All categories

1 year ago

Uranus (mass = 8.68 x 1025 kg) and

Physics

1 answer:

Pepsi [2]1 year ago

5 0

Answer:129,398,203.7 m

Explanation:

According to Newton's law of Universal Gravitation, the force $F$ exerted between two bodies of masses $M$ and $m$ and separated by a distance $d$ is equal to the product of their masses and inversely proportional to the square of the distance:

$F=G\frac{Mm}{d^2}$ (1)

Where:

$F=2.28(10)^{19} N$ is the gravitational force

$G=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}$ is the gravitational constant

$M=8.68(10)^{25} kg$ is the mass of Uranus

$m=6.59(10)^{19} kg$ is the mass of Uranu's moon, Mirana

$d$ is the distance between Uranus and its moon

Isolating $d$ :

$d=\sqrt{\frac{GMm}{F}}$ (2)

$d=\sqrt{\frac{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(8.68(10)^{25} kg)(6.59(10)^{19} kg)}{2.28(10)^{19} N}}$ (3)

Finally:

$d=129,398,203.7 m$

You might be interested in

During a compaction test in the lab a cylindrical mold with a diameter of 4in and a height of 4.58in was filled. The compacted s

Ray Of Light [21]

Answer:

part a : The dry unit weight is 0.0616  $lb/in^3$ 

part b : The void ratio is 0.77

part c : Degree of Saturation is 0.43

part d : Additional water (in lb) needed to achieve 100% saturation in the soil sample is 0.72 lb

Explanation:

Part a

Dry Unit Weight

The dry unit weight is given as

$\gamma_{d}=\frac{\gamma}{1+\frac{w}{100}}$

Here

$\gamma_d$ is the dry unit weight which is to be calculated
γ is the bulk unit weight given as

$\gamma =weight/Volume \\\gamma= 4 lb / \pi r^2 h\\\gamma= 4 lb / \pi (4/2)^2 \times 4.58\\\gamma= 4 lb / 57.55\\\gamma= 0.069 lb/in^3$

w is the moisture content in percentage, given as 12%

Substituting values

$\gamma_{d}=\frac{\gamma}{1+\frac{w}{100}}\\\gamma_{d}=\frac{0.069}{1+\frac{12}{100}} \\\gamma_{d}=\frac{0.069}{1.12}\\\gamma_{d}=0.0616 lb/in^3$

The dry unit weight is 0.0616  $lb/in^3$ 

Part b

Void Ratio

The void ratio is given as

$e=\frac{G_s \gamma_w}{\gamma_d} -1$

Here

e is the void ratio which is to be calculated

$\gamma_d$ is the dry unit weight which is calculated in part a
$\gamma_w$ is the water unit weight which is 62.4 $lb/ft^3$ or 0.04 $lb/in^3$

G is the specific gravity which is given as 2.72

Substituting values

$e=\frac{G_s \gamma_w}{\gamma_d} -1\\e=\frac{2.72 \times 0.04}{0.0616} -1\\e=1.766 -1\\e=0.766$

The void ratio is 0.77

Part c

Degree of Saturation

Degree of Saturation is given as

$S=\frac{G w}{e}$

Here

e is the void ratio which is calculated in part b

G is the specific gravity which is given as 2.72

w is the moisture content in percentage, given as 12% or 0.12 in fraction

Substituting values

$S=\frac{G w}{e}\\S=\frac{2.72 \times .12}{0.766}\\S=0.4261$

Degree of Saturation is 0.43

Part d

Additional Water needed

For this firstly the zero air unit weight with 100% Saturation is calculated and the value is further manipulated accordingly. Zero air unit weight is given as

$\gamma_{zav}=\frac{\gamma_w}{w+\frac{1}{G}}$

Here

$\gamma_{zav}$ is the zero air unit weight which is to be calculated

$\gamma_w$ is the water unit weight which is 62.4 $lb/ft^3$ or 0.04 $lb/in^3$

G is the specific gravity which is given as 2.72

w is the moisture content in percentage, given as 12% or 0.12 in fraction

$\gamma_{zav}=\frac{\gamma_w}{w+\frac{1}{G}}\\\gamma_{zav}=\frac{0.04}{0.12+\frac{1}{2.72}}\\\gamma_{zav}=\frac{0.04}{0.4876}\\\gamma_{zav}=0.08202 lb/in^3\\$

Now as the volume is known, the the overall weight is given as

$weight=\gamma_{zav} \times V\\weight=0.08202 \times 57.55\\weight=4.72 lb$

As weight of initial bulk is already given as 4 lb so additional water required is 0.72 lb.

4 0

2 years ago

A projectile is launched horizontally east at a speed of 29.4 M/s towards a wall 88.2 m away. What is the velocity of the projec

Drupady [299]

Time before projectile hits wall

= 88.2 m / 29.4 m/s = 3 seconds

Vertical velocity of projectile after three seconds

= 3*9.8 = 29.4 m/s

Horizontal velocity of projectile after three seconds, assuming no air resistance

= 29.4 m/s (given)

Conclusion:

velocity of projectile when it hits the wall

= < 29.4, -29.4> m/s

= sqrt(29.4^2+29.4^2) m/s east-bound at 45 degrees below horizontal

= 41.58 m/s east-bound at 45 degrees below horizontal.

6 0

2 years ago

A chair of mass 30.0 kg is at rest on a horizontal floor. The floor is not frictionless. You push on the chair with a force of 8

miv72 [106K]

First make sure you draw a force diagram. You should have Fn going up, Fg going down, Ff going left and another Fn going diagonally down to the right. The angle of the diagonal Fn (we'll call it Fn2) is 35° and Fn2 itself is 80N. Fn2 can be divided into two forces: Fn2x which is horizontal, and Fn2y which is vertical. Right now we only care about Fn2y.

To solve for Fn2y we use what we're given and some trig. Drawing out the actual force of Fn2 along with Fn2x and Fn2y we can see it makes a right triangle, with 80 as the hypotenuse. We want to solve for Fn2y which is the opposite side, so Sin(35)=y/80. Fn2y= 80sin35 = 45.89N

Next we solve for Fg. To do this we use Fg= 9.8 * m. Mass = 30kg, so Fg = 9.8 * 30 = 294N.

Since the chair isn't moving up or down, we can set our equation equal to zero. The net force equation in the vertical direction will be Fn + Fn2y -Fg = 0. If we plug in what we know, we get Fn + 45.89 -294 = 0. Then solve this algebraically.

Fn +45.89 -294 = 0
Fn +45.89 = 294
Fn = 248.11 N

You'll get a more accurate answer if you don't round Fn2y when solving for it, it would be something along the lines of 45.88611 etc

7 0

1 year ago

Read 2 more answers

An electric dipole with dipole moment p⃗ is in a uniform electric field E⃗ . A. Find all the orientation angles of the dipole me

Tasya [4]

Answer:

Explanation:

A) When a dipole is placed in an electric field , it experiences a torque equal to the following

torque = p x E = p E sinθ , where θ is angle between direction of p and E .

It will be zero if θ = 0

or if both p and E are oriented in the same direction.

It is the stable orientation of dipole.

If θ = 180° ,

Torque = 0

In this case both p and E are oriented in opposite direction .

It is the unstable orientation of the dipole because if we deflect the dipole by even small angle , it goes back to most stable orientation due to torque acting on it by electric field.

3 0

2 years ago

suppose that 273 g of one of the substances listed above displaces 26 mL of water. What is the substance?

guajiro [1.7K]

The unknown substance is silver. I don't see a list of available substances, but let's see if there's something reasonable available that will match. First, let's calculate the density of the unknown substance. Density is mass per volume, so 273 g / 26 mL = 10.5 g/mL Looking up a list of elements sorted by density, I see the following: 10.07 Actinium 10.22 Molybdenum 10.5 Silver 11.35 Lead And silver at 10.5 g/ml is a very nice match for the unknown substances' density of 10.5 g/ml.

6 0

2 years ago

Read 2 more answers