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Goryan [66]
4 years ago
11

The density of water is 1g/cm3. What is this in kg/m3? The density of water is 1g/cm3. What is this in kg/m3? Follow 3 answers 3

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Physics
1 answer:
Lelu [443]4 years ago
8 0
1 gram per cubic centimeter is equal to 1000 kilograms per cubic meter. Just multiply it with a 1000 and you have your solution.
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Which one doesn't belong in the group? oxygen, sulfur, selenium,
vlada-n [284]
It’s gonna be Oxygen ....
3 0
3 years ago
Is (2, 2) a solution of y &lt; 4x-6<br> Help
vesna_86 [32]

Answer:

B) No.

Explanation:

Okay,so,

this is equation is y=mx +b

mx represents the slope

and b represents the y-intercept

in order to figure this out you need to plot the y-intercept first

that makes its (0,-6) because the 6 is negative in the equation

4x is also equal to 4/1 since we dont know what x is

we have to do rise over run for this

you go up 4 spots on the y intercept from -6 because 4 is positive

then you go to the right 1 time because 1 is positive.

this leaves you at (1,-2)

so, (2,2) is NOT a solution

7 0
3 years ago
Tim’s cow is anemic. The cow is lacking which type of nutrient?
mixer [17]

Answer:

Iron deficiency

Explanation:

or more scientifically explained as decreased hemoglobin levels in your blood but still caused by lack of iron.

6 0
3 years ago
Read 2 more answers
Water flows over the edge of a waterfall at a rate of 1.2 x 10^6 kg/s. There are 50.0 m between the top and bottom of the waterf
Anit [1.1K]

Answer:

5.88×10⁸ W

Explanation:

Power = change in energy / time

P = mgh / t

P = (m/t) gh

P = (1.2×10⁶ kg/s) (9.8 m/s²) (50.0 m)

P = 5.88×10⁸ W

4 0
3 years ago
Read 2 more answers
A girl rolls a ball up an incline and allows it to re- turn to her. For the angle and ball involved, the acceleration of the bal
zalisa [80]

Answer:

3.28 m

3.28 s

Explanation:

We can adopt a system of reference with an axis along the incline, the origin being at the position of the girl and the positive X axis going up slope.

Then we know that the ball is subject to a constant acceleration of 0.25*g (2.45 m/s^2) pointing down slope. Since the acceleration is constant we can use the equation for constant acceleration:

X(t) = X0 + V0 * t + 1/2 * a * t^2

X0 = 0

V0 = 4 m/s

a = -2.45 m/s^2 (because the acceleration is down slope)

Then:

X(t) = 4*t - 1.22*t^2

And the equation for speed is:

V(t) = V0 + a * t

V(t) = 4 - 2.45 * t

If we equate this to zero we can find the moment where it stops and begins rolling down, that will be the highest point:

0 = 4 - 2.45 * t

4 = 2.45 * t

t = 1.63 s

Replacing that time on the position equation:

X(1.63) = 4 * 1.63 - 1.22 * 1.63^2 = 3.28 m

To find the time it will take to return we equate the position equation to zero:

0 = 4 * t - 1.22 * t^2

Since this is a quadratic equation it will have to answers, one will be the moment the ball was released (t = 0), the other will eb the moment when it returns:

0 = t * (4 - 1.22*t)

t1 = 0

0 = 4 - 1.22*t2

1.22 * t2 = 4

t2 = 3.28 s

7 0
3 years ago
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