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3 years ago

What causes competition among organisms?

Physics

1 answer:

dimaraw [331]3 years ago

5 0

Both organisms attempt to use the same limited sources

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A negative ion of charge -2e is located at the origin and a second negative ion of charge -3e is located nearby at x = 3.8 nm ,

Rus_ich [418]

Answer:

$\vec{F}_{21}=-5.63\times 10^{-11}N\\\\\vec{F}_{21}=\\$

Explanation:

Given that

$Q_1 = -2e\, C\\\\Q_2=-3e\,C\\\\x= 3.8 \times 10^{-9}\,m\\\\y= 3.2 \times 10^{-9}\,m\\\\r=\sqrt{x^2+y^2}\\\\r= 4.96\times 10^{-9} m\\$

As both charges are negative so there exist force of repulsion in direction as shown in figure.

$F_{12}=\frac{kQ_1Q_2}{r^2}\\\\F_{12}= \frac{(9\times 10^9)(6)(1.602\times 10^{-19})^2}{(4.96\times 10^{-9})^2}\\\\F_{12}=5.63\times 10^{-11}N$

Angle at which force F12 is acting is

$\theta=tan^{-1}\frac{3.2}{3.8}\\\\\theta=tan^{-1}\frac{y}{x}\\\\\theta= 40.1^o$

$F_{x}=F_{12}cos\theta\\\\F_{x}=(5.63\times 10^{-11})cos(40.1)\\\\F_{x}=4.306\times 10^{-11}N\\\\F_{y}=F_{12}sin\theta\\\\F_{y}=(5.63\times 10^{-11})sin(40.1)\\\\F_{y}=3.62\times 10^{-11}N\\\\$

$\vec{F}_{12}=\vec{F}_{x}+\vec{F}_y\\\\\vec{F}_{12}=4.30\times 10^{-11}\,\hat{i} + 3.62\times 10^{-11}\,\hat{j}\\\\\vec{F}_{12}=$

Force exerted on charge -2e is equal in magnitude to F12 but is in opposite direction

$F_{21}=-5.63\times 10^{-11}N$

$\vec{F}_{21}=$

7 0

4 years ago

Water drops fall from the edge of a roof at a steady rate. a fifth drop starts to fall just as the first drop hits the ground. a

Alchen [17]

The height of the roof is <u>3.57m</u>

Let the drops fall at a rate of 1 drop per t seconds. The first drop takes 5t seconds to reach the ground. The second drop takes 4t seconds to reach the bottom of the 1.00 m window, while the 3rd drop takes 3t s to reach the top of the window.

Calculate the distances traveled by the second and the third drops s₂ and s₃, which start from rest from the roof of the building.

$s_2=\frac{1}{2} g(4t)^2=8gt^2\\ s_3=\frac{1}{2} g(3t)^2=(4.5)gt^2$

The length of the window s is given by,

$s=s_2-s_3\\ (1.00 m)=8gt^2-4.5gt^2=3.5gt^2\\ t^2=\frac{1.00 m}{(3.5)(9.8m/s^2)} =0.02915s^2$

The first drop is at the bottom and it takes 5t seconds to reach down.

The height of the roof h is the distance traveled by the first drop and is given by,

$h=\frac{1}{2} g(5t)^2=\frac{25t^2}{2g} =\frac{25(0.02915s^2)}{2(9.8m/s^2)} =3.57 m$

the height of the roof is 3.57 m

8 0

4 years ago

Read 2 more answers

Determine the ratio of the resistivity of pure water to silver?

shutvik [7]

Silver is a very good conductor, this means its resistivity is very low (from table, we can check the precise value, which is $\rho_s = 1.6 \cdot 10^{-8} \Omega m$ ).

Pure water, instead, is a very bad conductor, this means its resistivity is very high, of order of $k \Omega \cdot m$ ( $10^3 \Omega m$ ). Even without knowing the precise value of the pure water resistivity, we can estimate the ratio between the pure water resistivity and the silver resistivity by comparing the two orders of magnitude:
$r= \frac{10^3 \Omega m}{10^{-8} \Omega m} \sim 10^{11}$

Therefore, we can say that the correct answer is
$3 \cdot 10^{11} : 1$

3 0

3 years ago

A box slides down a ramp inclined at 24◦ to the horizontal with an acceleration of 1.7 m/s 2 . The acceleration of gravity is 9.

dsp73

Answer:

Explanation:

In friction, the coefficient of friction formula is expressed as;

$\mu = \frac{F_f}{R}$

Ff is the frictional force = Wsinθ

R is the reaction = Wcosθ

Substitute inti the equation;

$\mu = \frac{Wsin \theta}{W cos\theta} \\\mu = \frac{sin \theta}{cos\theta} \\\mu = tan \theta$

Given

θ = 24°

$\mu = tan 24^0\\\mu = 0.445\\$

Hence the coefficient of kinetic friction between the box and the ramp is 0.445

3 0

3 years ago

How much force is needed to stop a 90-kg soccer player if he decelerates at 15 m/s^2?

svet-max [94.6K]

We have: F = m×a
Here, m = 90 Kg
a = 15 m/s²

Substitute their values into the expression:
F = 90 × 15
F = 1350 N

In short, Your Answer would be Option D

Hope this helps!

8 0

4 years ago