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2 years ago

Which method should be used to determine which type of natural event produces the greatest number of sand dunes?

Physics

1 answer:

jeka942 years ago

5 0

Answer:

Stabilizing dunes involves multiple actions. Planting vegetation reduces the impact of wind and water. Wooden sand fences can help retain sand and other material needed for a healthy sand dune ecosystem. Footpaths protect dunes from damage from foot traffic.

Explanation:

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You are on vacation in San Francisco and decide to take a cable car to see the city. A 5800-kgkg cable car goes 260 mm up a hill

Stella [2.4K]

Answer:

$4.325\times10^6J$

Explanation:

Mass of the cable car, m = 5800 kg

It goes 260 m up a hill, along a slope of $\theta=17^o$

Therefore vertical elevation of the car = $260sin\theta=260sin17^o=76.0166m$

Now, when you get into the cable car, it's velocity is zero, that is, initial kinetic energy is zero (since K.E. = $\frac{1}{2} mv^2$ ). Similarly as the car reaches the top, it halts and hence final kinetic energy is zero.

Therefore the only possible change in the cable car system is the change in it's gravitational potential energy.

Hence, total change in energy = mgh = $5800\times9.81\times76.0166J=4.325\times10^6J$

where, g = acceleration due to gravity

h = height/vertical elevation

4 0

2 years ago

A thin beam of light enters a thick plastic sheet from air at an angle of 32.0° with the normal and continues in the sheet at an

Cloud [144]

Answer:

1.36

Explanation:

$n_{air}$ = Index of refraction of air = 1

$n_{plastic}$ = Index of refraction of plastic = ?

i = angle of incidence in air = 32.0° deg

r = angle of refraction in plastic = 23.0° deg

Using Snell's law

$n_{air}$ Sini = $n_{plastic}$ Sinr

(1) SIn32.0 = $n_{plastic}$ Sin23.0

$n_{plastic}$ = 1.36

5 0

2 years ago

A box with a mass of 100.0 kg slides down a ramp with a 50 degree angle. What is the weight of the box? N What is the value of t

nadya68 [22]

1) weight of the box: 980 N

The weight of the box is given by:

$W=mg$

where m=100.0 kg is the mass of the box, and $g=9.8 m/s^2$ is the acceleration due to gravity. Substituting in the formula, we find

$W=(100.0 kg)(9.8 m/s^2)=980 N$

2) Normal force: 630 N

The magnitude of the normal force is equal to the component of the weight which is perpendicular to the ramp, which is given by

$N=W cos \theta$

where W is the weight of the box, calculated in the previous step, and $\theta=50^{\circ}$ is the angle of the ramp. Substituting, we find

$N=(980 N)(cos 50^{\circ})=630 N$

3) Acceleration: $7.5 m/s^2$

The acceleration of the box along the ramp is equal to the component of the acceleration of gravity parallel to the ramp, which is given by

$a_p = g sin \theta$

Substituting, we find

$W_p = (9.8 m/s^2)(sin 50^{\circ})=7.5 m/s^2$

5 0

2 years ago

The image shows positions of the earth and the moon in which region would an astronaut feel the lightest

trapecia [35]

Answer:

The moon region

Explanation:

This is because there is little to no gravity on the moon. That is where the astronaut would feel the lightest.

5 0

1 year ago

Read 2 more answers

A tennis player hits a ball 2.0 m above the ground. The ball leaves his racquet with a speed of 20 m/s at an angle 5.0 ∘ above t

ipn [44]

Answer:

ball clears the net

Explanation:

$v_{o}$ = initial speed of launch of the ball = 20 ms^{-1}

$\theta$ = angle of launch = 5 deg

Consider the motion of the ball along the horizontal direction

$v_{ox}$ = initial velocity = $v_{o} Cos\theta = 20 Cos5 = 19.92 ms^{-1}$

$t$ = time of travel

$X$ = horizontal displacement of the ball to reach the net = 7 m

Since there is no acceleration along the horizontal direction, we have

$X = v_{ox} t \\7 = v_{ox} t\\t = \frac{7}{v_{ox}}$ Eq-1

Consider the motion of the ball along the vertical direction

$v_{oy}$ = initial velocity = $v_{o} Sin\theta = 20 Sin5 = 1.74 ms^{-1}$

$t$ = time of travel

$Y_{o}$ = Initial position of the ball at the time of launch = 2 m

$Y$ = Final position of the ball at time "t"

$a_{y}$ = acceleration in down direction = - 9.8 ms⁻²

Along the vertical direction , position at any time is given as

$Y = Y_{o} + v_{oy} t + (0.5) a_{y} t^{2}\\Y = 2 + (20 Sin5) (\frac{7}{20 Cos5}) + (0.5) (- 9.8) (\frac{7}{20 Cos5})^{2}\\Y = 2.00758 m\\$

Since Y > 1 m

hence the ball clears the net

7 0

2 years ago