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mixas84 [53]
3 years ago
7

What is the approximate wavelength of a light whose second-order dark band forms a diffraction angle of 15.0° when it passes thr

ough a diffraction grating that has 250.0 lines per mm? 26 nm 32 nm 414 nm 518 nm
Physics
2 answers:
anzhelika [568]3 years ago
7 0
414nm just took the test
devlian [24]3 years ago
4 0
414 is the right answer. I just took the test and got an 100%
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An object originally at rest, is accelerated uniformly along a straight line to a speed of 8m/s in 2s. What is the acceleration
Anna11 [10]

Answer:

4m/s²

Explanation:

Initial velocity (u) = 0 m/s

Final velocity (v) = 8 m/s

Time taken (t) = 2 sec

Acceleration (a) = ?

We know

a  =  \frac{v - u}{t}  \\  =  \frac{8 - 0}{2} \\  =  \frac{8}{2}  \\  = 4 \: m |s ^{2}

Hope it will help :)

3 0
3 years ago
What are two uses of ethanol?
Deffense [45]
Gas and solar power I think
5 0
3 years ago
When organic polymers joined together on early earth, they formed ______? Protons Protobionts Monomers Plankton
Sidana [21]
I think the answer would be protons
4 0
3 years ago
Read 2 more answers
La distancia por carretera de Chitré a Parita es de 12 km; exprese en pies ésta distanciaLa distancia por carretera de Chitré a
densk [106]

Answer:

La distancia por carretera de Chitré a Parita es de 12 km o 39370.08 pies.

Explanation:

La regla de tres es una forma de resolver problemas de proporcionalidad entre tres valores conocidos y un valor desconocido, estableciendo una relación de proporcionalidad entre todos ellos.

Si la relación entre las magnitudes es directa, es decir, cuando una magnitud aumenta, también lo hace la otra (o cuando una magnitud disminuye, también lo hace la otra), se debe aplicar la regla directa de tres. Para resolver una regla directa de tres, se debe seguir la siguiente fórmula, siendo a, b y c los valores conocidos y x el valor a determinar:

a ⇒ b

c ⇒ x

Entonces x=\frac{c*b}{a}

La regla directa de tres es la regla que se aplica en este caso donde hay un cambio de unidades. Para realizar esta conversión de unidades, primero debes saber que 1 km = 3280,84 pies. Entonces, si 1 km son 3280,84 pies, ¿cuántos pies son 12 km?

1 km ⇒ 3280.84 pies

12 km ⇒ x

x=\frac{12 km*3280.84 pies}{1 km}

x= 39370.08 pies

<u><em>La distancia por carretera de Chitré a Parita es de 12 km o 39370.08 pies.</em></u>

7 0
3 years ago
Coulomb’s law and static point charge ensembles (15 points). A test charge of 2C is located at point (3, 3, 5) in Cartesian coor
fenix001 [56]

Answer:

a) F_{r}= -583.72MN i + 183.47MN j + 6.05GN k

b) E=3.04 \frac{GN}{C}

Step-by-step explanation.

In order to solve this problem, we mus start by plotting the given points and charges. That will help us visualize the problem better and determine the direction of the forces (see attached picture).

Once we drew the points, we can start calculating the forces:

r_{AP}^{2}=(3-0)^{2}+(3-0)^{2}+(5+0)^{2}

which yields:

r_{AP}^{2}= 43 m^{2}

(I will assume the positions are in meters)

Next, we can make use of the force formula:

F=k_{e}\frac{q_{1}q_{2}}{r^{2}}

so we substitute the values:

F_{AP}=(8.99x10^{9})\frac{(1C)(2C)}{43m^{2}}

which yields:

F_{AP}=418.14 MN

Now we can find its components:

F_{APx}=418.14 MN*\frac{3}{\sqrt{43}}i

F_{APx}=191.30 MNi

F_{APy}=418.14 MN*\frac{3}{\sqrt{43}}j

F_{APy}=191.30MN j

F_{APz}=418.14 MN*\frac{5}{\sqrt{43}}k

F_{APz}=318.83 MN k

And we can now write them together for the first force, so we get:

F_{AP}=(191.30i+191.30j+318.83k)MN

We continue with the next force. The procedure is the same so we get:

r_{BP}^{2}=(3-1)^{2}+(3-1)^{2}+(5+0)^{2}

which yields:

r_{BP}^{2}= 33 m^{2}

Next, we can make use of the force formula:

F_{BP}=(8.99x10^{9})\frac{(4C)(2C)}{33m^{2}}

which yields:

F_{BP}=2.18 GN

Now we can find its components:

F_{BPx}=2.18 GN*\frac{2}{\sqrt{33}}i

F_{BPx}=758.98 MNi

F_{BPy}=2.18 GN*\frac{2}{\sqrt{33}}j

F_{BPy}=758.98MN j

F_{BPz}=2.18 GN*\frac{5}{\sqrt{33}}k

F_{BPz}=1.897 GN k

And we can now write them together for the second, so we get:

F_{BP}=(758.98i + 758.98j + 1897k)MN

We continue with the next force. The procedure is the same so we get:

r_{CP}^{2}=(3-5)^{2}+(3-4)^{2}+(5-0)^{2}

which yields:

r_{CP}^{2}= 30 m^{2}

Next, we can make use of the force formula:

F_{CP}=(8.99x10^{9})\frac{(7C)(2C)}{30m^{2}}

which yields:

F_{CP}=4.20 GN

Now we can find its components:

F_{CPx}=4.20 GN*\frac{-2}{\sqrt{30}}i

F_{CPx}=-1.534 GNi

F_{CPy}=4.20 GN*\frac{2}{\sqrt{30}}j

F_{CPy}=-766.81 MN j

F_{CPz}=4.20 GN*\frac{5}{\sqrt{30}}k

F_{CPz}=3.83 GN k

And we can now write them together for the third force, so we get:

F_{CP}=(-1.534i - 0.76681j +3.83k)GN

So in order to find the resultant force, we need to add the forces together:

F_{r}=F_{AP}+F_{BP}+F_{CP}

so we get:

F_{r}=(191.30i+191.30j+318.83k)MN + (758.98i + 758.98j + 1897k)MN + (-1.534i - 0.76681j +3.83k)GN

So when adding the problem together we get that:

F_{r}=(-0.583.72i + 0.18347j +6.05k)GN

which is the answer to part a), now let's take a look at part b).

b)

Basically, we need to find the magnitude of the force and divide it into the test charge, so we get:

F_{r}=\sqrt{(-0.583.72)^{2} + (0.18347)^{2} +(6.05)^{2}}

which yields:

F_{r}=6.08 GN

and now we take the formula for the electric field which is:

E=\frac{F_{r}}{q}

so we go ahead and substitute:

E=\frac{6.08GN}{2C}

E=3.04\frac{GN}{C}

7 0
4 years ago
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