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Anit [1.1K]
2 years ago
8

four objects are situated along the y axis as follows: a 2.00kg object is at +3.00m. a 3.00kg object is at +2.50m, a 2.50kg obje

ct is at the origin ,and a 4.00kg object is at -0.500m. where is the center of mass of these objects?​
Physics
1 answer:
Veseljchak [2.6K]2 years ago
6 0

Answer: 1,348 meters

Explanation: Although the sign is missing from the location of the 4.00 kg object, it is assumed to be positive. The net moment of all the objects about the center of mass must be zero. Let the center of mass be on the y axis at a point  c . Adding the four moments together, we get:

(2.00)(3.00−c)+(3.00)(2.50−c)+(2.50)(0−c)+(4.00)(0.500−c)=0

6.00−2.00c+7.50−3.00c+0−2.50c+2.00−4.00c=0

11.5c=15.50

c=  1.348 metres

The center of mass is on the y axis at  y  = 1.348 metres.

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An empty rubber balloon has a mass of 12.5 g. The balloon is filled with helium at a density of 0.181 kg/m3. At this density the
Eduardwww [97]

Answer: 1.14 N

Explanation :

As any body submerged in a fluid, it receives an upward force equal to the weight of the fluid removed by the body, which can be expressed as follows:

Fb = δair . Vb . g = 1.29 kg/m3 . 4/3 π (0.294)3  m3. 9.8 m/s2

Fb = 1.34 N

In the downward direction, we have 2 external forces acting upon the balloon: gravity and the tension in the line, which sum must be equal to the buoyant force, as the balloon is at rest.

We can get the gravity force as follows:

Fg = (mb +mhe) g  

The mass of helium can be calculated as the product of the density of the helium times the volume of the balloon (assumed to be a perfect sphere), as follows:

MHe = δHe . 4/3 π (0.294)3 m3 = 0.019 kg

Fg = (0.012 kg + 0.019 kg) . 9.8 m/s2 = 0.2 N

Equating both sides of Newton´s 2nd Law in the vertical direction:

T + Fg = Fb

T = Fb – Fg = 1.34 N – 0.2 N = 1.14 N

6 0
2 years ago
Derive the formula for the moment of inertia of a uniform, flat, rectangular plate of dimensions l and w, about an axis through
Ad libitum [116K]

Answer:

A uniform thin rod with an axis through the center

Consider a uniform (density and shape) thin rod of mass M and length L as shown in (Figure). We want a thin rod so that we can assume the cross-sectional area of the rod is small and the rod can be thought of as a string of masses along a one-dimensional straight line. In this example, the axis of rotation is perpendicular to the rod and passes through the midpoint for simplicity. Our task is to calculate the moment of inertia about this axis. We orient the axes so that the z-axis is the axis of rotation and the x-axis passes through the length of the rod, as shown in the figure. This is a convenient choice because we can then integrate along the x-axis.

We define dm to be a small element of mass making up the rod. The moment of inertia integral is an integral over the mass distribution. However, we know how to integrate over space, not over mass. We therefore need to find a way to relate mass to spatial variables. We do this using the linear mass density of the object, which is the mass per unit length. Since the mass density of this object is uniform, we can write

λ = m/l (orm) = λl

If we take the differential of each side of this equation, we find

d m = d ( λ l ) = λ ( d l )

since  

λ

is constant. We chose to orient the rod along the x-axis for convenience—this is where that choice becomes very helpful. Note that a piece of the rod dl lies completely along the x-axis and has a length dx; in fact,  

d l = d x

in this situation. We can therefore write  

d m = λ ( d x )

, giving us an integration variable that we know how to deal with. The distance of each piece of mass dm from the axis is given by the variable x, as shown in the figure. Putting this all together, we obtain

I=∫r2dm=∫x2dm=∫x2λdx.

The last step is to be careful about our limits of integration. The rod extends from x=−L/2x=−L/2 to x=L/2x=L/2, since the axis is in the middle of the rod at x=0x=0. This gives us

I=L/2∫−L/2x2λdx=λx33|L/2−L/2=λ(13)[(L2)3−(−L2)3]=λ(13)L38(2)=ML(13)L38(2)=112ML2.

4 0
2 years ago
How can i calculate distances between objects by using the concepts of echo location
dangina [55]
Send wave from your location to the object and wait until echo is back.
Measure the time taken.

If you know the speed of wave (say sound wave), than just multiply by half time taken wave to return
5 0
3 years ago
You obtain a 100-W light bulb and a 50-W light bulb. Instead of connecting them in the normal way, you devise a circuit that pla
lesantik [10]

Answer:

When they are connected in series

     The  50 W bulb glow more than the 100 W bulb

Explanation:

From the question we are told that

     The power rating  of the first bulb is P_1  = 100 \ W

      The power rating of the second bulb is  P_2  =  50 \ W

     

Generally the power rating of the first bulb is mathematically represented as

      P_1  =  V^2 R

Where  V is the normal household voltage which is constant for both bulbs

  So  

        R_1  =  \frac{V^2}{P_1 }

substituting values

        R_1  =  \frac{V^2}{100}

Thus the resistance of the second bulb would be evaluated as

       R_2  =  \frac{V^2}{50}

From the above calculation we see that

        R_2  >  R_1

This power rating of the first bulb can also be represented mathematically as  

        P_  1  =  I^2_1  R_1

This power rating of the first bulb can also be represented mathematically as    

       P_  2  =  I^2_2 R_2

Now given that they are connected in series which implies that the same current flow through them so

       I_1^2 =  I_2^2

This means  that

       P \ \alpha  \  R

So  when they are connected in series

     P_2  >  P_1

This means that the 50 W bulb glows more than the 100 \ W bulb

3 0
3 years ago
Thermal escape of an atmosphere is most pronounced on worlds where the gravity is low and the temperature is high.
Alona [7]
The answer is false
7 0
3 years ago
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