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3 years ago

What is the difference between abiotic and biotic factors?

Physics

2 answers:

butalik [34]3 years ago

6 0

Abiotic is not living and biotic is alive

lapo4ka [179]3 years ago

3 0

Answer:

Abiotic referfers to non-living and biotic factors are living or once living

Explanation:

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ELEN [110]

Answer:

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3 years ago

Does an object travel farther on a smooth or slippery surface or on a rough surface? Why?

Vitek1552 [10]

Answer:

There is much more friction on the rough surface than there is on the smooth surface.

Explanation:

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3 years ago

A satellite orbits a planet of unknown mass in a circular orbit of radius 2.3 x 104 km. The gravitational force on the satellite

sladkih [1.3K]

Answer:

The kinetic energy is $KE = 7.59 *10^{10} \ J$

Explanation:

From the question we are told that

The radius of the orbit is $r = 2.3 *10^{4} \ km = 2.3 *10^{7} \ m$

The gravitational force is $F_g = 6600 \ N$

The kinetic energy of the satellite is mathematically represented as

$KE = \frac{1}{2} * mv^2$

where v is the speed of the satellite which is mathematically represented as

$v = \sqrt{\frac{G M}{r^2} }$

=> $v^2 = \frac{GM }{r}$

substituting this into the equation

$KE = \frac{ 1}{2} *\frac{GMm}{r}$

Now the gravitational force of the planet is mathematically represented as

$F_g = \frac{GMm}{r^2}$

Where M is the mass of the planet and m is the mass of the satellite

Now looking at the formula for KE we see that we can represent it as

$KE = \frac{ 1}{2} *[\frac{GMm}{r^2}] * r$

=> $KE = \frac{ 1}{2} *F_g * r$

substituting values

$KE = \frac{ 1}{2} *6600 * 2.3*10^{7}$

$KE = 7.59 *10^{10} \ J$

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3 years ago

During a canoe race, a camper paddles a distance of 406 meters in 70 seconds. What is the average speed in this race?

faltersainse [42]

5.8 a seccond!!!!!!!!!!!!!!!

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4 years ago

Read 2 more answers

The electric field strength in the space between two closely spaced parallel disks is 1.0 105 N/C. This field is the result of t

balu736 [363]

Answer:

$D=2.996\times 10^{-2} m$

Explanation:

*Assume the parallel disks have equal diameters.

Given the electric strength as $1.0\times 10^5 N/C.$ transferring $3.9\times 10^9$ electrons, the disk's Area can be calculated using the formula:

$E=\frac{\eta}{\epsilon_o}=\frac{Q}{A\epsilon_o}\\\\A=\frac{Q}{E\epsilon_o}\\\\=\frac{(3.9\times 10^9)\times (1.6\times10^{-19})}{(1.0\times 10^5 )\times (8.85\times10^{-12})}\\\\A=7.0508\times 10^{-4} \ m^2$

#We now calculate the disks diameter:

$A=\pi(D/2)^2\\\\2\sqrt{\frac{A}{\pi}}=D\\\\=2\sqrt{7.0508\times 10^{-4}/\pi}\\\\D=2.996\times 10^{-2} \ m$

Hence, the diameter of the disks is $D=2.996\times 10^{-2} m$

8 0

3 years ago