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il63 [147K]
3 years ago
13

Which action packs down the soil and depletes vegetation? A. wind erosion B. overgrazing C. sedimentation D. sand dune formation

Physics
2 answers:
Deffense [45]3 years ago
4 0

Answer:

The answer is B! (overgrazing)

Explanation:

I took this test, got it right. It is 100% overgrazing :)

Troyanec [42]3 years ago
3 0
Hey there,
The answer is A, wind erosion

Wind erosion reduces the capacity of the soil.

Hope this helps :))

<em>~Top♥</em>
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Inertia keeps your vehicle moving until a force slows it down, like __________.
GrogVix [38]
Technically all 3 of those would be correct answers. But applying the brakes would probably be the correct answer if all of the above isn't a answer.
5 0
3 years ago
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7. A spring balance reads force in Newton. The scale is 20 cm long and reads from 0 to
GarryVolchara [31]

Answer:

The potential  energy when it reads 40 N is PE  = 5.33 \ J

Explanation:

From the question we are told that

   The lowest reading of the spring balance is  0 N and this is at  0 cm = 0 m

   The height reading of the spring balance is 60 N  and this is at 20 cm =  0.20 m

   Generally the length corresponding to the reading of 40 N is mathematically represented as

       d = \frac{40 * 0.20 }{60 }

=>    d = 0.133 \  m

Generally the potential  energy is mathematically represented as

      PE = m * g * d

Here

     F = mg  =  40 N

So  

     PE = 40 * 0.133

=>  PE  = 5.33 \ J

7 0
3 years ago
imagine a bouncing ball that does not lose any energy as it bounces. could it ever bounce to a greater height than it was droppe
iren [92.7K]
Depends on if it was thrown or dropped
8 0
3 years ago
Read 2 more answers
Which of the following can penetrate the deepest (Please explain)
harkovskaia [24]

Answer: 3MeV electron

Explanation:

m_e={9.1\times 10^{-31}      m_α=4\times m_e    m_a={9.1\times 10^{-31}

m_p=1.67\times 10^{-27}

(a)  K.E. Energy of electron =\frac{1}{2}\times{m_e}\times{v_{e} ^{2}}=3MeV

v_{e} ^{2}=\frac{2\times3\times1.6\times10^{-19}\times10^{6}  }{9.1\times 10^{-31} }=1.05\times10^{18}

v_e=\sqrt{1.05\times10^{18} } = 1.025\times10^{9}\frac{m}{s}

(b) K.E. Energy of alpha particle =\frac{1}{2}\times{m_\alpha}\times{v_{\alpha} ^{2}}=10MeV

v_{\alpha} ^{2}= \frac{2\times10\times1.6\times10^{-19}\times10^{6}  }{4\times9.1\times 10^{-31} }=0.88\times10^{18}

v_\alpha=\sqrt{0.88\times10^{18} } =.94\times10^{9}\frac{m}{s}

(c) K.E. Energy of auger particle =\frac{1}{2}\times{m_a}\times{v_{a} ^{2}}=0.1MeV

v_{a} ^{2}=\frac{2\times0.1\times1.6\times10^{-19}\times10^{6}  }{9.1\times 10^{-31} }=0.035\times10^{18}

v_a=\sqrt{0.035\times10^{18} } =.19\times10^{9}\frac{m}{s}

(d)  K.E. Energy of proton particle =\frac{1}{2}\times{m_p}\times{v_{p} ^{2}}=400keV

v_{p} ^{2}=\frac{2\times400\times1.6\times10^{-19}\times10^{3}  }{1.67\times 10^{-27} }=0.766\times10^{14}

v_p=\sqrt{0.766\times10^{14} } =0.875\times10^{7}[tex]\frac{m}{s}

from (a),(b),(c),and (d) we can clearly say that the velocity of the electron is more so the penetration of the electron will be deepest.

6 0
3 years ago
You kick a soccer ball with a speed of 18 m/s at an angle of 43. How long does it take the the ball to reach the top of its traj
Sati [7]

Answer:

1.25 s

Explanation:

The motion of the soccer ball is an example of a projectile motion.

So, time taken by the ball to reach the top of the trajectory is exactly half of the total time of flight of the soccer ball.

In a projectile motion, the projectile is thrown at an angle to the ground with some initial velocity. The motion can be divided into two independent direction motions; one in the horizontal direction and the other in the vertical direction.

The time of flight is the time taken by the projectile to reach the ground again.

The formula for time of flight is given as:

T=\frac{2u\sin \theta}{g}\\Where,T\to \textrm{Time of flight}\\u\to \textrm{Initial velocity of ball}\\\theta \to \textrm{Angle of projection}\\g\to \textrm{Acceleration due to gravity}

Here, u=18\ m/s,\theta=43\°,g=9.8\ m/s^2. Therefore,

T=\frac{2\times 18\times \sin 43\°}{9.8}=2.505\ s

Now, the time taken to reach the top is half of the total time. Therefore,

T_{top}=\frac{T}{2}=\frac{2.505}{2}=1.25\ s

Therefore, it takes 1.25 s for the ball to reach the top of its trajectory.

5 0
3 years ago
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