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4 years ago

Which layer is thinnest under the oceans and thickest in the mountains

Physics

2 answers:

pogonyaev4 years ago

8 0

Crust is thinnest under the oceans and thickest under continental mountainranges.

Rasek [7]4 years ago

7 0

The answer is earth's crust

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A man hits a 50 grams golf ball such that it leaves the tee at an angle of 40o with the horizontal and strikes the ground at the

lukranit [14]

Answer:

Explanation:

Range of projectile R = 20 m

formula of range

R = u² sin2θ / g

u is initial velocity , θ is angle of projectile

putting the values

20 = u² sin2x 40 / 9.8

u² = 199

u = 14.10 m /s

At the initial point

vertical component of u

= u sin40 = 14.1 x sin 40

= 9.06 m/s

Horizontal component

= u cos 30

At the final point where the ball strikes the ground after falling , its speed remains the same as that in the beginning .

Horizontal component of velocity

u cos 30

Vertical component

= - u sin 30

= - 9.06 m /s

So its horizontal component remains unchanged .

change in vertical component = 9.06 - ( - 9.06 )

= 18.12 m /s

change in momentum

mass x change in velocity

= .050 x 18.12

= .906 N.s

Impulse = change in momentum

= .906 N.s .

3 0

4 years ago

This is the 3rd time I reposted this question lol.

Mamont248 [21]

Answer:

Blood absorbs nutrients and the waste products of cells. It carries waste away from cells and pumps nutrients through the whole body.

Explanation:

8 0

3 years ago

Waves<br> Please need help fast

mixas84 [53]

I believe its the third answer

6 0

2 years ago

2) A skier stands at rest and begins to ski downhill with an acceleration of 3.0 m/s² {downhill). What is

Finger [1]

Answer:

337.5m

Explanation:

<u>Kinematics</u>

Under constant acceleration, the kinematic equation holds:

$s=\frac{1}{2}at^2+v_ot+s_o$ , where "s" is the position at time "t", "a" is the constant acceleration, " $v_o$ " is the initial velocity, and $s_o$ is the initial position.

<u>Defining Displacement</u>

Displacement is the difference in positions: $s-s_o$ or $\Delta s$
$s=\frac{1}{2}at^2+v_ot+s_o$

$s-s_o=\frac{1}{2}at^2+v_ot$

$\Delta s=\frac{1}{2}at^2+v_ot$

<u>Using known information</u>

Given that the initial velocity is zero ("skier stands at rest"), and zero times anything is zero, and zero plus anything remains unchanged, the equation simplifies further to the following:

$\Delta s=\frac{1}{2}at^2+v_ot$

$\Delta s=\frac{1}{2}at^2+(0)*t$

$\Delta s=\frac{1}{2}at^2+0$

$\Delta s=\frac{1}{2}at^2$

So, to find the displacement after 15 seconds, with a constant acceleration of 3.0 m/s², substitute the known values, and simplify:

$\Delta s=\frac{1}{2}at^2$

$\Delta s=\frac{1}{2}(3.0[\frac{m}{s^2}])(15.0[s])^2$

$\Delta s=337.5[m]$

5 0

2 years ago

monochromatic light from a distant source is incident on a slit 0.75 mm wide. on screen 2 m away, the distance from the central

hjlf

Displacement from the center line for minimum intensity is 1.35 mm , width of the slit is 0.75 so Wavelength of the light is 506.25.

<h3>How to find Wavelength of the light?</h3>

When a wave is bent by an obstruction whose dimensions are similar to the wavelength, diffraction is observed. We can disregard the effects of extremes because the Fraunhofer diffraction is the most straightforward scenario and the obstacle is a long, narrow slit.

This is a straightforward situation in which we can apply the

Fraunhofer single slit diffraction equation:

y = mλD/a

Where:

y = Displacement from the center line for minimum intensity = 1.35 mm

λ = wavelength of the light.

D = distance

a = width of the slit = 0.75

m = order number = 1

Solving for λ

λ = y + a/ mD

Changing the information that the issue has provided:

λ = 1.35 * 10^-3 + 0.75 * 10^-3 / 1*2

=5.0625 *10^-7 = 506.25

Wavelength of the light 506.25.

To learn more about Wavelength of the light refer to:

brainly.com/question/15413360

#SPJ4

5 0

1 year ago