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3 years ago

• How does wind shape Earth’s surface?

Physics

1 answer:

ehidna [41]3 years ago

6 0

Lamina and turbulent flow

Explanation:

mentioning about lamina and turbulent flow we could say that both form in different period of time

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A woman does 260 J of work to move a table 1.9 m across the floor. What is the magnitude of the force applied by the woman to th

balandron [24]

Answer:136.8N

Explanation:

Work=260joules

Distance=1.9m force=?

Force=work/distance

Force=260/1.9

Force=136.8N

4 0

3 years ago

The forces acting on a leaf are

otez555 [7]

Affected plants

Explanation:

8 0

3 years ago

In young’s double slit experiment, the measured fringe width is 0.5 mm for a Sodium light of 589 nm at a distance of 1.5 m. A br

Levart [38]

Answer:

(A). The order of the bright fringe is 6.

(B). The width of the bright fringe is 3.33 μm.

Explanation:

Given that,

Fringe width d = 0.5 mm

Wavelength = 589 nm

Distance of screen and slit D = 1.5 m

Distance of bright fringe y = 1 cm

(A) We need to calculate the order of the bright fringe

Using formula of wavelength

$\lambda=\dfrac{dy}{mD}$

$m=\dfrac{d y}{\lambda D}$

Put the value into the formula

$m=\dfrac{1\times10^{-2}\times0.5\times10^{-3}}{589\times10^{-9}\times1.5}$

$m=5.65 = 6$

(B). We need to calculate the width of the bright fringe

Using formula of width of fringe

$\beta=\dfrac{yd}{D}$

Put the value in to the formula

$\beta=\dfrac{1\times10^{-2}\times0.5\times10^{-3}}{1.5}$

$\beta=3.33\times10^{-6}\ m$

$\beta=3.33\ \mu m$

Hence, (A). The order of the bright fringe is 6.

(B). The width of the bright fringe is 3.33 μm.

3 0

3 years ago

Each of the following diagrams shows a spaceship somewhere along the way between Earth and the Moon (not to scale); the midpoint

djyliett [7]

Answer:

$F_5 >F_4>F_1 >F_2>F_3$

Where $F_i$ represent the force for each of the 5 cases $i -1,2,3,4,5$ presented on the figure attached.

Explanation:

For this case the figure attached shows the illustration for the problem

We have an inverse square law with distance for the force, so then the force of gravity between Earth and the spaceship is lower when the spaceship is far away from Earth.

Th formula is given by:

$F = G \frac{m_{Earth} m_{Spaceship}}{r^2}$

Where G is a constant $G = 6.674 x10^{-11} m^2/ (ks s^2)$

$m_{Earth}$ represent the mass for the earth

$m_{spaceship}$ represent the mass for the spaceship

$r$ represent the radius between the earth and the spaceship

For this reason when the distance between the Earth and the Spaceship increases the Force of gravity needs to decrease since are inversely proportional the force and the radius, and for the other case when the Earth and the spaceship are near then the radius decrease and the Force increase.

Based on this case we can create the following rank:

$F_5 >F_4>F_1 >F_2>F_3$

Where $F_i$ represent the force for each of the 5 cases $i -1,2,3,4,5$ presented on the figure attached.

6 0

3 years ago

(20) A rocket is launched vertically. At time t = 0 seconds, the rocket’s engine shuts down. At the time, the rocket has reached

Diano4ka-milaya [45]

Answer:

Explanation:

Given that,

h(t) = -9.8t² / 2 + 125t + 500

for t > 0.

At t = 0, the rocket is at height h = 500m, at a velocity of Vo = 125m/s.

We want to find the maximum height reached by rocket

Using mathematics maxima and minima

let find the turning point when dh/dt = 0

dh/dt = -9.8t + 125

dh / dt = 0 = -9.8t + 125

9.8t = 125

t = 125 / 9.8

t = 12.76s

Let find the turning point to know if this time t = 12.76 is maximum or minimum point

Let find d²h / dt²

d²h / dt² = -9.8

Since, d²h/dt² < 0, then, at t = 12.76s is the maximum points.

Then, the maximum height reached is

h = -9.8t² / 2 + 125t + 500

h = -9.8(12.76)² / 2 + 125(12.76) + 500

h = -797.80 + 1595 + 500

h = 1297.2 m

The maximum height reached is 1297.2 m

From the attachment, the maximum height is 1297.2m at t = 12.76sec.

Comment, the result are the same for both the analysis aspect and the graphical aspect.

3 0

3 years ago