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3 years ago

Why isn't earth overcrowded with animals

1 answer:

3 0

The Earth isn't crowded with animals because Humans have overcrowded Earth, but also keep a constant limit with the animal population. However, many species are an endangered species due to the lack of balancing between certain creatures, such as turtles.
I hope this helps!

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Urgent!

Masja [62]

mass of iron block given as

$m_1 = 1.90 kg$

density of iron block is

$\rho = 7860 kg/m^3$

now the volume of the iron piece is given as

$V = \frac{m}{\rho}$

$V = \frac{1.90}{7860} = 2.42* 10^{-4} m^3$

Now when this iron block is complete submerged in oil inside the beaker the buoyancy force on the iron block will be given as

$F_b = \rho_L V g$

here we know that

$\rho_L$ = density of liquid = 916 kg/m^3

$F_b = 916* 2.42 * 10^{-4} * 9.8$

$F_b = 2.17 N$

Now for the reading of spring balance we can say the spring force and buoyancy force on the block will counter balance the weight of the block at equilibrium

$F_s + F_b = mg$

$F_s + 2.17 = 1.90* 9.8$

$F_s = 16.45 N$

So reading of spring balance will be 16.45 N

Now for other scale which will read the normal force of the surface we can write that normal force on the container will balance weight of liquid + container and buoyancy force on block

$F_n = F_g + F_b$

$F_n = (1 + 2.50)*9.8 + 2.17$

$F_n = 34.3 + 2.17 = 36.47 N$

So the other scale will read 36.47 N

3 0

3 years ago

Fill in the blank: An object is most likely to sink in water if _________________.

drek231 [11]

High density
random words to fill up 20 character minimum for answering questions :P

4 0

2 years ago

Read 2 more answers

Un the way to the moon, the Apollo astro-

kherson [118]

Answer:

Distance = 345719139.4[m]; acceleration = 3.33*10^{19} [m/s^2]

Explanation:

We can solve this problem by using Newton's universal gravitation law.

In the attached image we can find a schematic of the locations of the Earth and the moon and that the sum of the distances re plus rm will be equal to the distance given as initial data in the problem rt = 3.84 × 108 m

$r_{e} = distance earth to the astronaut [m].\\r_{m} = distance moon to the astronaut [m]\\r_{t} = total distance = 3.84*10^8[m]$

Now the key to solving this problem is to establish a point of equalisation of both forces, i.e. the point where the Earth pulls the astronaut with the same force as the moon pulls the astronaut.

Mathematically this equals:

$F_{e} = F_{m}\\F_{e} =G*\frac{m_{e} *m_{a}}{r_{e}^{2} } \\$

$F_{m} =G*\frac{m_{m}*m_{a} }{r_{m} ^{2} } \\where:\\G = gravity constant = 6.67*10^{-11}[\frac{N*m^{2} }{kg^{2} } ] \\m_{e}= earth's mass = 5.98*10^{24}[kg]\\ m_{a}= astronaut mass = 100[kg]\\m_{m}= moon's mass = 7.36*10^{22}[kg]$

When we match these equations the masses cancel out as the universal gravitational constant

$G*\frac{m_{e} *m_{a} }{r_{e}^{2} } = G*\frac{m_{m} *m_{a} }{r_{m}^{2} }\\\frac{m_{e} }{r_{e}^{2} } = \frac{m_{m} }{r_{m}^{2} }$

To solve this equation we have to replace the first equation of related with the distances.

$\frac{m_{e} }{r_{e}^{2} } = \frac{m_{m} }{r_{m}^{2} } \\\frac{5.98*10^{24} }{(3.84*10^{8}-r_{m} )^{2} } = \frac{7.36*10^{22} }{r_{m}^{2} }\\81.25*r_{m}^{2}=r_{m}^{2}-768*10^{6}* r_{m}+1.47*10^{17} \\80.25*r_{m}^{2}+768*10^{6}* r_{m}-1.47*10^{17} =0$

Now, we have a second-degree equation, the only way to solve it is by using the formula of the quadratic equation.

$r_{m1,2}=\frac{-b+- \sqrt{b^{2}-4*a*c } }{2*a}\\ where:\\a=80.25\\b=768*10^{6} \\c = -1.47*10^{17} \\replacing:\\r_{m1,2}=\frac{-768*10^{6}+- \sqrt{(768*10^{6})^{2}-4*80.25*(-1.47*10^{17}) } }{2*80.25}\\\\r_{m1}= 38280860.6[m] \\r_{m2}=-2.97*10^{17} [m]$

We work with positive value

rm = 38280860.6[m] = 38280.86[km]

<u>Second part</u>

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The distance between the Earth and this point is calculated as follows:

re = 3.84 108 - 38280860.6 = 345719139.4[m]

Now the acceleration can be found as follows:

$a = G*\frac{m_{e} }{r_{e} ^{2} } \\a = 6.67*10^{11} *\frac{5.98*10^{24} }{(345.72*10^{6})^{2} } \\a=3.33*10^{19} [m/s^2]$

6 0

3 years ago

An earthquake sends a seismic wave through the ground. What is the<br> medium? ANSWER QUICK!

konstantin123 [22]

Answer:

THE GROUND IS THE MEDIUM OF SEISMIC WAVES

6 0

2 years ago

Sharon throws a 0.20 Kg with an acceleration of 10 m/s/s.

OlgaM077 [116]

Force=A×M
10m/s×0.20kg
=2Newton

4 0

3 years ago