Which is larger? 1 L or 1 m ³
Answers
1 L = 0.001 m^3
So 1m^3 is larger
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The excitation of an electron on the surface of a photocell required 5.0 x 10 –27 J of energy. Calculate the wavelength of light that was needed to excite the electron in an atom on the photocell
what can accurately be said about a resultant wave that displays both reinforcement and interference?
Answers
For this case, the first thing we must do is define the scientific notation.
We have then:
Scientific notation is a way of writing numbers too big or small in a conventional way.
A number written in scientific notation has the following form:
Where,
m: it is called matinsa
e: it is the order of magnitude
Therefore, for the given number we have that the scientific notation is:
Answer:
the power of 10 when 0.000028 is written in scientific notation is:
-5
Answers
Explanation:
Atomic number is defined as the total number of protons present in an element.
Each element of the periodic table has different atomic number because each of them have different number of protons.
For example, atomic number of Na is 11, and atomic number of Ca is 20.
On the other hand, atomic mass is the sum of total number of protons and neutrons present in an atom.
For example, atomic mass of nitrogen is 14 that is, it contains 7 protons and 7 neutrons.
Thus, we can conclude that all atoms of the same element must have the same number of protons.
Answer: Protons
Explanation: The number of protons corresponds to the atomic number.
tied together
Answers
Answer:
a=6m/sec²
Explanation:
f=5 N
a1=8m/sec²
a2=24m/sec
m1=f/a1=5/8kg
m2=f/a2=5/24kg
New, f=(m1+m2)a
a=5/5/8+5/24=5/15+5/24=24*5/20
a=6m/sec²
B. meniscusity.
C. viscosity.
D. capillarity.
Answers
the correct answer is
D. capillarity
Answers
Answers
This is a perfect opportunity to stuff all that data into the general equation for the height of an object that has some initial height, and some initial velocity, when it is dropped into free fall.
H(t) = (H₀) + (v₀ T) + (1/2 a T²)
Height at any time 'T' after the drop =
(initial height) +
(initial velocity) x (T) +
(1/2) x (acceleration) x (T²) .
For the balloon problem ...
-- We have both directions involved here, so we have to define them:
Upward = the positive direction
Initial height = +150 m
Initial velocity = + 3 m/s
Downward = the negative direction
Acceleration (of gravity) = -9.8 m/s²
Height when the bag hits the ground = 0 .
H(t) = (H₀) + (v₀ T) + (1/2 a T²)
0 = (150m) + (3m/s T) + (1/2 x -9.8 m/s² x T²)
-4.9 T² + 3T + 150 = 0
Use the quadratic equation:
T = (-1/9.8) [ -3 plus or minus √(9 + 2940) ]
= (-1/9.8) [ -3 plus or minus 54.305 ]
= (-1/9.8) [ 51.305 or -57.305 ]
T = -5.235 seconds or 5.847 seconds .
(The first solution means that the path of the sandbag is part of
the same path that it would have had if it were launched from the
ground 5.235 seconds before it was actually dropped from balloon
while ascending.)
Concerning the maximum height ... I don't know right now any other
easy way to do that part without differentiating the big equation.
So I hope you've been introduced to a little bit of calculus.
H(t) = (H₀) + (v₀ T) + (1/2 a T²)
H'(t) = v₀ + a T
The extremes of 'H' (height) correspond to points where h'(t) = 0 .
Set v₀ + a T = 0
+3 - 9.8 T = 0
Add 9.8 to each side: 3 = 9.8 T
Divide each side by 9.8 : T = 0.306 second
That's the time after the drop when the bag reaches its max altitude.
Oh gosh ! I could have found that without differentiating.
- The bag is released while moving UP at 3 m/s .
- Gravity adds 9.8 m/s of downward speed to that every second.
So the bag reaches the top of its arc, runs out of gas, and starts
falling, after
(3 / 9.8) = 0.306 second .
At the beginning of that time, it's moving up at 3 m/s.
At the end of that time, it's moving with zero vertical speed).
Average speed during that 0.306 second = (1/2) (3 + 0) = 1.5 m/s .
Distance climbed during that time = (average speed) x (time)
= (1.5 m/s) x (0.306 sec)
= 0.459 meter (hardly any at all)
But it was already up there at 150 m when it was released.
It climbs an additional 0.459 meter, topping out at 150.459 m,
then turns and begins to plummet earthward, where it plummets
to its ultimate final 'plop' precisely 5.847 seconds after its release.
We can only hope and pray that there's nobody standing at
Ground Zero at the instant of the plop.
I would indeed be remiss if were to neglect, in conclusion,
to express my profound gratitude for the bounty of 5 points
that I shall reap from this work. The moldy crust and tepid
cloudy water have been delicious, and will not soon be forgotten.