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elixir [45]
3 years ago
8

The volume of an object as a function of time calculated by V=At^3+B/t,where t is time measured in seconds and V is in cubic met

ers. Determine the dimension of the constants A and B
Physics
1 answer:
S_A_V [24]3 years ago
8 0

Answer:

A = L^3 T^-3  ,  B = L^3 T

Explanation:

Given: volume ( V )  = At^3 + B/t  ------ ( 1 )

dimension of volume = L^3

and the Dimension of time = T

back to equation ( 1 )

L^3 = A * T^3   ------- ( 2 )

also L^3 = B/T ------- ( 3 )

from equation ( 2 )

A = L^3 T^-3

from equation ( 3 )

B = L^3 T

<u>The dimensions of the constants A and B </u>

A = L^3 T^-3  ,  B = L^3 T

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Two particles are located on the x axis. particle 1 has a mass m and is at the origin. particle 2 has a mass 2m and is at x = +l
wlad13 [49]

The solution would be like this for this specific problem:

<span>
The force on m is:</span>

<span>
GMm / x^2 + Gm(2m) / L^2 = 2[Gm (2m) / L^2] -> 1

The force on 2m is:</span>

<span>
GM(2m) / (L - x)^2 + Gm(2m) / L^2 = 2[Gm (2m) / L^2] -> 2

From (1), you’ll get M = 2mx^2 / L^2 and from (2) you get M = m(L - x)^2 / L^2 

Since the Ms are the same, then 

2mx^2 / L^2 = m(L - x)^2 / L^2 

2x^2 = (L - x)^2 

xsqrt2 = L - x 

x(1 + sqrt2) = L 

x = L / (sqrt2 + 1) From here, we rationalize. 

x = L(sqrt2 - 1) / (sqrt2 + 1)(sqrt2 - 1) 

x = L(sqrt2 - 1) / (2 - 1) 


x = L(sqrt2 - 1) </span>

 

= 0.414L

 

<span>Therefore, the third particle should be located the 0.414L x axis so that the magnitude of the gravitational force on both particle 1 and particle 2 doubles.</span>

8 0
3 years ago
Why is Jupiter so much larger than Earth? Check all that apply
Liono4ka [1.6K]

Answer:

Moreover, Boss says that even if Jupiter is proven to have a core, the planet still could have formed that core through disk instability. Enough dust could have collected and cemented together in the dense gas to form a core many times larger than the size of the Earth.

Explanation:

The same is true of most other objects in the solar system — except Jupiter. The gas giant is so big that it pulls the center of mass between it and the sun, also known as the barycenter, some 1.07 solar radii from the star's center — which is about 30,000 miles above the sun's surface.

69,911 km

69,911 kmJupiter/Radius

3 0
3 years ago
Read 2 more answers
What is the velocity of the object?
dmitriy555 [2]
<h2>Hey There!</h2><h2>_____________________________________</h2><h2>Answer:</h2><h2 /><h2>\huge\boxed{\text{V = 9.5 m/s}}</h2><h2>_____________________________________</h2>

<h2>DATA:</h2>

mass = m = 2kg

Distance = x = 6m

Force = 30N

TO FIND:

Work = W = ?

Velocity = V = ?

<h2>SOLUTION:</h2>

According to the object of mass 2 kg travels a distance when the force was exerted on it. The graph between the Force and position was plotted which shows that 30 N of force was used to push the object till the distance of 6.0m.

To find the work, I will use the method of determining the area of the plotted graph. As the graph is plotted in the straight line between the Force and work, THE PICTURE ATTCHED SHOWS THE AREA COVERED IN BLUE AS WORK DONE AND HEIGHT AS 30m AND DISTANCE COVERED AS 6m To solve for the area(work) of triangle is given as,

{\Longrightarrow}\qquad \qquad \qquad W\ =\ \frac{1}{2}\;(Base)\:(Height)

Base is the x-axis of the graph which is Position i.e. 6m

Height is the y-axis of the graph which is Force i.e. 30N

So,

                           W\ =\ \frac{1}{2}\:6\:x\:30

                           W   =  90 J

The work done is 90 J.

According to the principle of work and kinetic energy (also known as the work-energy theorem) states that the work done by the sum of all forces acting on a particle equals the change in the kinetic energy of the particle.

{\Longrightarrow}\qquad \qquad \qquad W\quad =\quad K.E\\\\{\Longrightarrow}\qquad \qquad \qquad W\quad =\quad \frac{1}{2}\ m\ V^2 \\\\{\Longrightarrow}\qquad \qquad \qquad W\quad =\quad \frac{1}{2}\ 2\ (V_f-V_i)^2\\\\{V_i\ is\ 0\ because\ the\ object\ was\ initially\ at\ rest}\\\\ {\Longrightarrow}\qquad \qquad \qquad W\quad\ =\ \frac{1}{2}\ x\ 2\ (V_f-0)^2 \\\\{\Longrightarrow}\qquad \qquad \qquad 90\quad = \frac{1}{2}\ x\ 2\ (V_f)^2

\\\\{\Longrightarrow}\qquad \qquad \qquad V_f\quad =\ \sqrt{\frac{2\ (90)\ }{2}}\\\\{\Longrightarrow}\qquad \qquad \qquad \boxed {V_f\quad =\ 9.48\ m/s}

\boxed{The\ Velocity\ of\ the\ Object\ of\ mass\ 2kg\ at\ 6\ meters\ of\ distance\ was\ 9.48\ m/s}

<h2>_____________________________________</h2><h2>Best Regards,</h2><h2>'Borz'</h2>

8 0
3 years ago
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mart [117]
G g j inj g f d t u u i
8 0
2 years ago
Looking at the bottom standing wave in the picture, how many complete waves are present?
Shtirlitz [24]

Answer:

the last one

Explanation:

4 0
2 years ago
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