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3 years ago

Why did one liter of have a greater mass than the other liter prepared by a different method

Physics

2 answers:

VikaD [51]3 years ago

4 0

Maybe the water wasnt stable enough and probably couldnt read the water level correctly

gavmur [86]3 years ago

4 0

Nobody can answer your question, because we have no idea
what belongs after the word "... of ...", so we don't know what
substance(s) you're talking about.

All we can say for sure is that the first method produces the substance
with greater density, because it has more mass in the same volume.

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State the equation of momentum impulse theorem.

Paladinen [302]

Answer:

J = Δp

Explanation:

The impulse-momentum theorem says that the impulse J is equal to the change in momentum p.

J = Δp

3 0

3 years ago

what was the acceleration of the cart with Low fan speed cm/s squared? what was the acceleration of the cart with medium fan spe

ArbitrLikvidat [17]

Explanation:

The attached figure shows data for the cart speed, distance and time.

For low fan speed,

Distance, d = 500 cm

Time, t = 7.4 s

Average velocity,

$v=\dfrac{d}{t}\\\\v=\dfrac{500}{7.4}\\\\v=67.56\ cm/s$

Acceleration,

$a=\dfrac{v}{t}\\\\a=\dfrac{67.56}{7.4}\\\\a=9.12\ cm/s^2$

For medium fan speed,

Distance, d = 500 cm

Time, t = 6.4 s

Average velocity,

$v=\dfrac{d}{t}\\\\v=\dfrac{500}{6.4}\\\\v=78.12\ cm/s$

Acceleration,

$a=\dfrac{v}{t}\\\\a=\dfrac{78.12}{6.4}\\\\a=12.2\ cm/s^2$

For high fan speed,

Distance, d = 500 cm

Time, t = 5.6 s

Average velocity,

$v=\dfrac{d}{t}\\\\v=\dfrac{500}{5.6}\\\\v=89.28\ cm/s$

Acceleration,

$a=\dfrac{v}{t}\\\\a=\dfrac{89.28}{5.6}\\\\a=15.94\ cm/s^2$

Hence, this is the required solution.

8 0

2 years ago

Read 2 more answers

As a marble rolls across the floor, it gradually slows to a stop due to friction. Which statement best describes the change in m

Olegator [25]

Answer:

I belive it would be "C"

Explanation:

If it was any of the other answers "B" it would instantly stop. "A" it would roll forever.

3 0

2 years ago

If it requires 8.0 J of work to stretch a particular spring by 2.0 cm from its equilibrium length, how much more work will be re

jenyasd209 [6]

Answer:

The amount of work done required to stretch spring by additional 4 cm is 64 J.

Explanation:

The energy used for stretching spring is given by the relation :

$E = \frac{1}{2}kx^{2}$ .......(1)

Here k is spring constant and x is the displacement of spring from its equilibrium position.

For stretch spring by 2.0 cm or 0.02 m, we need 8.0 J of energy. Hence, substitute the suitable values in equation (1).

$8 = \frac{1}{2}\timesk\times k \times(0.02)^{2}$

k = 4 x 10⁴ N/m

Energy needed to stretch a spring by 6.0 cm can be determine by the equation (1).

Substitute 0.06 m for x and 4 x 10⁴ N/m for k in equation (1).

$E = \frac{1}{2}\times4\times10^{4}\times (0.06)^{2}$

E = 72 J

But we already have 8.0 J. So, the extra energy needed to stretch spring by additional 4 cm is :

E = ( 72 - 8 ) J = 64 J

7 0

2 years ago

You are heading toward an island in your speed boat when you see a friend standing on the shore of the island. You sound the boa

monitta

Answer:

Greater than

Explanation:

The Wavelength will be higher than what will be heard without any motion on the boat due to the Doppler Effect, which is the change in the frequency of a sound wave whenever there's a relative motion between the source of the wave and observer. The amount of shift in frequency depends on the speed of the source towards the observer; the higher the velocity of the source, the higher the shift.

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3 years ago