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3 years ago

energy in the form of electromagnetic radiation. Most often associated with a wavelength that is visible to the eye

Physics

2 answers:

vovangra [49]3 years ago

3 0

light energy

Energy in the form of electromagnetic radiation. Most often associated with a wavelength that is visible to the eye.

sweet-ann [11.9K]3 years ago

3 0

Answer:

Light energy

Explanation:

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Brilliant_brown [7]

Answer:

now it lets me. B

Explanation:

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6 0

3 years ago

Light from a laser strikes a diffraction grating that has 5500 lines per centimeter. The central and first-order maxima are sepa

WINSTONCH [101]

.Answer:

491.4 nm

Explanation:

The distance between central and first maxima is,

$y=0.455m$

And the distance between screen abnd grating is,

$L=1.62 m$

Now the angle can be find as,

$tan\theta=\frac{y}{L} \\\theta=tan^{-1}(\frac{0.455}{1.62}) \\\theta=15.68^{\circ}$

Now the grating distance is,

$d=\frac{1}{5500} cm\\d=1.82\times 10^{-6}m$

Now with m=1 condition will become,

$\lambda=dsin\theta$

So,

$\lambda=1.82\times 10^{-6}m\times sin(15.68^{\circ})\\\lambda=1.82\times 10^{-6}m\times 0.270\\\lambda=491.4\times 10^{-9}m\\\lambda=491.4 nm$

Therefore the wavelength of laser light is 491.4 nm.

3 0

3 years ago

How many significant figures does 56030 have? 3 4. 5 6

Gre4nikov [31]

The correct answer is 4

8 0

4 years ago

Read 2 more answers

Can someone help me?

Gwar [14]

24- series
25- parallel
26- no, because they’re connected in series
27- yes, because they’re connected in parallel

7 0

3 years ago

If the average pitcher is releasing the ball from a height of 1.8 m above the ground, and the pitcher's mound is 0.2 m higher th

mina [271]

The catcher can catch the ball at a height of 0.96 m from the ground.

The distance between the pitcher's mound and the catcher's box is about 60'6", which translates to 18.44 m. An average pitcher can pitch with speeds ranging from 88 mph to 97 mph, which is from 39.3 m/s to 43.4 m/s.

Assume the pitcher pitches a ball horizontally with a speed of 40 m/s. If the catcher catches the ball in a time t, then the ball travels a horizontal distance x of 18.44 m and at the same time falls through a height y.

The horizontal motion of the ball is uniform motion since no force acts on the ball ( assuming no air resistance) and hence the acceleration of the ball along the horizontal direction is zero.

Therefore,

$x=ut$

Calculate the time t by substituting 18.44 m for x and 40 m/s for u.

$t=\frac{x}{u} \\ =\frac{18.44 m}{40 m/s} \\ =0.461s$

The ball is acted upon by the earth's gravitational attraction and hence it accelerates downwards with an acceleration equal to the acceleration due to gravity g.

Since a horizontal projection is assumed, the ball has no component of velocity in the downward direction.

Therefore, for vertical motion, which is an accelerated motion, the distance y, the ball falls in the time t taken by it to reach the catcher's box is given by the equation,

$y=\frac{1}{2} gt^2$

Substitute 9.8 m/s² for g and 0.461 s for t.

$y=\frac{1}{2} gt^2\\ y=\frac{1}{2}(9.8 m/s^2)(0.461s)^2=1.04 m$

The pitcher releases the ball at a height of 1.8 m from a mound which is at a height of 0.2 m. Thus, the ball is released at a height of 2.0 m from the ground. It falls through a distance of 1.04 m in the time it takes to reach the catcher.

Therefore, the height at which the catcher needs to keep his glove so as to catch the ball is given by, $(2.0 m)-(1.04 m)=0.96 m$

The catcher needs to hold his glove at a height of 0,96 m from the ground.

8 0

3 years ago