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adell [148]
3 years ago
14

Help please!

Physics
1 answer:
Marina CMI [18]3 years ago
6 0

It's the old-but-important concept:  "Speed" depends on who's measuring it and how THEY're moving.  Different observers may very well observe different speeds, and they're all correct.  There's no such thing as "the REAL speed".

When you were a little kid, did you ever get on a moving escalator or walkway, and when you got to the middle, you turned around and walked the <em>opposite</em> way, so that somebody watching you from the outside would see you not moving at all ?

Say you're on a school bus that's driving 10 mph along the street pavement, and you get up out of your seat and run forward up the aisle at 10 mph.  Somebody outside the bus sees you passing them at 20 mph !

If instead, you run toward the BACK of the bus at 10 mph, somebody outside the bus sees you bobbing up and down but not moving forward or backward at all.

All this  problem is saying is:  The bus is driving along at 15 m/s.  A passenger on the bus puts his little baseball down on the floor on its little feet, and it runs backwards down the aisle, toward the back of the bus, at 15 m/s.  Somebody is standing outside looking into the bus as it passes by and the little baseball is scurrying toward the back of the bus.  How will HE describe the motion of the baseball ?    

Have you got it now ?

"Relative to the Earth" just  means how fast an object is passing the stores and telephone poles and people standing still ... things that are attached to the Earth.  "Relative to the bus" would mean how fast an object is passing by people sitting on the bus.

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8. two +1 C charges are separated by 3000m. What is the magnitude of the electric force between them?
Sidana [21]

Answer:

1000 N

Explanation:

The magnitude of the electrostatic force between two charged object is given by

F=k\frac{q_1 q_2}{r^2}

where

k is the Coulomb constant

q1, q2 is the magnitude of the two charges

r is the distance between the two objects

Moreover, the force is:

- Attractive if the two forces have opposite sign

- Repulsive if the two forces have same sign

In this problem:

q_1=q_2=+1C are the two charges

r = 3000 m is their separation

Therefore, the electric force between the charges is:

F=(9\cdot 10^9)\frac{(1)(1)}{3000^2}=1000 N

8 0
3 years ago
ALWAYS use significant figure rules. Remember that these rules apply to all numbers that are measurements. You travel 20.0 mph f
timurjin [86]
The actual answer is 165 miles, but using significant figure rules the answer is 200. This is because the sig fig rules are as follows ...
<span>1. Non-zero digits are always significant.
2. Any zeros between two significant digits are significant.
<span>3. A final zero or trailing zeros in the decimal portion ONLY are significant.
</span></span>
So the zeroes in a number like 20 or 23,000 are NOT significant. When you add numbers you must find the addend with the lowest amount of significant figures and round the answer to that. In this case most of the addends only have one sig fig, so you round 165 to 200 to make it only have one sig fig.
5 0
4 years ago
Read 2 more answers
What is a gravitational force...??
o-na [289]
The simplest answer is that gravity is the field created by a mass distribution in the spacetime around it; gravitational force is the force exterted by the field on a test mass in the field.
5 0
3 years ago
Please help! I will give brainlist.
ki77a [65]

Answer:

Either temperature or energies but I am pretty sure its temperature

Explanation:

4 0
3 years ago
Three identical particles, q1, q2, and q3, each with charge q = 5.00 μC, are placed along a circle of radius r = 2.00 m at angle
wariber [46]

Answer:

11250 N/C

Direction: 0 deg counterclockwise from positive x-axis

Explanation:

q = magnitude of charge on each particle = 5 μC =  5 x 10⁻⁶ C

r = distance of each particle from center of circle = 2 m

E = Magnitude of electric field at the center by each particle

Magnitude of electric field at the center by each particle is given as

E = \frac{kq}{r^{2} }

inserting the values

E = \frac{(9\times10^{9} )(5\times10^{-6})}{2^{2} }\\E = 11.25\times10^{3} NC^{-1}

From the diagram , we see that being equal and opposite, the electric fields due to charge q₁ and q₃ cancel out.

So net electric field at center is only due to charge q₂ direction towards positive x-direction

So

E_{res} = Resultant electric field = 11250 N/C

Direction: 0 deg counterclockwise from positive x-axis

7 0
3 years ago
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