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3 years ago

What is 1.2 kg converted into mg. I need to know for step by step please?

1 answer:

4 0

Here’s a good photo to reference when converting in the metric system.

Each time you move down a step you move the decimal to the right, each time you move up a step you move the decimal to the left.

We are going from 1.2 kg or kilograms, which is at the very top left of the ladder. To get to mg or milligrams, we would have to make six jumps, so we’d move the decimal over six times.

1.2 > 12. > 120. > 1200. > 12000. > 120000. > 1200000.

So our final answer would be 1,200,000mg.

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3 years ago

A car travels from point A to point B, moving in the same direction but with a non-constant speed. The first half of the distanc

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Answer:

Explanation:

From A to B

distance traveled with velocity $v_1$ in time $t_1$

$\frac{d}{2}=v_1t_1----1$

from B to C

distance traveled is 0.5 d with $v_2$ and $v_3$ velocity for half-half time

$\frac{d}{2}=\frac{v_2t_2}{2}+\frac{v_3t_3}{2}----2$

divide 1 and 2 we get

$\frac{1}{1}=\frac{2v_1t_1}{v_2t_2+v_3t_3}$

$\frac{t_1}{t_2}=\frac{v_2+v_3}{2v_1}$

Now average velocity is given by

$v_{avg}=\frac{d}{t_1+t_2}$

taking $t_1$ common

$v_{avg}=\frac{2v_1t_1}{t_1(1+\frac{t_2}{t_1})}$

$v_{avg}=\frac{2v_1}{1+\frac{2v_1}{v_2+v_3}}$

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3 years ago

A -3.00 nc point charge is at the origin, and a second -5.50 nc point charge is on the x-axis at x = 0.800 m. find the electric

Liula [17]

The electric field produced by a single-point charge is given by

$E(r)=k\frac{q}{r^2}$

where

k is the Coulomb's constant

q is the charge

r is the distance from the charge

To find the electric field at x=0.200 m, we need to find the electric field produced by each charge at that point, and then find their resultant.

1) The first charge is $q=-3.00 nC=-3.00 \cdot 10^{-9} C$ , and it is located at x=0, so its distance from the point x=0.200 m is

$r=0.200 m-0=0.2 m$

Therefore, the electric field is

$E_1=(8.99 \cdot 10^9 Nm^2C^{-2})\frac{(3.0 \cdot 10^{-9} C)}{(0.2 m)^2}=675 N/C$

And since the charge is negative, the direction of the field is toward the charge, so toward negative x direction.

2) The second charge is $q=-5.50 nC=-5.5 \cdot 10^{-9}C$ and it is located at x=0.800 m, so its distance from the point is

$r=0.800 m-0.200 m=0.6 m$

Therefore, the electric field is

$E_2 = (8.99 \cdot 10^9 Nm^2C^{-2})\frac{(5.5 \cdot 10^{-9} C)}{(0.6 m)^2}=137.5 N$

And since the charge is negative, the direction of the field is toward the charge, so toward positive x-direction.

3) The total electric field at x=0.200 m will be given by the difference between the two fields (because they are in opposite directions). Taking the x-positive direction as positive direction, we have

$E=E_2 -E_1 =137.5 N/C/C-675 N/C=-537.5 N/C$