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madreJ [45]
3 years ago
7

Which of the following is an impact of trampling of beach grass by humans and vehicles?

Physics
2 answers:
kap26 [50]3 years ago
4 0
I think the answer is B.
ololo11 [35]3 years ago
4 0

the answer is b becaue what comes out when u turn it on smoke that no good for popullation.

You might be interested in
A plane starting from rest accelerates to 40 m/s in 10 s. How far did the plane travel during this time?
andrew-mc [135]

Answer:

400 meters

Explanation:

using the equation:

  • speed = distance/time
  • it would look like this when plugged in: 40=d/10
  • d (distance) is the unknown variable, so you have to solve for that
  • multiply both sides by 10 since you want to separate d to solve its number.
  • 10d/10=40 × 10
  • the two tens on the d part of the equation would cancel out, leaving only d.
  • Therefore, d = 400

Hopefully this helps.

5 0
3 years ago
The average height of an apple tree is 4.00 meters. How long would it take an apple falling from that height to reach the ground
Elodia [21]
We know, s = ut + 1/2gt²
In free-fall, u = 0
so, t = √2s/g

Now, substitute the values, 
t = √2*4 / 9.8   [ -ve sign shows opposite direction only ]
t = √8/ 9.8
t = 0.90 sec

In short, Your Answer would be 0.90 sec

Hope this helps!
8 0
4 years ago
Read 2 more answers
A 250 GeV beam of protons is fired over a distance of 1 km. If the initial size of the wave packet is 1 mm, find its final size
Margarita [4]

Answer:

The final size is approximately equal to the initial size due to a very small relative increase of 1.055\times 10^{- 7} in its size

Solution:

As per the question:

The energy of the proton beam, E = 250 GeV =250\times 10^{9}\times 1.6\times 10^{- 19} = 4\times 10^{- 8} J

Distance covered by photon, d = 1 km = 1000 m

Mass of proton, m_{p} = 1.67\times 10^{- 27} kg

The initial size of the wave packet, \Delta t_{o} = 1 mm = 1\times 10^{- 3} m

Now,

This is relativistic in nature

The rest mass energy associated with the proton is given by:

E = m_{p}c^{2}

E = 1.67\times 10^{- 27}\times (3\times 10^{8})^{2} = 1.503\times 10^{- 10} J

This energy of proton is \simeq 250 GeV

Thus the speed of the proton, v\simeq c

Now, the time taken to cover 1 km = 1000 m of the distance:

T = \frac{1000}{v}

T = \frac{1000}{c} = \frac{1000}{3\times 10^{8}} = 3.34\times 10^{- 6} s

Now, in accordance to the dispersion factor;

\frac{\delta t_{o}}{\Delta t_{o}} = \frac{ht_{o}}{2\pi m_{p}\Delta t_{o}^{2}}

\frac{\delta t_{o}}{\Delta t_{o}} = \frac{6.626\times 10^{- 34}\times 3.34\times 10^{- 6}}{2\pi 1.67\times 10^{- 27}\times (10^{- 3})^{2} = 1.055\times 10^{- 7}

Thus the increase in wave packet's width is relatively quite small.

Hence, we can say that:

\Delta t_{o} = \Delta t

where

\Delta t = final width

3 0
3 years ago
What is average velocity
Sphinxa [80]

Answer:

Average velocity is the displacement of an object over time that also in a specified direction...

Explanation:

brainliest plz

6 0
3 years ago
If 20N force produces an acceleration of 5ms^-2 In a body then the mass of the body will be:
meriva
F = ma
m = F/a
m = 20N / 5m/s^2
m = 4kg
5 0
3 years ago
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