PLEASE Please HELP ME...

Physics

1 answer:

cupoosta [38]3 years ago

7 0

Answer:

girl this easy ask yo teacher for help lol

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How far must you stretch a spring with k = 1600 N/m to store 260 J of energy?

Olenka [21]

Energy stored in a spring = 1/2*k*x^2

Where, x = increase in length of the spring

Therefore,
260 = 1/2*1600*x^2 =800x^2 => x^2 = 260/800 = 0.325 => x = sqrt (0.325) = 0.57 m

The length of the spring must increase by 0.57 m to store 260 J

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Consider a series rlc circuit where r = 45.0 ω, c = 15.5 μf, and l = 0.0940 h, driven at a frequency of 50.0 hz. determine the p

givi [52]

$R = 45 \ \Omega. \newline C = 15.5 \ \mu F = 15.5 \cdot 10^{-6} \ F.\newline l = 0.0940 \ H.\newline \nu = 50 \ Hz. \newline \newline x_{l} = 2 \pi \nu l = 2 \pi \cdot 50 \cdot 0.0940 = 29.5309709437 \approx 29.5 \ \Omega. \newline x_{c} = \frac{1}{2 \pi C \nu} = \frac{1}{2 \pi \cdot 15.5 \cdot 10^{-6} \cdot 50} = 0.00486946861 \approx 0.005 \ \Omega. \newline Z = \sqrt{R^{2}+(x_{l}-x_{c})^{2}} = \sqrt{45^{2}+(29.5-0.005)^{2}} = 53.8047862648 \approx 54 \ \Omega. \newline \newline \cos(\theta) = \frac{R}{Z} = \frac{45}{54} = 0.8\bar{3} \Rightarrow \arccos(0.8\bar{3}) = 33.5573098 \approx 33.5 \textdegree.$

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4 years ago

The distance between the first and fifth minima of a single-slit diffraction pattern is 0.500 mm with the screen 37.0 cm away fr

WARRIOR [948]

In the single-slit experiment, the displacement of the minima of the diffraction pattern on the screen is given by
$y_n= \frac{n \lambda D}{a}$ (1)
where
n is the order of the minimum
y is the displacement of the nth-minimum from the center of the diffraction pattern
$\lambda$ is the light's wavelength
D is the distance of the screen from the slit
a is the width of the slit

In our problem,
$D=37.0 cm=0.37 m$
$\lambda=530 nm=5.3 \cdot 10^{-7} m$
while the distance between the first and the fifth minima is
$y_5-y_1 = 0.500 mm=0.5 \cdot 10^{-3} m$ (2)

If we use the formula to rewrite $y_5, y_1$ , eq.(2) becomes
$\frac{5 \lambda D}{a} - \frac{1 \lambda D}{a} =\frac{4 \lambda D}{a}= 0.5 \cdot 10^{-3} m$
Which we can solve to find a, the width of the slit:
$a= \frac{4 \lambda D}{0.5 \cdot 10^{-3} m}= \frac{4 (5.3 \cdot 10^{-7} m)(0.37 m)}{0.5 \cdot 10^{-3} m}= 1.57 \cdot 10^{-3} m=1.57 mm$

7 0

4 years ago