ASTRONOMY section

How many orders of magnitude (powers of ten) is the distance to the moon greater than the distance across the Atlantic ocean?

kupik [55]

Answer:

The distance to the moon greater than the distance across the Atlantic ocean by an order of 1.

Explanation:

We have the average distance across the Atlantic ocean, $d_i=3700\ mi$

the average distance to the moon, $d_m=238,900\ mi$

The factor by which the distance to moon is greater than the width of Atlantic ocean:

$f=\frac{d_m}{d_i}$

$f=\frac{238,900}{3700}$

$f=64.57$

i.e.

the distance to the moon greater than the distance across the Atlantic ocean by an order of 1.

6 0

3 years ago

The electric potential inside a charged spherical conductor of radius R is is given by V = keQ/R, and the potential outside is g

Bumek [7]

Answer:

For outer points of shell

$E = \frac{k_eQ}{r^2}$

Now for inner point of shell

$E = 0$

Explanation:

As we know that out side the shell electric potential is given as

$V = \frac{K_e Q}{r}$

inside the shell the electric potential is given as

$V = \frac{K_e Q}{R}$

now we know the relation between electric potential and electric field as

$E = - \frac{dV}{dr}$

so we can say for outer points of the shell

$E = -\frac{dV}{dr}$

$E = - \frac{d}{dr}(\frac{K_eQ}{r})$

$E = \frac{k_eQ}{r^2}$

Now for inner point again we can use the same

$E = - \frac{dV}{dr}$

$E = - \frac{d}{dr}(\frac{K_eQ}{R})$

$E = 0$

6 0

3 years ago

What does Atomic Collisions mean

stiks02 [169]

Answer:

Atomic and molecular collision processes are the physical interactions of atoms and molecules when they are brought into close contact with each other and with electrons, protons, neutrons or ions. This includes energy-conserving elastic scattering and inelastic scattering.

4 0

3 years ago

An 12 N force is applied to a 1 kg object. What is the magnitude of the objects acceleration?

Sauron [17]

Answer:

a=12 m/s²

Explanation:

Newton's second law of motion states that the acceleration of a body is directly proportional to the force applied and takes place in the direction of force.

This can be summarized as: F=ma, where m is the mass of the object on which force F acts. a is the acceleration due to the force applied.

12N= 1kg×a

a=12N/1kg

a=12m/s²

6 0

4 years ago

The car has a constant deceleration of 4.20 m/s^2. If its initial velocity was 24.0 m/s, how long does it take to come to a stop

nalin [4]

Answer:

The time is 5.71 sec.

Explanation:

Given that,

Acceleration $a= -4.20 m/s^2$

Initial velocity = 24.0 m/s

We need to calculate the time

Using equation of motion

v = u+at[/tex]

Where, v = final velocity

u = inital velocity

t = time

a = acceleration

Put the value into the formula

$0 =24.0 +(-4.20)\times t$

$t = \dfrac{-24.0}{-4.20}$

$t=5.71\ sec$

Hence, The time is 5.71 sec.

5 0

4 years ago