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2 years ago

What is the difference between density and specific gravity

1 answer:

4 0

Density is defined as mass per unit volume. It has the SI unit kg m-3 or Kg/m-3 and is an absolute quantity. Specific gravity is the ratio of a material's density with that of water at 4 °C (where it is most dense and is taken to have the value 999.974 kg·m-3). It is therefore a relative quantity with no units.

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What is the best explanation for the observation that the electric charge on the stem became positive as the charged bee approac

topjm [15]

Answer:

Option B seems to be the appropriate choice,

Explanation:

The given question is incomplete. Please find the attachment of the complete question.

From either the principle or law of the Coulombs, there would have been either pressure (force) through attachment or repulsion amongst charges.
It is evident from the analysis presented that perhaps the bee becomes positively (+) charged and would have been drawn either by negatively (-) charged roots of the plants or stem.

As a consequence, the vegetation is electric current polarized because as activated bee pursued.

3 0

2 years ago

Collisions between galaxies are thought to:

dimaraw [331]

The bigger galaxy survives the smaller one gets eaten up

3 0

3 years ago

A 66 kg person is parachuting and experiencing a downward acceleration of 2.1 m/s2. The mass of the parachute is 4.0 kg. (a) Wha

UkoKoshka [18]

Answer:

539N

Explanation:

Parameters given:

Mass of person = 66kg

Mass of parachute = 4kg

Downward acceleration = 2.1 m/s²

The total force acting on the system is the sum of the force of gravity and the force due to air acting opposite gravity:

F = mg - F(a)

Where g = acceleration due to gravity

Force is given as:

F = ma

=> ma = mg - F(a)

=> F(a) = mg - ma

F(a) = m(g - a)

m is the mass of the entire sustem

m = 66 + 4 = 70kg

=> F(a) = 70(9.8 - 2.1)

F(a) = 70 * 7.7

F(a) = 539N

The upward force on the open parachute from the air is 539N

6 0

3 years ago

A satellite in a circular orbit of radius R around planet X has an orbital period T. If Planet X had one-fourth as much mass, th

Iteru [2.4K]

<h2>Answer: 2T</h2>

According to the Third Kepler’s Law of Planetary motion <em>“The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.</em>

In other words, this law states a relation between the orbital period $T$ of a body (moon, planet, satellite) orbiting a greater body in space with the size $R$ of its orbit.

This Law is originally expressed as follows (in the case of planet X and assuming we have a circular orbit):

$T^{2}=\frac{4\pi^{2}}{GM}R^{3}$ (1)

Where:

$G$ is the Gravitational Constant

$M=1.9(10)^{27}kg$ is the mass of planet X

$R$ is the radius of the orbit of the satellite around planet X

If we want to find the period, we have to express equation (1) as written below and substitute all the values:

$T=2\pi\sqrt{\frac{R^{3}}{GM}}$ (2)

Now, we are asked to find the period when tha mass of the planet is $\frac{1}{4}M$ . In order to do this, we have to rewrite equation (2) with this new value:

$T=2\pi\sqrt{\frac{R^{3}}{G(\frac{1}{4}M)}}$ (3)

Solving:

$T=4\pi\sqrt{\frac{R^{3}}{G(\frac{1}{4}M)}}$ (4)

On the other hand, if we multiply both sides of equation (2) by 2, we have:

$2T=4\pi\sqrt{\frac{R^{3}}{GM}}$ (5)

As we can see, (5) is equal to (4). This means the orbital period is twice the orignal period.

Hence, the answer is:

If Planet X had <u>one-fourth </u>as much mass, the <u>orbital period</u> of this satellite in an orbit of the same radius would be <u>2T.</u>