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tester [92]
2 years ago
14

R=1(1149×109meter)2−(1298×109meter)2

Physics
1 answer:
kvv77 [185]2 years ago
7 0

Answer:

R=-3.64603\times 10^{23}

Explanation:

It is required to simplify R. We have,

R=(1149\times 10^9\ m)^2-(1298\times 10^9\ m)^2

It can be calculated step by step such that,

(1149)^2=1320201\\\\\text{and}\\\\(1298)^2=1684804

Plugging these values,

R=[(1149)^2\times (10^9)^2]-[(1298)^2\times (10^9)^2)]\\\\R=1320201\times 10^{18}-1684804\times 10^{18}\\\\R=-3.64603\times 10^{23}

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Ne4ueva [31]
Your answer is A. If velocity decreases, your kinetic energy would also decrease.
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A single slit diffraction experiment performed with an argon laser of wavelength 454.6 nm produces a pattern on a screen with da
tankabanditka [31]

To develop this problem it is necessary to apply the concepts related to Interference and the Two slit experiment.

Young's equation defines the separation between fringes this phenomenon as,

d = \frac{\lambda D}{a}

Where,

\lambda = Wavelength

d = Separation between fringes

a =  Slit width

D = Distance between the slits and screen

Then for the two case we have

d_1 = \frac{\lambda D}{a_1}

d_2 = \frac{\lambda D}{a_2}

Calculating the new separation between the fringes would be

\frac{d_2}{d_1} = \frac{\frac{\lambda D}{a_1}}{\frac{\lambda D}{a_2}}

\frac{d_2}{d_1} = \frac{\lambda_1}{\lambda_2}

d_2= \frac{\lambda_1}{\lambda_2} d_1

We have then,

d_2 = \frac{632.9*10^{-8}}{454.6*10^{-9}} 10*10^{-3}

d_2 = 13.9mm

Therefore the correct answer is c.

7 0
2 years ago
The velocity of a 3.00 kg parti- cle is given by :v = (8.00tiˆ + 3.00t2jˆ) m/s, with time t in seconds. At the instant the net f
FromTheMoon [43]

Answer:

Part a)

Direction of net force is

\theta = 46.7 degree

Part b)

Direction of the velocity is given as

\phi = 27.9 degree

Explanation:

As we know that the velocity of the particle is given as

v = 8.00 t \hat i + 3.00 t^2\hat j

now the acceleration is given as

a = \frac{dv}{dt}

a = 8.00 \hat i + 6.00 t\hat j

now magnitude of net acceleration is given as

a = \sqrt{64 + 36t^2}

F = 3a

35 = 3\sqrt{64 + 36t^2}

t = 1.41 s

Part a)

Now direction of net force is given as

tan\theta = \frac{F_y}{F_x}

tan\theta = \frac{6t}{8}

tan\theta = \frac{6(1.41)}{8}

\theta = 46.7 degree

Part b)

Direction of the velocity is given as

tan\phi = \frac{v_y}{v_x}

tan\phi = \frac{3.00 t^2}{8.00 t}

tan\phi = \frac{3.00(1.41)}{8.00}

\phi = 27.9 degree

8 0
2 years ago
A standard 14.16-inch (0.360-meter) computer monitor is 1024 pixels wide and 768 pixels tall. Each pixel is a square approximate
Vlad1618 [11]

Answer:

0.0031 m

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d = Pupil diameter

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The pupil diameter is 0.0031 m

4 0
2 years ago
The buoyant force on an object fully submerged in a liquid depends on (select all that apply)
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The buoyant force on an object fully submerged in a liquid depends on
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of the object depends on the object's volume and the object's mass.

So the only item on this list that it DOESN't depend on is the mass of
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I guess that means that the buoyant force on a fully submerged object is
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8 0
2 years ago
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