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2 years ago

How would the period of a planet depend on its orbital distance from a star? in this comparison,

1 answer:

6 0

The answer would change to Spec because the orbital period depends only on the distance from star and the star's mass, so if Spec became the closer planet, it would orbit in the shorter time.

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You are trying to determine the specific gravity of a solid object that floats in water. If m is the mass of your object, mS is

Alisiya [41]

Answer:

Specific Gravity = m/[m(s)-m(os)]

Explanation:

Specific gravity, also called relative density, is the ratio of the density of a substance to the density of a reference substance. By this definition we need to find out the ratio of density of the object of mass m to the density of the surrounding liquid.

m = mass of the object

Weight in air

W (air) = mg, where g is the gravitational acceleration

Weight with submerged with only one mass

m(s)g + Fb = mg + m(b)g, consider this to be equation 1

where Fb is the buoyancy force

Weight with submerged with both masses

m(os)g + Fb’ = mg + m(b)g, consider this to be equation 2

equation 1 – equation 2 would give us

m(s)g – m(os)g = Fb’ – Fb

where Fb = D x V x g, where D is the density of the liquid the object is submerged in, g is the force of gravity and V is the submerged volume of the object

m(s)g – m(os)g = D(l) x V x g

m(s) – m(os) = D(l) x V

we know that Mass = Density x V, which in our case would be, D(b) x V, which also means

V = Mass/D(b), where D(b) is the density of the mass

Substituting V into the above equation we get

m(s) – m(os) = [D(l) x m)/ D(b)]

Rearranging to get the ratio of density of object to the density of liquid

D(b)/D(l) = m/[m(s)-m(os)], where D(b)/D(l) denotes the specific gravity

8 0

2 years ago

You have a rod with a length of 146.4 cm. You prop up one end on a brick which is 3.8 cm thick. Your uncertainty in measuring th

AfilCa [17]

Answer:

$\partial \theta = 0.003$

Explanation:

we know that

$sin\theta = \frac{3.8}{146.4}$

$\theta = sin^{-1} \frac{3.8}{146.4}$

$\theta = 1.484°$

$\theta = 1.484° *\frac{\pi}{180} = 0.0259 radians$

as we see that $sin\theta = \theta$

relative error $\frac{\partial \theta}{\theta} = \frac{\partial X}{X_1} +\frac{\partial X}{X_2}$

Where X_1 IS HEIGHT OF ROCK

$X_2$ IS THE HEIGHT OF ROAD

$\partial X$ = uncertainity in measuring distance

$\partial X = 0.05$

Putting all value to get uncertainity in angle

$\frac{\partial \theta}{0.0259} = \frac{0.05}{3.8} +\frac{0.05}{146.4}$

solving for $\partial \theta$ we get

$\partial \theta = 0.003$

3 0

2 years ago

A skateboarder wants to cross a large playground and notices that there are large shapes painted on its asphalt surface. One sha

Elden [556K]

<h2>Diagonal of circle </h2>

Explanation:

As the skateboarder wants to cross the play ground . The surface is rough .

As we know , the force of friction is non-conservative force . Thus work is required against this force .

We have formula:

work done = Force x distance (in one direction )

Te force applied cannot be changed , so he is to decrease the distance .

In case of circle , diameter is the minimum distance . Thus he is supposed to move along it .

6 0

3 years ago

I need help plz!!! Please help...

RSB [31]

Answer: 60

Explanation:

4 0

2 years ago

Q1 is located at the origin, Q2 is located at x = 2.50 cm and Q3 is located at x = 3.50 cm. Q1 has a charge of +4.92μC and Q3 ha

Inessa05 [86]

Answer:

$+1.11\mu C$

Explanation:

A charge located at a point will experience a zero electrostatic force if the resultant electric field on it due to any other charge(s) is zero.

$Q_1$ is located at the origin. The net force on it will only be zero if the resultant electric field intensity due to $Q_2$ and $Q_3$ at the origin is equal to zero. Therefore we can perform this solution without necessarily needing the value of $Q_1$ .