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3 years ago

A 58.5 kg sprinter starts a race with an acceleration of 4.40 m/s2. What is the net external force on him?

Physics

1 answer:

mote1985 [20]3 years ago

4 0

Answer:

External force on him = 257.40 N

Equation is; F = ma (where 'f' is force, 'm' is mass and 'a' is acceleration)

Explanation:

The mass of the sprinter is 58.5 kg

His acceleration is 4.40 m/s²

According to Newton's second law of motion; F = ma

External force on the sprinter = 58.5 kg × 4.40 m/s² = 257.40 N

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What kind of stars make up the galactic nucleus?

Hatshy [7]

I think the correct answer would be old and metal poor stars are found in the galactic nucleus. This nucleus us a region in the center of a galaxy which contains a higher luminosity than other parts. It produces very high amounts of energy. Hope this helps.

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3 years ago

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Two identical cylinders at the same pressure contain the same gas. if cylinder a contains three times as much gas as cylinder b,

Alborosie

By ideal gas theory, cylinder b has the higher temperature.

We need to know about the ideal gas theory to solve this problem. The ideal gas can be represented by

P . V = n . R . T

where P is the pressure, V is volume, n is the number of molecules, R is the ideal gas constant and T is temperature.

From the question above, we know that

Pa = Pb = P

na = 3nb

Find the temperature of the cylinder a

P . V = n . R . Ta

Ta = P . V /( na . R )

Substitute na

Ta = P . V /( (3nb) . R )

Ta = (1/3) x (P . V /( (nb . R ))

Find the temperature of the cylinder b

P . V = n . R . Tb

Tb = P . V /( nb . R )

The cylinder a temperature is 3 times smaller than the temperature in cylinder b.

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2 years ago

Near the top of the Citigroup Center building in New York City, there is an object with mass of 4.00 ✕ 105 kg on springs that ha

jasenka [17]

Explanation:

It is given that,

Mass of an object, $m=4\times 10^5\ kg$

(a) Time period of oscillation, T = 2.4 s

The formula for the time period of spring is given by :

$T=2\pi \sqrt{\dfrac{m}{k}}$

Where

k is the force constant

$k=\dfrac{4\pi ^2 m}{T^2}$

$k=\dfrac{4\pi ^2 \times 4\times 10^5}{(2.4)^2}$

$k=2.74\times 10^6\ N/m$

(b) Displacement in the spring, x = 2.2 m

Energy stored in the spring is given by :

$U=\dfrac{1}{2}kx^2$

$U=\dfrac{1}{2}\times 2.74\times 10^6\ N/m\times (2.2\ m)^2$

$U=6.63\times 10^6\ J$

Hence, this is the required solution.

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3 years ago

Precipitation tends to be _______ on the _______ side of a mountain because water vapor _______ as it rises there.

Lera25 [3.4K]

Answer:

Higher, Windward side, Condenses

Explanation:

The Windward side refers to that side of a mountain that faces the direction from which the wind is blowing. In this direction, the moisture containing hot air blowing from a distant place moves upward and strikes the mountain at a greater height, where the air mass is thin and the temperature is relatively cold. As the temperature and pressure decrease with altitude, the hot uprising air cools and gradually condenses. This results in the occurrence of high precipitation over this region i.e. the windward side of the mountain.

Therefore, the precipitation is always higher on the windward side of a mountain as the hot air undergoes condensation at greater height as it rises upward.

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3 years ago

A resin tube and a PVC rod are both rubbed against styrofoam. Do

Ivan

The concepts of electrostatics allow us to find the result for the question about the electric forces in the two tubes is:

The tubes are attracted by having charges of different signs.

<h3>Electrostatics</h3>

Electrostatics studies the transfer of charge between bodies when they rub together, in generating the charges of the insulating bodies they are not mobile, so to transfer them from one body to another they must be in contact.

The force electric between charges is of two types:

Attractive. If the charges are of different signs.

Repulsive. If the charges are of the same sign.

The transferred charge depends on the bodies involved in the tables showns that the resin is positively charged when rubbed, the PVC acquires very little charge and polystyrene acquires a negative charge due to rubbing.

They indicate that the resin is rubbed with polystyrene, therefore the resin must acquire a positive charge and the polystyrene a negative charge.

Then the PVC is rubbed with the polystyrene, which initially we will assume neutral, the PVC does not acquire a charge and the polystyrenes acquires a negative charge, if the polystyrene is isolated from the ground, it shares this negative charge with the PVC, therefore the two materials remain with half of the negative charge.

Finally, we bring the resin, which is positively charged, closer to the PVC, which has a slight negative charge, therefore the two bodies must attract each other.

In conclusion using the concepts of electrostatics we can find the result for the question about the electric forces in the two tubes is:

The tubes are attracted by having charges of different signs.

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2 years ago