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3 years ago

List two ways that waves erode land

2 answers:

8 0

Waves erode through forcing air bubbles into tiny cracks in rocks (hydraulic action), and also by abrasion, whereby it causes rocks to be hurled against rock faces.

I hope this helps!

Veseljchak [2.6K]3 years ago

8 0

Answer:

Waves erode by abrasion and deposition

Explanation:

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Me ajudem, Por favor!!!!!!

zzz [600]

Answer:

a) $a=4\,\frac{m}{s^2}$

b) $V(t)=4\,t\,+3$

c) $V(1)=7 \,\frac{m}{s} \\$

d) Displacement = 22 m

e) Average speed = 11 m/s

Explanation:

Notice that the acceleration is the derivative of the velocity function, which in this case, being a straight line is constant everywhere, and which can be calculated as:

$slope= \frac{15=3}{3-0} =4\,\frac{m}{s^2}$

Therefore, acceleration is $a=4\,\frac{m}{s^2}$

b) the functional expression for this line of slope 4 that passes through a y-intercept at (0, 3) is given by:

$y=m\,x+b\\V(t)=4\,t\,+3$

c) Since we know the general formula for the velocity, now we can estimate it at any value for 't", for example for the requested t = 1 second:

$V(t)= 4\,t+3\\V(1)=4\,(1)+3\\V(1)=7 \,\frac{m}{s}$

d) The displacement between times t = 1 sec, and t = 3 seconds is given by the area under the velocity curve between these two time values. Since we have a simple trapezoid, we can calculate it directly using geometry and evaluating V(3) (we already know V(1)):

Displacement = $\frac{(7+15)\,2}{2} = 22\,\,m$

e) Recall that the average of a function between two values is the integral (area under the curve) divided by the length of the interval:

Average velocity = $\frac{22}{2} = 11\, \,\frac{m}{s}$

3 0

3 years ago

Find the horizontal component and the vertical component

aleksandr82 [10.1K]

Answer:

v=3.66,h-3.66

Explanation:

vertical = 10sin60 - 10sin 30

horizontal =10cos60 + 10cos 30

v = 10×0.8660-10×0.5

h = 10×0.5 + 10 × 0.8660

v=8.660-5.0 = 3.66

h= 5.0-8.660 = -3.66

8 0

2 years ago

A horizontal circular platform rotates counterclockwise about its axis at the rate of 0.945 rad/s. You, with a mass of 69.7 kg,

pickupchik [31]

Answer:

317.22

Explanation:

Given

Circular platform rotates ccw 93.1kg, radius 1.93 m, 0.945 rad/s

You 69.7kg, cw 1.01m/s, at r

Poodle 20.2 kg, cw 1.01/2 m/s, at r/2

Mutt 17.7 kg, 3r/4

You

Relative

ω = v/r

= 1.01/1.93

= 0.522

Actual

ω = 0.945 - 0.522

= 0.42

I = mr^2

= 69.7*1.93^2

= 259.6

L = Iω

= 259.6*0.42

= 109.4

Poodle

Relative

ω = (1.01/2)/(1.93/2)

= 0.5233

Actual

ω = 0.945- 0.5233

= 0.4217

I = m(r/2)^2

= 20.2*(1.93/2)^2

= 18.81

L = Iω

= 18.81*0.4217

= 7.93

Mutt

Actual

ω = 0.945

I = m(3r/4)^2

= 17.7(3*1.93/4)^2

= 37.08

L = Iω

= 37.08*0.945

= 35.04

Disk

I = mr^2/2

= 93.1(1.93)^2/2

= 173.39

L = Iω

= 173.39*0.945

= 163.85

Total

L = 109.4+ 7.93+ 36.04+ 163.85

= 317.22 kg m^2/s

4 0

3 years ago

To drive your car from Auburn to Atlanta (173.8 km) takes a significant amount of work. Assuming the force of friction between t

Daniel [21]

Answer:

1788.402 MJ

Explanation:

Work done = Force (N) x distance (m)

First we have to convert distance into metres:

173.8 x 1000 = 173,800 m

Then plug these values into the equation above:

173,800 x 10290 = 1788402000 J

The reason it's Joules (the unit for energy) is because work done = energy transferred

Now we have to convert Joules into Mega Joules:

1 J = 1/1000000 MJ

1788402000 J = 1788402000/ 1000000 = 1788.402 MJ

5 0

3 years ago

a body builder uses a chest expander with 5 strings.it takes a force of 200n to pull one string by 15cm.how much force will be n

wolverine [178]

Answer:

1000N

Explanation:

It takes 200N to expand 15cm of one string.

For 5 strings, force required =

200 x 5 = 1000N

5 0

2 years ago