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3 years ago

PLEASE help!!

Physics

1 answer:

sergeinik [125]3 years ago

5 0

Answer:

in which standard you are, i am typing your answer till please reply.

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Determine how would the frequency of the pendulum change if it was taken to the moon by finding the ratio of its frequency on th

andrezito [222]

Answer:

you lady boo

Explanation:

7 0

3 years ago

Irina finds an unlabeled box of fine needles, and wants to determine how thick they are. A standard ruler will not do the job, a

Vsevolod [243]

Answer:

$d=\frac{1.22\times 640\times 10^{-9}\times 21.7}{7.35\times 10^{-2}}=2305.21\times 10^{-7}m$

Explanation:

The expression which represent the first diffraction minima by a circular aperture is given by $d sin\Theta =1.22\lambda$ --------eqn 1

The angle through which the first minima is diffracted is given by $tan\Theta =\frac{y_1}{D}$ ---------eqn 2

As $\Theta$ is very small so we can write $sin\Theta =tan\Theta$

So from eqn 1 and eqn 2 we can write

$y_1=\frac{1.22\lambda D}{d}$ --------eqn 3

Here $y_1$ is the position of first maxima D is the distance of screen from the circular aperture d is the diameter of aperture

It is given that diameter of circular aperture is 14.7 cm so $y_1=\frac{14.7}{2}=7.35 \ cm$

Now putting all these value in eqn 3

$d=\frac{1.22\lambda D}{y_1}$

$d=\frac{1.22\times 640\times 10^{-9}\times 21.7}{7.35\times 10^{-2}}=2305.21\times 10^{-7}m$

8 0

3 years ago

In 2007, michael carter (u.s.) set a world record in the shot put with a throw of 24.77 m. what was the initial speed of the sho

Serga [27]

The initial speed of the shot is 15.02 m/s.

The Shot put is released at a height y from the ground with a speed u. It is released at an angle θ to the horizontal. In a time t, the shot put travels a distance R horizontally.

Pl refer to the attached diagram.

Resolve the velocity u into horizontal and vertical components, u ₓ=ucosθ and uy=u sinθ. The horizontal component remains constant in the absence of air resistance, while the vertical component varies due to the action of the gravitational force.

Write an expression for R.

$R=u_xt=(ucos \theta)t$

Therefore,

$t=\frac{R}{ucos\theta} .......(1)$

In the time t, the net displacement of the shotput is y in the downward direction.

Use the equation of motion,

$y=u_yt-\frac{1}{2}gt^2=(usin\theta) t-\frac{1}{2}gt^2$

Substitute the value of t from equation (1).

$y=(ucos\theta)(\frac{R}{ucos\theta} )-\frac{1}{2} g(\frac{R}{ucos\theta} )^2\\ =Rtan\theta-(\frac{gR^2}{2u^2cos^2\theta} )$

Substitute -2.10 m for y, 24.77 m for R and 38.0° for θ and solve for u.

$y=Rtan\theta-(\frac{gR^2}{2u^2cos^2\theta} )\\ (-2.10m)=(24.77 m)(tan38.0^o)-\frac{(9.8 m/s^2)(24.77m)^2}{2u^2(cos38.0^o)^2} \\ u^2=225.71(m/s)^2\\ u=15.02m/s$