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3 years ago

Is it true the atoms in sulphur liquid are the same as the atoms in solid sulphur?

1 answer:

3 0

Yes, it is true that the atoms in liquid sulphur are the same as the atoms
in solid sulphur.

Also, if you can heat it past 832°F, it'll boil, and you'll have sulphur vapor.
Those atoms are also the same.

In any one of its three physical phases, you can also spell it as "sulfur"
without changing the nature of its atoms.

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What is the velocity of an object that has a momentum of 4000 kg-m/s and a mass of 115 kg? Round to the nearest hundredth.

insens350 [35]

The answer is C. You divide 4000 kg/s by 115 kg.

5 0

3 years ago

A construction crane lowers a beam into place at constant speed. Consider the work Wg done by gravity on the beam and the work W

aalyn [17]

Answer:

Wg is positive and WT negative.

(Letters in options are all wrongly written).

Explanation:

Remember that the work of a force is the internal product between the force and the displacement $W=F.d$ .

Since the displacement is downwards like the weight, the work done by gravity is positive, while the work done by the tension is negative since it points upwards.

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3 years ago

How does the earth orbit the sun?

scoray [572]

Answer:

The Sun's gravity pulls on the planets, just as Earth's gravity pulls down anything that is not held up by some other force and keeps you and me on the ground.

Explanation:

Hope that helps

6 0

2 years ago

(a) What is the potential between two points situated 10 cm and 20 cm from a 3.0-μC point charge? (b) To what location should th

julia-pushkina [17]

Answer:

(a) 135 kV

(b) The charge chould be moved to infinity

Explanation:

(a)

The potential at a distance of r from a point charge, Q, is given by

$V = -\dfrac{kQ}{r}$

where $k = 9\times 10^9 \text{ F/m}$

Difference in potential between the points is

$kQ\left[-\dfrac{1}{0.2\text{ m}} -\left( -\dfrac{1}{0.1\text{ m}}\right)\right] = \dfrac{kQ}{0.2\text{ m}} = \dfrac{9\times10^9\text{ F/m}\times3\times10^{-6}\text{ C}}{0.2\text{ m}}$

$PD = 135\times 10^3\text{ V} = 135\text{ kV}$

(b)

If this potential difference is increased by a factor of 2, then the new pd = 135 kV × 2 = 270 kV. Let the distance of the new location be x.

$270\times10^3 = kQ\left[-\dfrac{1}{x}-\left(-\dfrac{1}{0.1\text{ m}}\right)\right]$

$10 - \dfrac{1}{x} = \dfrac{270000}{9\times10^9\times3\times10^{-6}} = 10$

$\dfrac{1}{x} = 0$

$x = \infty$

The charge chould be moved to infinity

7 0

3 years ago

An object with velocity 141 ft/s has a kinetic energy of 1558.71 ftâˆ™lbf, on a planet whose gravity is 31.5 ft/s2. What is its

Sidana [21]

Answer:

The mass of the object is 5.045 lbm.

Explanation:

Given;

kinetic energy of the object, K.E = 1558.71 ft.lbf

velocity of the object, V = 141 ft/s

The kinetic energy of the object is calculated as;

$K.E = \frac{1}{2} mV^2\\\\mV^2 = 2K.E\\\\m = \frac{2K.E}{V^2} \\\\1 \ lbf = 32.174 \ lbm.ft/s^2\\\\m = \frac{2 \ \times \ 1558.71 \ ft.lbf \ \times \ 32.174 \ lbm.ft/s^2 }{(141 \ ft/s)2 \ \ \times \ \ \ \ 1 \ lbf\ }$

$m = \frac{(2 \ \times \ 1558.71 \ \times \ 32.174) \ lbm.ft^2/s^2 }{(141 )^2\ ft^2/s^2 }\\\\m = \frac{(2 \ \times \ 1558.71 \ \times \ 32.174) \ lbm }{(141 )^2 }\\\\m = 5.045 \ lbm$

Therefore, the mass of the object is 5.045 lbm.

6 0

2 years ago