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Artyom0805 [142]

3 years ago

5

The earth's worst mass extinction was:

2 answers:

Ronch [10]3 years ago

7 0

<span>The Permian–Triassic (P–Tr) extinction event, colloquially known as the Great Dying, the End ...... Jump up ^ St. Fleur, Nicholas (16 February 2017). "After Earth's Worst Mass Extinction, Life Rebounded Rapidly, Fossils Suggest". New York Times ...</span><span>

</span>

Volgvan3 years ago

4 0

Permian-Triassic ..........................................................

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Carbonic acid _____.

artcher [175]

Dissolves limestone and other rocks.

3 0

3 years ago

Read 2 more answers

What is the precision for each measurement for the online simulation. What would be the precision for force

zimovet [89]

Answer:

Unit of precision for force is the Newton.

Explanation:

It is the official unit used to describe force in science and mostly abbreviated with the symbol N.

4 0

3 years ago

Zero, a hypothetical planet, has a mass of 5.3 x 1023 kg, a radius of 3.3 x 106 m, and no atmosphere. A 10 kg space probe is to

Andrej [43]

(a) $3.1\cdot 10^7 J$

The total mechanical energy of the space probe must be constant, so we can write:

$E_i = E_f\\K_i + U_i = K_f + U_f$ (1)

where

$K_i$ is the kinetic energy at the surface, when the probe is launched

$U_i$ is the gravitational potential energy at the surface

$K_f$ is the final kinetic energy of the probe

$U_i$ is the final gravitational potential energy

Here we have

$K_i = 5.0 \cdot 10^7 J$

at the surface, $R=3.3\cdot 10^6 m$ (radius of the planet), $M=5.3\cdot 10^{23}kg$ (mass of the planet) and m=10 kg (mass of the probe), so the initial gravitational potential energy is

$U_i=-G\frac{mM}{R}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{3.3\cdot 10^6 m}=-1.07\cdot 10^8 J$

At the final point, the distance of the probe from the centre of Zero is

$r=4.0\cdot 10^6 m$

so the final potential energy is

$U_f=-G\frac{mM}{r}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{4.0\cdot 10^6 m}=-8.8\cdot 10^7 J$

So now we can use eq.(1) to find the final kinetic energy:

$K_f = K_i + U_i - U_f = 5.0\cdot 10^7 J+(-1.07\cdot 10^8 J)-(-8.8\cdot 10^7 J)=3.1\cdot 10^7 J$

(b) $6.3\cdot 10^7 J$

The probe reaches a maximum distance of

$r=8.0\cdot 10^6 m$

which means that at that point, the kinetic energy is zero: (the probe speed has become zero):

$K_f = 0$

At that point, the gravitational potential energy is

$U_f=-G\frac{mM}{r}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{8.0\cdot 10^6 m}=-4.4\cdot 10^7 J$

So now we can use eq.(1) to find the initial kinetic energy:

$K_i = K_f + U_f - U_i = 0+(-4.4\cdot 10^7 J)-(-1.07\cdot 10^8 J)=6.3\cdot 10^7 J$

3 0

3 years ago

a typical top fuel dragstar has a mass of 900 kg what is the force applied to a dragstar by the static friction of the strip the

AnnZ [28]

Wonderful.
You've given us ' m '.
Now, if we only had ' a ', we could come back at you with ' f '.

6 0

4 years ago

The 630-nm light from a helium-neon laser irradiates a grating. The light then falls on a screen where the first bright spot is

dimaraw [331]

Answer:

464.8 nm

Explanation:

The second wavelength of light can be calculated using the next equation:

$\lambda = \frac{x*d}{L}$

<u>Where:</u>

<em>λ : is the wavelength of light</em>

<em>x: is the distance from the central maximum</em>

<em>d: is the distance between the spots </em>

<em>L: is the lenght from the screen to the bright spot</em>

For the first wavelength of light we have:

$\lambda_{1} = \frac{x_{1}*d}{L}$

$630 \cdot 10^{-9} m = \frac{0.61 m*d}{L}$

$\frac{d}{L} = \frac{630 \cdot 10^{-9} m}{0.61 m} = 1.033 \cdot 10^{-6}$ (1)

For the second wavelength of light we have:

$\lambda_{2} = \frac{x_{2}*d}{L}$

$\lambda_{2} = 0.45 m*\frac{d}{L}$ (2)

By entering equation (1) into equation (2) we have:

$\lambda_{2} = 0.45 m* 1.033 \cdot 10^{-6} = 4.648 \cdot 10^{-7} m = 464.8 nm$

Therefore, the second wavelength is 464.8 nm

I hope it helps you!

3 0

3 years ago