n^2=\frac{1}{\frac{1}{3^2}-\frac{1}{1880*10^{-9}m(1.09737*10^7m^{-1})}}\\n^2=15.96\\n=4$

6 0

4 years ago

At t1 = 2.00 s, the acceleration of a particle in counterclockwise circular motion is 6.00 i + 4.00 j m/s2 . It moves at constan

Jet001 [13]

The particle moves with constant speed in a circular path, so its acceleration vector always points toward the circle's center.

At time $t_1$ , the acceleration vector has direction $\theta_1$ such that

$\tan\theta_1=\dfrac{4.00}{6.00}\implies\theta_1=33.7^\circ$

which indicates the particle is situated at a point on the lower left half of the circle, while at time $t_2$ the acceleration has direction $\theta_2$ such that

$\tan\theta_2=\dfrac{-6.00}{4.00}\implies\theta_2=-56.3^\circ$

which indicates the particle lies on the upper left half of the circle.

Notice that $\theta_1-\theta_2=90^\circ$ . That is, the measure of the major arc between the particle's positions at $t_1$ and $t_2$ is 270 degrees, which means that $t_2-t_1$ is the time it takes for the particle to traverse 3/4 of the circular path, or 3/4 its period.

Recall that

$\|\vec a_{\rm rad}\|=\dfrac{4\pi^2R}{T^2}$

where $R$ is the radius of the circle and $T$ is the period. We have

$t_2-t_1=(5.00-2.00)\,\mathrm s=3.00\,\mathrm s\implies T=\dfrac{3.00\,\rm s}{\frac34}=4.00\,\mathrm s$

and the magnitude of the particle's acceleration toward the center of the circle is

$\|\vec a_{\rm rad}\|=\sqrt{\left(6.00\dfrac{\rm m}{\mathrm s^2}\right)^2+\left(4.00\dfrac{\rm m}{\mathrm s^2}\right)^2}=7.21\dfrac{\rm m}{\mathrm s^2}$

So we find that the path has a radius $R$ of

$7.21\dfrac{\rm m}{\mathrm s^2}=\dfrac{4\pi^2R}{(4.00\,\mathrm s)^2}\implies\boxed{R=2.92\,\mathrm m}$

8 0

3 years ago

Only two horizontal forces act on a 3.0 kg body that can move over a frictionless floor. one force is 9.0 n, acting due east, an

Zolol [24]

The first thing you should do for this case is to find the horizontal and vertical components of the forces acting on the body.
We have then:
Horizontal = 9-9.2cos (58) = 4.124742769 N.
Vertical = 9.2sin (58) = 7.802042485 N
Then, the resulting net force is:
F = √ ((4.124742769) ^ 2 + (7.802042485) ^ 2) = 8.825268826 N
Then by definition:
F = m * a
Clearing the acceleration:
a = F / m
a = (8.825268826) / (3.0) = 2.941756275 m / s ^ 2
answer:
The magnitude of the body's acceleration is
2.941756275 m / s ^ 2

6 0

3 years ago

I will mark brainlist pls help and. Give right answers

Vera_Pavlovna [14]

I think its b. Answer:

Explanation:

7 0

3 years ago

Read 2 more answers

An object is launched at a velocity of 20 m/s in a direction making an angle of 25° upward with the horizontal. (Given gravity g

MrMuchimi

Answer:

Explanation:

a) The formulas for the components Vx and Vy of the velocity and components x and y of the displacement are given by x = V0 cos(θ) Vy = V0 sin(θ) - g t

x = V0 cos(θ) t y = V0 sin(θ) t - (1/2) g t2

In the problem V0 = 20 m/s, θ = 25° and g = 9.8 m/s2.

The height of the projectile is given by the component y, and it reaches its maximum value when the component Vy is equal to zero. That is when the projectile changes from moving upward to moving downward.(see figure above) and also the animation of the projectile.

Vy = V0 sin(θ) - g t = 0

solve for t

t = V0 sin(θ) / g = 20 sin(25°) / 9.8 = 0.86 seconds

Find the maximum height by substituting t by 0.86 seconds in the formula for y

maximum height y (0.86) = 20 sin(25°)(0.86) - (1/2) (9.8) (0.86) 2 = 3.64 meters

b) The time of flight is the interval of time between when projectile is launched: t1 and when the projectile touches the ground: t2. At t = t1 and t = t2, y = 0 (ground). Hence

V0 sin(θ) t - (1/2) g t2 = 0

Solve for t

t(V0 sin(θ) - (1/2) g t) = 0

two solutions

t = t1 = 0 and t = t2 = 2 V0 sin(θ) / g

Time of flight = t2 - t1 = 2 (20) sin(θ) / g = 1.72 seconds.

c) In part c) above we found the time of flight t2 = 2 V0 sin(θ) / g. The horizontal range is the horizontal distance given by x at t = t2.

range = x(t2) = V0 cos(θ) t2 = 2 V0 cos(θ) V0 sin(θ) / g = V02 sin(2θ) / g = 202 sin (2(25°)) / 9.8 = 31.26 meters

d) The object hits the ground at t = t2 = 2 V0 sin(θ) / g (found in part b above)

The components of the velocity at t are given by

Vx = V0 cos(θ) Vy = V0 sin(θ) - g t

The components of the velocity at t = 2 V0 sin(θ) / g are given by

Vx = V0 cos(θ) = 20 cos(25°) Vy = V0 sin(25°) - g (2 V0 sin(25°) / g) = - V0 sin(25°)

The magnitude V of the velocity is given by

V = √[ Vx2 + Vy2 ] = √[ (20 cos(25°))2 + (- V0 sin(25°))2 ] = V0 = 20 m/s

7 0

3 years ago