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statuscvo [17]
3 years ago
13

I need help!!! 60 POINTS!!!!

Physics
2 answers:
galina1969 [7]3 years ago
8 0

The answer is "A and B are different elements, while C is an isotope of B".

Scrat [10]3 years ago
3 0
Correct one is b
Good luck
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Assuming that the limits of the visible spectrum are approximately 380 and 700 nm, find the angular range of the first-order vis
ch4aika [34]

Answer:

angular range is ( 0.681 rad , 0.35 rad )

Explanation:

given data

wavelength λ = 380 nm = 380 × 10^{-9} m

wavelength λ  = 700 nm =  700 × 10^{-9} m

to find out

angular range of the first-order

solution

we will apply here slit experiment equation that is

d sinθ = m λ    ...........1

here m is 1 for single slit and d is = \frac{1}{900*10^3 m}

so put here value in equation 1 for 380 nm

we get

d sinθ = m λ

\frac{1}{900*10^3} sinθ = 1 × 380 × 10^{-9}

θ = 0.35 rad

and for 700 nm

we get

d sinθ = m λ

\frac{1}{900*10^3} sinθ = 1 × 700 × 10^{-9}

θ = 0.681 rad

so angular range is ( 0.681 rad , 0.35 rad )

3 0
3 years ago
Which best describes electromagnetic waves moving from gamma rays to radio waves along the electromagnetic spectrum
kondor19780726 [428]

Answer:

Explanation:

No

9 0
3 years ago
What is the change in length of a 1400. m steel, (12x10^-6)/(C0) , pipe for a temperature change of 250.0 degrees Celsius? Remem
8090 [49]

Answer:

\Delta L = 4.2 m

Explanation:

As per the formula of thermal expansion we know that

L = L_o(1 + \alpha\Delta T)

so here we will have

L_o = 1400 m

\alpha = 12 \times 10^{-6} per ^oC

\Delta T = 250 degree C

so here change in the length of the rod is given as

\Delta L = L - L_o

\Delta L = L_o \alpha \Delta T

\Delta L = 1400 (12 \times 10^{-6})(250)

\Delta L = 4.2 m

8 0
3 years ago
A car having an initial velocity of 12 meters per second speed up uniformly to 22 meters per second. This change is made in 4 se
Sonbull [250]

(22-12)/4

10/4

2.5 meters per second sqaured

7 0
3 years ago
Suppose you wish to fabricate a uniform wire out of 1.10 g of copper. If the wire is to have a resistance R = 0.390 Ω, and if al
Citrus2011 [14]

To solve this problem we will apply the concepts related to volume, as a function of length and area, as of mass and density. Later we will take the same concept of resistance and resistivity, equal to the length per unit area. Once obtained from the known constants it will be possible to obtain the area by matching the two equations:

Mass of copper wire(m) = 1.10g = 1.10*10^{-3} kg

Density (\rho)= 8.92*10^3kg/m^3

Resistively of copper (\gamma) = 1.7*10^{-8}\Omega \cdot m

Resistance (R) = 0.390\Omega

Volume is defined as,

V= lA \text{ and }\frac{m}{\rho}

lA= \frac{1.10*10^{-3}}{8.92*10^3}

lA = 1.233*10^{-7} m^3 (1)

We know that,

\frac{l}{A} = \frac{R}{\gamma}

\frac{l}{A}= \frac{0.390\Omega}{1.7*10^{-8}\Omega m}

\frac{l}{A} = 2.2941*10^7 m^{-1} (2)

Multiplying equation we have

l^2 = (1.233*10^{-7})( 2.2941*10^7)

l^2 = 2.8286m^2

l =\sqrt{2.8286m^2}

l = 1.68m

Therefore the length of the wire is 1.68m

6 0
3 years ago
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