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Mariulka [41]
2 years ago
5

Which term best describes those processes that move weathered rock materials and soils downslope?

Physics
1 answer:
mote1985 [20]2 years ago
5 0

Answer:

mass wasting

Explanation:

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A 800 kg safe is 2.1 m above a heavy-duty spring when the rope holding the safe breaks. The safe hits the spring and compresses
stellarik [79]

Answer:

k = 17043.5 N/m = 17.04 KN/m

Explanation:

First we need to find the force applied by safe pn the spring:

F = Weight of Safe

F = mg

where,

F = Force Applied by the safe on the spring = ?

m = mass of safe = 800 kg

g = 9.8 m/s²

Therefore,

F = (800 kg)(9.8 m/s²)

F = 7840 N

Now, using Hooke's Law:

F = kΔx

where,

K = Spring Constant = ?

Δx = compression = 46 cm = 0.46 m

Therefore,

7840 N = k (0.46 m)

k = 7840 N/0.46 m

<u>k = 17043.5 N/m = 17.04 KN/m</u>

5 0
3 years ago
When beavers inhabit an area, they cut down trees and construct beaver dams, which slow the flow of rivers or streams. Which of
Airida [17]

Answer: H

Explanation:

When beavers dam a stream they slow the movement of water. Behind the beaver dam, a pond of still water is formed. This pond is then colonized by animals and plants that typically live in lakes rather than streams.

6 0
2 years ago
A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.6 m/s at ground level.
kow [346]

Before the engines fail, the rocket's altitude at time <em>t</em> is given by

y_1(t)=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2

and its velocity is

v_1(t)=80.6\dfrac{\rm m}{\rm s}+\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t

The rocket then reaches an altitude of 1150 m at time <em>t</em> such that

1150\,\mathrm m=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2

Solve for <em>t</em> to find this time to be

t=11.2\,\mathrm s

At this time, the rocket attains a velocity of

v_1(11.2\,\mathrm s)=124\dfrac{\rm m}{\rm s}

When it's in freefall, the rocket's altitude is given by

y_2(t)=1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2

where g=9.80\frac{\rm m}{\mathrm s^2} is the acceleration due to gravity, and its velocity is

v_2(t)=124\dfrac{\rm m}{\rm s}-gt

(a) After the first 11.2 s of flight, the rocket is in the air for as long as it takes for y_2(t) to reach 0:

1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2=0\implies t=32.6\,\mathrm s

So the rocket is in motion for a total of 11.2 s + 32.6 s = 43.4 s.

(b) Recall that

{v_f}^2-{v_i}^2=2a\Delta y

where v_f and v_i denote final and initial velocities, respecitively, a denotes acceleration, and \Delta y the difference in altitudes over some time interval. At its maximum height, the rocket has zero velocity. After the engines fail, the rocket will keep moving upward for a little while before it starts to fall to the ground, which means y_2 will contain the information we need to find the maximum height.

-\left(124\dfrac{\rm m}{\rm s}\right)^2=-2g(y_{\rm max}-1150\,\mathrm m)

Solve for y_{\rm max} and we find that the rocket reaches a maximum altitude of about 1930 m.

(c) In part (a), we found the time it takes for the rocket to hit the ground (relative to y_2(t)) to be about 32.6 s. Plug this into v_2(t) to find the velocity before it crashes:

v_2(32.6\,\mathrm s)=-196\frac{\rm m}{\rm s}

That is, the rocket has a velocity of 196 m/s in the downward direction as it hits the ground.

3 0
3 years ago
If the radius of a blood vessel drops to 84.0% of its original radius because of the buildup of plaque, and the body responds by
erma4kov [3.2K]

To develop this problem it is necessary to apply the equations concerning Bernoulli's law of conservation of flow.

From Bernoulli it is possible to express the change in pressure as

\Delta P = \frac{1}{2}\rho (v_1^2-v_2^2)+ \rho g (h_1h_2)

Where,

v_i =Velocity

\rho = Density

g = Gravitational acceleration

h = Height

From the given values the change of flow is given as

R = r^4P

Therefore between the two states we have to

\frac{R_2}{R_1} = \frac{r_2^4 P_2}{r_1^4 P_1} *100\%

\frac{R_2}{R_1} = \frac{84^4 (110)}{100^4*(100)} *100\%

\frac{R_2}{R_1} = 54.77\%

The flow rate will have changed to 54.77 % of its original value.

4 0
2 years ago
A certain ocean wave has a frequency of 0.07 hertz and a wavelength of 10 meters. what is the wave's speed? a 0.07 m/s
Elis [28]
V=f*wavelength
v=0.07hz*10m
v=0.7 m/s
8 0
3 years ago
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