All categories

2 years ago

Find the moments Mx and My and the center of mass of the system, assuming that the particles have equal mass m.

1 answer:

5 0

If we plot the points on the coordinate system we can see that for x, equal masses are occupying x = 1 through 4. with an equal density value. because they are all uniform we can intuit the moment to be the center of that distribution so Mx = 2.5.
Looking at y we have two equal masses at y = 1 and another set of masses at y = 3. This means that we have a moment in between those distributions and our My = 2.
The center of mass is the combination of the two, and it is at pt (2.5, 2). Because the masses considered are in general and also equal to each other.

You might be interested in

a partially inflated weather balloon has a volume of 1.56 * 10^3 L and a pressure of 98.9 lPa. What is the volume of the balloon

e-lub [12.9K]

To solve the problem, we enumerate all the given first. Then the required and lastly the solution.

Given:

V1= 1.56x10^3 L = 1560 L P2 = 44.1 kPa
P1 = 98.9 kPa

Required: V2

Solution:

Assuming the gas is ideal. Ideal gas follows Boyle's Law which states that at a given temperature the product of pressure and volume of a gas is constant. In equation,

PV = k

Applying to the problem, we have

P1*V1 = P2*V2
(98.9 kPa)*(1560 L) = (44.1 kPa)*V2
V2 = 3498.5 L

ANSWER: V2 = 3498.5 L

7 0

2 years ago

Milk containing 3.7% fat and 12.8% total solids is to be evaporated to produce a product containing 7.9% fat. What is the yield

ratelena [41]

Answer:

the yield of product is YP=46.835 % and the concentration of solids is

Cs = 27.33%

Explanation:

Assuming that all the solids and fats remains in the milk after the evaporation, then the mass of product mP will be

Mass of fat in 100 kg of milk = 100 kg* 0.037 = mP* 0.079

mP = 100 kg* 0.037/0.079 = 46.835 kg

then the yield YP of the product is

YP= mP / 100 kg = 46.835 kg / 100 kg = 46.835 %

YP= 46.835 %

the concentration of solids Cs is

Mass of solids in 100 kg of milk = 100 kg* 0.128 = 46.835 kg * Cs

Cs = 100 kg* 0.128 / 46.835 kg = 0.2733 = 27.33%

Cs = 27.33%

3 0

2 years ago

The moon’s orbital speed around Earth is 3.680 × 10^3 km/h. Suppose the moon suffers a perfectly elastic collision with a comet

Naily [24]

Answer:

Speed of comet before collision is

$v_{2_{i}}=-2.5\times10^{3}\quad km/h$

Explanation:

Correction: (As stated after collision comet moves away from moon so velocity of moon and moon and comet must be opposite in direction. as spped of moon after collision is −4.40 × 10^2km/h so that comet's must be 5.740 × 10^3km/h instead of -5.740 × 10^3km/h)

Solution:

$mass \quad of\quad moon = m_{1}\\\\mass\quad of \quad comet = m_{2} = 0.5 m_{1}\\\\speed\quad of\quad moon\quad before\quad collision = v_{1_{i}}=3.680\times 10^3 km/h\\\\speed \quad of\quad moon\quad after\quad collision=v_{1_{f}} = -4.40 \times 10^2 km/h\\\\speed\quad of\quad comet\quad after\quad collision =v_{2_{f}} =5.740 \times 10^3 km/h$

Case is considered as partially inelastic collision, by conservation of momentum

$m_{1}v_{1_{i}}+m_{2}v_{2_{i}}=m_{1}v_{1_{f}}+m_{2}v_{2_{f}}\\\\m_{1}v_{1_{i}}+0.5m_{1}v_{2_{i}}=m_{1}v_{1_{f}}+0.5m_{1}v_{2_{f}}\\\\v_{1_{i}}+0.5v_{2_{i}}=v_{1_{f}}+0.5v_{2_{f}}\\\\v_{2_{i}}=2(v_{1_{f}}+0.5v_{2_{f}}-v_{1_{i}})\\\\v_{2_{i}}=2(-4.40 \times 10^2+0.5(5.740 \times 10^3)-3.680 \times 10^3 )\\\\v_{2_{i}}=-2.5\times10^{3}\quad km/h$

8 0

2 years ago

Four students measured the acceleration of gravity. The accepted value for their location is 9.78m/s2. Which student’s measureme

Lisa [10]

On comparing values , we see that student which has the largest percent error is A. Student 4: 9.61 m/s2 .

Explanation:

Here, we have Four students measured the acceleration of gravity. The accepted value for their location is 9.78m/s2. Let's calculate which student’s measurement has the largest percent error :

A. Student 4: 9.61 m/s2

Percentage of error = $\frac{9.78-9.61}{9.78} (100) = 1.738%$ %.

B. Student 3: 9.88 m/s2

Percentage of error = $\frac{9.88-9.78}{9.78} (100) = 1.022%$ %.

C. Student 2: 9.79 m/s2

Percentage of error = $\frac{9.79-9.78}{9.78} (100) = 0.1022%$ % .

D. Student 1: 9.78 m/s2

Percentage of error = $\frac{9.78-9.61}{9.78} (100) = 0%$ % .

On comparing values , we see that student which has the largest percent error is A. Student 4: 9.61 m/s2 .

4 0

2 years ago

Two vectors are presented as a=3.0i +5.0j and b=2.0i+4.0j find (a) a x b, ab (c) (a+b)b and (d) the component of a along the dir

Svet_ta [14]

Let's ask this question step by step:
Part A)
a x b = (3.0i + 5.0j) x (2.0i + 4.0j) = (12-10) k = 2k
ab = (3.0i + 5.0j). (2.0i + 4.0j) = 6 + 20 = 26
Part (c)
(a + b) b = [(3.0i + 5.0j) + (2.0i + 4.0j)]. (2.0i + 4.0j)
(a + b) b = (5.0i + 9.0j). (2.0i + 4.0j)
(a + b) b = 10 + 36
(a + b) b = 46
Part (d)
comp (ba) = (a.b) / lbl
a.b = (3.0i + 5.0j). (2.0i + 4.0j) = 6 + 20 = 26
lbl = root ((2.0) ^ 2 + (4.0) ^ 2) = root (20)
comp (ba) = 26 / root (20)
answer
2k
26
46
26 / root (20)

3 0

2 years ago