All categories

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.

You might be interested in

Which are properties of a liquid? Check all that apply.

Strike441 [17]

Answer:

the property of liquid are

1 they can flow from one place to another if surface is slanted

2 it cannot be compressed

6 0

2 years ago

Explanation's of E=MC²

natita [175]

Answer:

E means energy

M= Mass

C=speed of light squared (the exponent means squared)

4 0

3 years ago

Read 2 more answers

Electromagnetic waves differ from other types of waves because they are

nikklg [1K]

I think it is D because u think of space and Electromagnetic Waves Travel Through Space Moving All Kinds of Particles From The Sun With It Which Creates The Northern and Southern Lights In The Poles :-)

8 0

2 years ago

A car travels a distance of 100 km. For the first 30 minutes it is driven at a constant speed of 80 km/hr. The motor begins to v

gregori [183]

Explanation:

First, we need to determine the distance traveled by the car in the first 30 minutes, $d_{\frac{1}{2}}$ .

Notice that the unit measurement for speed, in this case, is km/hr. Thus, a unit conversion of from minutes into hours is required before proceeding with the calculation, as shown below

$d_{\frac{1}{2}\text{h}} \ = \ \text{speed} \ \times \ \text{time taken} \\ \\ \\ d_{\frac{1}{2}\text{h}} \ = \ 80 \ \text{km h}^{-1} \ \times \ \left(\displaystyle\frac{30}{60} \ \text{h}\right) \\ \\ \\ d_{\frac{1}{2}\text{h}} \ = \ 80 \ \text{km h}^{-1} \ \times \ 0.5 \ \text{h} \\ \\ \\ d_{\frac{1}{2}\text{h}} \ = \ 40 \ \text{km}$

Now, it is known that the car traveled 40 km for the first 30 minutes. Hence, the remaining distance, $d_{\text{remain}}$ , in which the driver reduces the speed to 40km/hr is

$d_{\text{remain}} \ = \ 100 \ \text{km} \ - \ 40 \ \text{km} \\ \\ \\ d_{\text{remain}} \ = \ 60 \ \text{km}$ .

Subsequently, we would also like to know the time taken for the car to reach its destination, denoted by $t_{\text{remian}}$ .

$t_{\text{remain}} \ = \ \displaystyle\frac{\text{distance}}{\text{speed}} \\ \\ \\ t_{\text{remain}} \ = \ \displaystyle\frac{60 \ \text{km}}{40 \ \text{km hr}^{-1}} \\ \\ \\ t_{\text{remain}} \ = \ 1.5 \ \text{hours}$ .

Finally, with all the required values at hand, the average speed of the car for the entire trip is calculated as the ratio of the change in distance over the change in time.

$\text{speed} \ = \ \displaystyle\frac{\Delta d}{\Delta t} \\ \\ \\ \text{speed} \ = \ \displaystyle\frac{100 \ \text{km}}{(0.5 \ \text{hr} \ + \ 1.5 \ \text{hr})} \\ \\ \\ \text{speed} \ = \ \displaystyle\frac{100 \ \text{km}}{2 \ \text{hr}} \\ \\ \\ \text{speed} \ = \ 50 \ \text{km hr}^{-1}$

Therefore, the average speed of the car is 50 km/hr.

8 0

2 years ago

A chinook salmon needs to jump a waterfall that is 1.5 m high. If the fish starts from a distance of 1.00 m from the base of the

____ [38]

<span>let the fsh jump with initial velocity (u) in direction (angle p) with horizontal

it can cross and reach top of trajectory if its top height h = 1.5m
and horizontal distance d = (1/2) Range
--------------------------------------...
let t be top height time
at top height, vertical component of its velocity =0
vy = 0 = u sin p - gt
t = u sin p/g
h = [u sin p]*t - 0.5 g[t[^2
1.5 = u^2 sin^2 p/g - u^2 sin^2 p/2g
u^2 sin^2 p/2g = 1.5
u^2 sin^2 p = 1.5*2*9.8 = 29.4
u sin p = 5.42 m/s >>>>>>>>>>>>>>> V-component
=====================
t = HALF the time of flight
d = (1/2) Range (R) = (1/2) [2 u^2 sin p cos p/g]
1 = u^2 sin p cos p/g
u sin p * u cos p = 9.8
5.42 * u cos p = 9.8
u cos p = 1.81 m/s >>>>>>>>>>>>> H-component
check>>
u = sqrt[u^2 cos^2 p + u^2 sin^2 p] = 5.71 m/s
u < less than fish's potential jump speed 6.26 m/s

so it will able to cross</span>

8 0

2 years ago