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Lubov Fominskaja [6]
3 years ago
14

Consider the aquatic phosphorus cycle. On land most phosphorus is found in rocks and minerals. In the oceans, phosphorus is depo

sited from decomposition, excretion from animals, run-off from the land, and finally ______________ of rocks and minerals.
A) deposition
B) erosion
C) reformation
D) weathering

Physics
2 answers:
Paraphin [41]3 years ago
8 0

Answer: D) weathering

Phosphorus is a chemical organic element that is found in numerous compounds of the Earth typically found in water, sediments and soils. The phosphorus in the soil is being added up by the chemical fertilizers. In the ocean or marine water the phosphorus is deposited as animal waste and by the weathering of the rocks and minerals.

The weathering is a process by which the rocks are broken down into small parts and sediments by the action of the physical agents such as water, wind and chemicals. In the given situation the weathering is the process by which the element phosphorus will be added to the bottom bed of the ocean.

dmitriy555 [2]3 years ago
4 0
D) weathering  im 100% sure i took the test and i pasted so i know 
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A cosmic-ray proton in interstellar space has an energy of 10.0 MeV and executes a circular orbit having a radius equal to that
kherson [118]

Answer:

= 7.88 × 10^-12 T

Explanation:

From the above question, we are told that:

Kinetic Energy of the proton is K. E = 10.0 MeV

Step 1

We convert 10.0 MeV to Joules

1 Mev = 1.602 × 10-13 Joules

10.0 MeV = 10.0 × 1.602 × 10^-13 Joules = 1.602 × 10^-12 J

Hence, the Kinectic energy of a proton = 1.602 × 10^-12 J

Step 2

Find the Speed of the Proton

The formula for Kinectic Energy =

K.E = 1/ 2 mv²

Where

m = mass of the proton

v = speed of the proton

K.E of the proton = 1.602 × 10^-12 J

Mass of the proton = 1.6726219 × 10^-27 kilograms

Speed of the proton = ?

1.602 × 10^-12J = 1/2 × 1.6726219 × 10^-27 × v²

1.602 × 10^-12J = 8.3631095 ×10^-28 × v²

v² = 1.602 × 10^-12/8.3631095 ×10^-28

v = √(1.602 × 10^-12/8.3631095 ×10^-28)

v = 43772331.227m/s

v = 4.3772331227 × 10^7m/s

Approximately = 4.4 × 10^7 m/s

Step 3

Find the Magnetic Field of that region of space

The formula for Magnetic Field =

B = m v / q r

We are told that the proton executes a circular orbit, hence,

mv = √2m(KE)

m = Mass of the proton = 1.6726219 × 10^-27 kg

K.E of the proton = 1.602 × 10^-12 J

v = speed of the proton = 4.4 × 10^7 m/s

q = Electric charge = 1.6 × 10^-19 C

r = radius of the orbit = 5.80Ã10^10 m

= 5.8 × 10^10m

Magnetic Field =

=√ (2 × 1.6726219 × 10^-27 kg × 1.602 × 10^-12 J) /( 1.6 × 10^-19 C × 5.80 × 10^10 m)

= 7.88 × 10^-12 T

The magnetic field in that region of space is approximately 7.88 × 10^-12 T

4 0
3 years ago
**100 points** PLEASE ANSWER IN 3 PARAGRAPHS
Deffense [45]

Answer:

In the previous section, we defined circular motion. The simplest case of circular motion is uniform circular motion, where an object travels a circular path at a constant speed. Note that, unlike speed, the linear velocity of an object in circular motion is constantly changing because it is always changing direction. We know from kinematics that acceleration is a change in velocity, either in magnitude or in direction or both. Therefore, an object undergoing uniform circular motion is always accelerating, even though the magnitude of its velocity is constant.

You experience this acceleration yourself every time you ride in a car while it turns a corner. If you hold the steering wheel steady during the turn and move at a constant speed, you are executing uniform circular motion. What you notice is a feeling of sliding (or being flung, depending on the speed) away from the center of the turn. This isn’t an actual force that is acting on you—it only happens because your body wants to continue moving in a straight line (as per Newton’s first law) whereas the car is turning off this straight-line path. Inside the car it appears as if you are forced away from the center of the turn. This fictitious force is known as the centrifugal force. The sharper the curve and the greater your speed, the more noticeable this effect becomes.

Figure 6.7 shows an object moving in a circular path at constant speed. The direction of the instantaneous tangential velocity is shown at two points along the path. Acceleration is in the direction of the change in velocity; in this case it points roughly toward the center of rotation. (The center of rotation is at the center of the circular path). If we imagine Δs becoming smaller and smaller, then the acceleration would point exactly toward the center of rotation, but this case is hard to draw. We call the acceleration of an object moving in uniform circular motion the centripetal acceleration ac because centripetal means center seeking.

hope it helps! stay safe and tell me if im wrong pls :D

(brainliest if you want, or if its right pls) :)

4 0
2 years ago
a spaceship is traveling at a speed of 15000 km/s from planet b toward planet a the spaceship sends out a signal with a waveleng
Bumek [7]

Answer: 4nmeter

Explanation: The two observer a and b will measure the same wavelength since the speed of the space craft is very small compared with the speed of light c. That is

V which is the speed of space craft 15000km/s = 15000000m/s

Comparing this with the speed of light c 3*EXP(8)m/s we have

15000000/300000000

= 0.05=0.1

Therefore the speed of the space craft V in terms of the speed of light c is 0.1c special relativity does not apply to object moving at such speed. So the wavelength would not be contracted it will remain same for both observers.

4 0
3 years ago
Read 2 more answers
A ball is kicked at an angle of 35° with the ground.a) What should be the initial velocity of the ball so that it hits a target
stiks02 [169]

Answer:

a.18.5 m/s

b.1.98 s

Explanation:

We are given that

\theta=35^{\circ}

a.Let v_0 be the initial velocity of the ball.

Distance,x=30 m

Height,h=1.8 m

v_x=v_0cos\theta=v_0cos35

v_y=v_0sin\theta=v_0sin35

x=v_0cos\theta\times t=v_0cos35\times t

t=\frac{30}{v_0cos35}

h=v_yt-\frac{1}{2}gt^2

Substitute the values

1.8=v_0sin35\frac{30}{v_0cos35}-\frac{1}{2}(9.8)(\frac{30}{v_0cso35})^2

1.8=30tan35-\frac{6574.6}{v^2_0}

\frac{6574.6}{v^2_0}=21-1.8=19.2

v^2_0=\frac{6574.6}{19.2}

v_0=\sqrt{\frac{6574.6}{19.2}}=18.5 m/s

Initial velocity of the ball=18.5 m/s

b.Substitute the value then we get

t=\frac{30}{18.5cos35}

t=1.98 s

Hence, the time for the ball to reach the target=1.98 s

7 0
3 years ago
A jet plane flying 600 m/s experiences an acceleration of 4g when pulling out of the dive. What is the radius of curvature of th
Marina CMI [18]

Answer:

The radius of the curve is 9,183.67 m.

Explanation:

Given;

velocity of the jet plane, v = 600 m/s

acceleration of the jet plane, a = 4g = 4 x 9.8 m/s² = 39.2 m/s²

The radius of the curve is calculated from centripetal acceleration formula as given below;

a = \frac{v^2}{r} \\\\r = \frac{v^2}{a} \\\\r = \frac{600^2}{39.2} \\\\r = 9,183.67 \ m

Therefore, the radius of the curve is 9,183.67 m.

3 0
3 years ago
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