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LenKa [72]
3 years ago
6

À stone is thrown

Physics
1 answer:
masya89 [10]3 years ago
3 0
A = -9.8

v = -9.8t -8

s = -4.9 t2 -8t +25

So… -5t^2 -8t + 25 =0, we’ll rearrange to 5t^2 + 8t - 25. We get two roots, one is positive and is 1.59 seconds

V = -9.8(1.59) - 8 = -23.6

So… it takes 1.59 seconds to hit the ground at -23.6 m/s.
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According to Kepler's Third Law, a solar-system planet that has an orbital radius of 4 AU would have an orbital period of about
NARA [144]

Answer:

Orbital period, T = 1.00074 years

Explanation:

It is given that,

Orbital radius of a solar system planet, r=4\ AU=1.496\times 10^{11}\ m

The orbital period of the planet can be calculated using third law of Kepler's. It is as follows :

T^2=\dfrac{4\pi^2}{GM}r^3

M is the mass of the sun

T^2=\dfrac{4\pi^2}{6.67\times 10^{-11}\times 1.989\times 10^{30}}\times (1.496\times 10^{11})^3    

T^2=\sqrt{9.96\times 10^{14}}\ s

T = 31559467.6761 s

T = 1.00074 years

So, a solar-system planet that has an orbital radius of 4 AU would have an orbital period of about 1.00074 years.

6 0
3 years ago
How might the velocity (speed) of wind or water affect the deposition of sediments?
Maksim231197 [3]
Well depending on the speed of both of those things is were the rock will be placed and it also determines how fast can an environment change
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6 0
3 years ago
What two energies does a bulldozer use​
sesenic [268]
Mechanical and electrical
3 0
3 years ago
you read a primary source and a secondary source that discuss the same experiment. There is a difference in the conclusion made
Nuetrik [128]
You should trust the primary source more.

This is because the primary source is make its conclusion from direct observation, while the secondary source is possibly making reference to another secondary source or to another primary.

The primary source should be trusted more because it is from direct observation.     
4 0
3 years ago
A flat uniform circular disk (r= 2.00m,
dusya [7]

Incomplete question.The Complete question is here

A flat uniform circular disk (radius = 2.00 m, mass = 1.00 ✕ 102 kg) is initially stationary. The disk is free to rotate in the horizontal plane about a friction less axis perpendicular to the center of the disk. A 40.0-kg person, standing 1.25 m from the axis, begins to run on the disk in a circular path and has a tangential speed of 2.00 m/s relative to the ground.

a.) Find the resulting angular speed of the disk (in rad/s) and describe the direction of the rotation.

b.) Determine the time it takes for a spot marking the starting point to pass again beneath the runner's feet.

Answer:

(a)ω = 1 rad/s

(b)t = 2.41 s

Explanation:

(a) initial angular momentum = final angular momentum  

0 = L for disk + L............... for runner

0 = Iω² - mv²r ...................they're opposite in direction

0 = (MR²/2)(ω²) - mv²r ................where is ω is angular speed which is required in part (a) of question

0 = [(1.00×10²kg)(2.00 m)² / 2](ω²) - (40.0 kg)(2.00 m/s)²(1.25 m)

0=200ω²-200

200=200ω²

ω = 1 rad/s

b.)

lets assume the "starting point" is a point marked on the disk.

The person's angular speed is  

v/r = (2.00 m/s) / (1.25 m) = 1.6 rad/s

As the person and the disk are moving in opposite directions, the person will run part of a revolution and the turning disk would complete the whole revolution.

(angle) + (angle disk turns) = 2π

(1.6 rad/s)(t) + ωt = 2π

t[1.6 rad/s + 1 rad/s] = 2π

t = 2.41 s

6 0
3 years ago
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