All categories

2 years ago

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

Physics

1 answer:

kupik [55]2 years ago

6 0

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.

You might be interested in

What step does the Rin PRICES stand for? Why is this step important?

posledela

Explanation:

The five-step process for treating a muscle or joint injury such as an ankle sprain is called "P.R.I.C.E." which is short for Protection, Rest, Ice, Compression, and Elevation).

6 0

3 years ago

A very long uniform line of charge has charge per unit length λ1 = 4.68 μC/m and lies along the x-axis. A second long uniform li

Kitty [74]

Answer:

$E_{net} = 6.44 \times 10^5 N/C$

Explanation:

As we know that electric field due to infinite line charge distribution at some distance from it is given as

$E = \frac{2k \lambda}{r}$

now we need to find the electric field at mid point of two wires

So here we need to add the field due to two wires as they are oppositely charged

Now we will have

$E_{net} = \frac{2k\lambda_1}{r} + \frac{2k\lambda_2}{r}$

now plug in all data

$\lambda_1 = 4.68 \muC/m$

$\lambda_2 = 2.48 \mu C/m$

$r = 0.200 m$

now we have

$E_{net} = \frac{2k}{r}(4.68 + 2.48)$

$E_{net} = \frac{2(9\times 10^9)}{0.200}(7.16 \times 10^{-6})$

$E_{net} = 6.44 \times 10^5 N/C$

8 0

2 years ago

Help fast please!!!!!

inysia [295]

We're told that the planets have EQUAL MASS.

If that's true, then the strength of the gravitational forces between
each planet and the star depends only on the distance between
them ... the farther a planet is from the star, the smaller the
gravitational forces are IF we're talking about planets with
equal masses.

Planet-X is closer to the star, and Planet-Y is farther from it.
From this we know that the gravitational forces between the
star and Planet-X are greater, and the forces between the star
and Planet-Y are smaller.

'A' says this.

'B' is totally absurd, because it talks about gravity repelling things.

'C' says exactly the opposite for the two planets.

'D' says that distance doesn't matter. We know this is absurd,
simply because we're never pulled toward Jupiter in our daily life.

6 0

3 years ago

Read 2 more answers

In a historical movie, two knights on horseback start from rest 86 m apart and ride directly toward each other to do battle. Sir

Harlamova29_29 [7]

Answer:

Relative to Sir George's starting point, the knights collide at a distance of 38.43 m from Sir George's starting point.

Explanation:

Let the distance covered by Sir George be $S_{1}$

and the distance covered by Sir Alfred be $S_{2}$

Since the knights collide, hence they must have traveled for the same amount of time just before collision

From one of the equations of motion for linear motion

$S = ut + \frac{1}{2}at^{2}$

Where $S$ is the distance traveled

$u$ is the initial velocity

$a$ is the acceleration

and $t$ is the time

For Sir George,

$S = S_{1}$

$u$ = 0 m/s (Since they start from rest)

$a =$ 0.21 m/s²

Hence,

$S = ut + \frac{1}{2}at^{2}$ becomes

$S_{1} = (0)t + \frac{1}{2}(0.21)t^{2}\\S_{1} = 0.105 t^{2}\\$

$t^{2} = \frac{S_{1}}{0.105}$

Now, for Sir Alfred

$S = S_{2}$

$u$ = 0 m/s (Since they start from rest)

$a =$ 0.26 m/s²

Hence,

$S = ut + \frac{1}{2}at^{2}$ becomes

$S_{2} = (0)t + \frac{1}{2}(0.26)t^{2}\\S_{2} = 0.13 t^{2}\\$

$t^{2} = \frac{S_{2}}{0.13}$

Since, they traveled for the same time, $t$ just before collision, we can write

$\frac{S_{1}}{0.105}= \frac{S_{2}}{0.13}$

Since, the two nights are 86 m apart, that is, the sum of the distances covered by the knights just before collision is 86 m. Then we can write that

$S_{1} + S_{2} = 86$ m

Then, $S_{2} = 86 - S_{1}$

Then,

$\frac{S_{1}}{0.105}= \frac{S_{2}}{0.13}$ becomes

$\frac{S_{1}}{0.105}= \frac{86 -S_{1}}{0.13}$

$0.13{S_{1}}= 0.105({86 -S_{1}})\\0.13{S_{1}}= 9.03 - 0.105S_{1}}\\0.13{S_{1}} + 0.105S_{1}}= 9.03 \\0.235{S_{1}} = 9.03\\{S_{1}} =\frac{9.03}{0.235}$

$S_{1} = 38.43$ m

∴ Sir George covered a distance of 38.43 m just before collision.

Hence, relative to Sir George's starting point, the knights collide at a distance of 38.43 m from Sir George's starting point.

6 0

3 years ago

B) If the actual counterweight fitted to this boom was

Lostsunrise [7]

Answer:

50 N

4.2 N

Explanation:

i) The force needed to balance the boom is 2400 N. If the weight of the counterbalance is 2350 N, then the downward force the park attendant must apply is 50 N.

ii) When the boom is resting on the end support, the normal force is:

∑τ = Iα

-W (0.50) + F (3.0) − N (6.0) = 0

-0.50 W + 3.0 F = 6.0 N

N = (-0.50 W + 3.0 F) / 6.0

N = (-0.50 × 2350 + 3.0 × 400) / 6.0

N ≈ 4.2

6 0

2 years ago