If a ball of mass 40 g moving with a speed of 200 m/s is brought to

Hello,Plz Help Which conditions are necessary for rain to form?

dimulka [17.4K]

Vas happenin!
The third one makes no since because the clouds carry the rain. It isn’t always cold when it’s going to rain

The fourth one is a good one

The second one again it’s not always cold when it’s raining

The first one could be it also

Hmmm I would go with the last one

Sorry if it’s wrong

5 0

3 years ago

Use Kepler’s third law and the orbital motion of Earth to determine the mass of the Sun. The average distance between Earth and

dexar [7]

Kepler’s
third law formula: T^2=4pi^2*r^3/(GM)

We’re trying to find M, so:

M=4pi^2*r^3/(G*T^2)

M=4pi^2*(1.496
× 10^11 m)^3/((6.674× 10^-11N*m^2/kg^2)*(365.26days)^2)

M=1.48× 10^40(m^3)/((N*m^2/kg^2)*days^2))

Let’s work
with the units:

(m^3)/((N*m^2/kg^2)*days^2))=

=(m^3*kg^2)/(N*m^2*days^2)

=(m*kg^2)/(N*days^2)

=(m*kg^2)/((kg*m/s^2)*days^2)

=(kg)/(days^2/s^2)

=(kg*s^2)/(days^2)

So:

M=1.48× 10^40(kg*s^2)/(days^2)

Now we need to convert days to seconds in order to cancel
them:

1 day=24 hours=24*60minutes=24*60*60s=86400s

M=1.48× 10^40(kg*s^2)/((86400s)^2)

M=1.48× 10^40(kg*s^2)/(
86400^2*s^2)

M=1.48× 10^40kg/86400^2

M=1.98x10^30kg

The
closest answer is 1.99
× 10^30

(it may vary
a little with rounding – the difference is less than 1%)

8 0

3 years ago

Read 2 more answers

Whats the answer???????????

Leviafan [203]

Answer: Less than 4 ohms

Explanation:

We have three resistors with the following resistance:

$R_{1}=4\Omega$

$R_{2}=6\Omega$

$R_{3}=8\Omega$

Now, when the resistors are connected in parallel, the total resistance $R$ is calculated as follows:

$\frac{1}{R}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}$

Isolating $R$ :

$R=\frac{R_{1}R_{2}R_{3}}{R_{3}(R_{1}+R_{2})+R_{1}R_{2}}$

Rewriting with th known values:

$R=\frac{(4\Omega)(6\Omega)(8\Omega)}{8\Omega(4\Omega+6\Omega)+(4\Omega)(6\Omega)}$

Finally:

$R=1.84 \Omega$

Hence, the correct option is less than 4 ohms.

4 0

2 years ago

I don't hang out with girls so if you are a boy and you want to be friend say hi to me

MatroZZZ [7]

Answer:

Uh No thanks but make me brainiest!

Explanation:

4 0

2 years ago

Read 2 more answers

A solid disk of mass 2 kg and radius 2 m is given a horizontal push of 20N at a point .3 m above its center. a. What is the mini

Margaret [11]

Answer:

$\mu_s=1.0205$

Explanation:

Given:

mass of solid disk, $m=2\ kg$

radius of disk, $r=2\ m$

force of push applied to disk, $F=20\ N$

distance of application of force from the center, $s=0.3\ m$

<em>For the condition of no slip the force of static friction must be greater than the applied force so that there is no skidding between the contact surfaces at the contact point.</em>

$\therefore F$

where:

$f_s$ = static frictional force

$\Rightarrow 20$

$\mu_s>1.0204$

7 0

2 years ago