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4 years ago

Is this a testable question?

2 answers:

8 0

This is a non testable question because it cannot be answered by doing an experiment. But it could be modified for example Dogs are more obedient then cats.

Stells [14]4 years ago

4 0

Not really; it's primarily a matter of opinion.

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When a low-pressure gas of hydrogen atoms is placed in a tube and a large voltage is applied to the end of the tube, the atoms w

FromTheMoon [43]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The value of n is $n =7$

Explanation:

From the question we are told that

The value of m = 2

For every value of $m, n = m+ 1, m+2,m+3,....$

The modified version of Balmer's formula is $\frac{1}{\lambda} = R [\frac{1}{m^2} - \frac{1}{n^2} ]$

The Rydberg constant has a value of $R = 1.097 *10^{7} m^{-1}$

The objective of this solution is to obtain the value of n for which the wavelength of the Balmer series line is smaller than 400nm

For m = 2 and n =3

The wavelength is

$\frac{1}{\lambda } = (1.097 * 10^7)[\frac{1}{2^2} - \frac{1}{3^2} ]$

$\lambda = \frac{1}{1523611.1112}$

$\lambda = 656nm$

For m = 2 and n = 4

The wavelength is

$\frac{1}{\lambda } = (1.097 * 10^7)[\frac{1}{2^2} - \frac{1}{4^2} ]$

$\lambda = \frac{1}{2056875}$

$\lambda = 486nm$

For m = 2 and n = 5

The wavelength is

$\frac{1}{\lambda } = (1.097 * 10^7)[\frac{1}{2^2} - \frac{1}{5^2} ]$

$\lambda = \frac{1}{2303700}$

$\lambda = 434nm$

For m = 2 and n = 6

The wavelength is

$\frac{1}{\lambda } = (1.097 * 10^7)[\frac{1}{2^2} - \frac{1}{6^2} ]$

$\lambda = \frac{1}{2422222}$

$\lambda = 410nm$

For m = 2 and n = 7

The wavelength is

$\frac{1}{\lambda } = (1.097 * 10^7)[\frac{1}{2^2} - \frac{1}{7^2} ]$

$\lambda = \frac{1}{2518622}$

$\lambda = 397nm$

So the value of n is 7

7 0

3 years ago

A person throws a stone from the corner edge of a building. The stone's initial velocity is 28.0 m/s directed at 43.0° above the

Naya [18.7K]

The stone's acceleration, velocity, and position vectors at time $t$ are

$\mathbf a(t)=-g\,\mathbf j$

$\mathbf v(t)=v_{i,x}\,\mathbf i+\left(v_{i,y}-gt\right)\,\mathbf j$

$\mathbf r(t)=v_{i,x}t\,\mathbf i+\left(y_i+v_{i,y}t-\dfrac g2t^2\right)\,\mathbf j$

where

$g=9.80\dfrac{\rm m}{\mathrm s^2}$

$v_{i,x}=\left(28.0\dfrac{\rm m}{\rm s}\right)\cos43.0^\circ\approx20.478\dfrac{\rm m}{\rm s}$

$v_{i,y}=\left(28.0\dfrac{\rm m}{\rm s}\right)\sin43.0^\circ\approx19.096\dfrac{\rm m}{\rm s}$

and $y_i$ is the height of the building and initial height of the rock.

(a) After 6.1 s, the stone has a height of 5 m. Set the vertical component $(\mathbf j)$ of the position vector to 5 m and solve for $y_i$ :

$5\,\mathrm m=y_i+\left(19.096\dfrac{\rm m}{\rm s}\right)(6.1\,\mathrm s)-\dfrac12\left(9.80\dfrac{\rm m}{\mathrm s^2}\right)(6.1\,\mathrm s)^2$

$\implies\boxed{y_i\approx70.8\,\mathrm m}$

(b) Evaluate the horizontal component $(\mathbf i)$ of the position vector when $t=6.1\,\mathrm s$ :

$\left(20.478\dfrac{\rm m}{\rm s}\right)(6.1\,\mathrm s)\approx\boxed{124.92\,\mathrm m}$

(c) The rock's velocity vector has a constant horizontal component, so that

$v_{f,x}=v_{i,x}\approx20.478\dfrac{\rm m}{\rm s}$

where $v_{f,x}$

For the vertical component, recall the formula,

${v_{f,y}}^2-{v_{i,y}}^2=2a\Delta y$

where $v_{i,y}$ and $v_{f,y}$ are the initial and final velocities, $a$ is the acceleration, and $\Delta y$ is the change in height.

When the rock hits the ground, it will have height $y_f=0$ . It's thrown from a height of $y_i$ , so $\Delta y=-y_i$ . The rock is effectively in freefall, so $a=-g$ . Solve for $v_{f,y}$ :

${v_{f,y}}^2-\left(19.096\dfrac{\rm m}{\rm s}\right)^2=2(-g)(-124.92\,\mathrm m)$

$\implies v_{f,y}\approx-53.039\dfrac{\rm m}{\rm s}$

(where we took the negative square root because we know that $v_{f,y}$ points in the downward direction)

So at the moment the rock hits the ground, its velocity vector is

$\mathbf v_f=\left(20.478\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(-53.039\dfrac{\rm m}{\rm s}\right)\,\mathbf j$

which has a magnitude of

$\|\mathbf v_f\|=\sqrt{\left(20.478\dfrac{\rm m}{\rm s}\right)^2+\left(-53.039\dfrac{\rm m}{\rm s}\right)^2}\approx\boxed{56.855\dfrac{\rm m}{\rm s}}$

(d) The acceleration vector stays constant throughout, so

$\mathbf a(t)=\boxed{-g\,\mathbf j}$

4 0

3 years ago

Which type of wave affects the surface of the land by causing it to rise and fall like waves on an ocean?

denis-greek [22]

It is a seismic wave that effects the surface of the land by causing it to rise and fall like waves in an ocean. Seismic waves can be broken down into a number of different types. Specifically, it is a Raleigh surface wave that behaves much like water waves since it moves the ground up and down as it travels. Seismic waves are produced in earthquakes.

5 0

3 years ago

Explain why friction makes it harder to push a box filled with groceries than an empty box

kupik [55]

It is harder to push a box with groceries in it because the mass is weighing the box down. Which makes it harder to push

5 0

3 years ago

A 4.40-m-long, 500 kg steel uniform beam extends horizontally from the point where it has been bolted to the framework of a new

icang [17]

Answer:

Torque=13798.4 N.m

Explanation:

Given data

Mass of beam m₁=500 kg

Mass of the person m₂=70 kg

length of steel r₁=4.40m

center of gravity of the beam is at r₂=r₁/2 =4.40/2 = 2.20m

To find

Torque

Solution

Torque due to beam own weight

$T_{1}=m_{1}gr_{1}\\ T_{1}=500*2.2*9.8\\T_{1}=10780N.m$

Torque due to person

$T_{2}=m_{2}r_{2}g\\T_{2}=70*(4.40)*(9.8)\\T_{2}=3018.4 N.m$

Now for total torque

$T_{total}=T_{1}+T_{2}\\T_{total}=10780+3018.4\\T_{total}=13798.4N.m$

4 0

3 years ago