What is 3*10^-6 divided by 2.5*10^6 expressed in standard notation?

1. Calcula la fuerza de atracción electrostática entre dos cuerpos de cargas q1 = -18 C y q2 = +5 mC, separados entre sí por una

romanna [79]

Answer:

A) F=-20.16×10⁹N

B) if the distance doubles, force is 4 times smaller.

Explanation:

q1=-28C

q2=5mC=0.005C

d=25cm=0.25m

Electrostatic force between charges: F=k×q1×q2/d², where k is a coefficient that has the value k=9 × 10⁹ N⋅m²⋅C^(-2) for air.

Thus:

F=9×10⁹×(-28)×0.005/0.25²

F=-20.16×10⁹N

The minus sign indicates attraction.

If distance doubles, d1=2×d, then we have 4d² at the denominator and the force is 4 times smaller.

6 0

3 years ago

Which of the following cannot be determined by analyzing the periodic table?

dybincka [34]

Your answer is C) The year the element was discovered :)

4 0

2 years ago

Read 2 more answers

Suppose that a comet that was seen in 550 A.D. by Chinese astronomers was spotted again in year 1941. Assume the time between ob

pickupchik [31]

To solve the problem it is necessary to apply the concepts related to Kepler's third law as well as the calculation of distances in orbits with eccentricities.

Kepler's third law tells us that

$T^2 = \frac{4\pi^2}{GM}a^3$

Where

T= Period

G= Gravitational constant

M = Mass of the sun

a= The semimajor axis of the comet's orbit

The period in years would be given by

$T= 1941-550\\T= 1391y(\frac{31536000s}{1y})\\T=4.3866*10^{10}s$

PART A) Replacing the values to find a, we have

$a^3= \frac{T^2 GM}{4\pi^2}$

$a^3 = \frac{(4.3866*10^{10})^2(6.67*10^{-11})(1.989*10^{30})}{4\pi^2}$

$a^3 = 6.46632*10^{39}$

$a = 1.86303*10^{13}m$

Therefore the semimajor axis is $1.86303*10^{13}m$

PART B) If the semi-major axis a and the eccentricity e of an orbit are known, then the periapsis and apoapsis distances can be calculated by

$R = a(1-e)$

$R = 1.86303*10^{13}(1-0.997)$

$R= 5.58*10^{10}m$

7 0

3 years ago

The current theory of the structure of the

IRISSAK [1]

1) The mass of the continent is $3.3\cdot 10^{21} kg$

2) The kinetic energy of the continent is 624 J

3) The speed of the jogger must be 4 m/s

Explanation:

1)

We start by finding the volume of the continent. We have:

$L = 5850 km = 5.85\cdot 10^6 m$ is the side

$t = 35 km = 3.5\cdot 10^4 m$ is the depth

So the volume is

$V=L^2 t = (5.85\cdot 10^6)^2 (3.5\cdot 10^4)=1.20\cdot 10^{18} m^3$

We also know that its density is

$d=2750 kg/m^3$

Therefore, we can find the mass by multiplying volume by density:

$m=dV=(2750)(1.20\cdot 10^{18})=3.3\cdot 10^{21} kg$

2)

The kinetic energy of the continent is given by:

$K=\frac{1}{2}mv^2$

where

$m=3.3\cdot 10^{21} kg$ is its mass

v = 3.2 cm/year is the speed

We have to convert the speed into m/s. We have:

3.2 cm = 0.032 m

$1 year = 1(365)(24)(60)(60)=3.15\cdot 10^7 s$

So, the speed is:

$v=\frac{0.032 m}{3.15 \cdot 10^7 s}=1.02\cdot 10^{-9} m/s$

So, we can now find the kinetic energy:

$K=\frac{1}{2}(1.20\cdot 10^{21})(1.02\cdot 10^{-9})^2=624 J$

3)

Here we have a jogger of mass

m = 78 kg

And the jogger has the same kinetic energy of the continent, so

K = 624 J

The kinetic energy of the jogger is given by

$K=\frac{1}{2}mv^2$

where v is the speed of the jogger.

Solving for v, we find the speed that the jogger must have:

$v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(624)}{78}}=4 m/s$

Learn more about kinetic energy:

brainly.com/question/6536722

#LearnwithBrainly

3 0

3 years ago

99 POINTS!!!!

Papessa [141]

The situation given above can be answered through the concept of the First Law of thermodynamics which states that the change in internal energy is equal to the difference between the work done and the heat added to the system. The work done by the object is negative and the heat added is positive.
change in internal energy = -500J + 1400 J = 900 J

3 0

3 years ago

What is 3*10^-6 divided by 2.5*10^6 expressed in standard notation?​

What is 310^-6 divided by 2.510^6 expressed in standard notation?