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Alina [70]
3 years ago
10

A 0.5 m3 container is filled with a fluid whose specific volume is 0.001 m3/kg. At standard gravitational acceleration, the cont

ents of this container weigh:
Physics
1 answer:
xenn [34]3 years ago
5 0

Answer: the contents of this container weighs 4905 kg.m/s²

Explanation:

Given that;

volume of a container V = 0.5 m³

we know that standard gravitational acceleration g = 9.81 m/s²

specific volume of liquid filled in the container v = 0.001 m³/kg

now we express the equation for weight of the container.

W = mg

W = (pV)g

W = Vg / ν

so we substitute

W =  (0.5 m³)(9.81 m/s ) / 0.001 m³/kg

W = 4.905 / 0.001

W = 4905 kg.m/s²

Therefore, the contents of this container weighs 4905 kg.m/s²

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What is the dependent variable in this
Whitepunk [10]

Answer:

"the plants had already started growing."

Explanation:

I think this is the answer because the definition of a dependent variable is the variable that is being affected by the change. Since the plants had already started growing BECAUSE of "They ran out of the farmer's compost, so

some of the plants received that compost

when the seeds were planted and other

plants got hardware store compost after

the plants had already started growing."

Sorry if I am wrong, I am just a 4th grader, pls don't hate on me, I am just trying to help :)

6 0
3 years ago
Suppose that a comet that was seen in 550 A.D. by Chinese astronomers was spotted again in year 1941. Assume the time between ob
pickupchik [31]

To solve the problem it is necessary to apply the concepts related to Kepler's third law as well as the calculation of distances in orbits with eccentricities.

Kepler's third law tells us that

T^2 = \frac{4\pi^2}{GM}a^3

Where

T= Period

G= Gravitational constant

M = Mass of the sun

a= The semimajor axis of the comet's orbit

The period in years would be given by

T= 1941-550\\T= 1391y(\frac{31536000s}{1y})\\T=4.3866*10^{10}s

PART A) Replacing the values to find a, we have

a^3= \frac{T^2 GM}{4\pi^2}

a^3 = \frac{(4.3866*10^{10})^2(6.67*10^{-11})(1.989*10^{30})}{4\pi^2}

a^3 = 6.46632*10^{39}

a = 1.86303*10^{13}m

Therefore the semimajor axis is 1.86303*10^{13}m

PART B) If the semi-major axis a and the eccentricity e of an orbit are known, then the periapsis and apoapsis distances can be calculated by

R = a(1-e)

R = 1.86303*10^{13}(1-0.997)

R= 5.58*10^{10}m

7 0
3 years ago
Please i need help! ill give brainliest toooo!
Igoryamba

Answer:

b

Explanation:

7 0
3 years ago
Read 2 more answers
Shondra takes notes in class.
sdas [7]

Answer:

I beileve its A because energy is what gives the ability to do work

Explanation:

4 0
3 years ago
A certain part of a computer chip has a length of 1.5 micrometers. How many meters is this ?
Alchen [17]

When you're working with units of measure, "micro" almost always means "millionth", or 1 x 10⁻⁶ .  And that's exactly the case in this example.

"1.5 micrometer" = 1.5 millionths of a meter, or <em>1.5 x  10⁻⁶ meter.</em>

4 0
3 years ago
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