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dexar [7]
3 years ago
5

In the mobile m1=0.42 kg and m2=0.47 kg. What must the unknown distance to the nearest tenth of a cm be if the masses are to be

balanced? The answer is 13.4 In the mobile what is the value for m3 to the nearest hundredth of a kilogram?

Physics
1 answer:
LuckyWell [14K]3 years ago
3 0

Complete Question

The complete question is shown on the first uploaded image

Answer:

Explanation:

From he question we are told that

    The first mass is   m_1 = 0.42kg

      The second mass is  m_2 = 0.47kg

From the question we can see that at equilibrium the moment about the point where the  string  holding the bar (where m_1 \ and \ m_2 are hanged ) is attached is zero  

   Therefore we can say that

               m_1 * 15cm  = m_2 * xcm

Making x the subject of the formula  

                x = \frac{m_1 * 15}{m_2}

                    = \frac{0.42 * 15}{0.47}

                     x = 13.4 cm

Looking at the diagram we can see that the tension T  on the string holding the bar where m_1  \  and   \ m_2 are hanged  is as a result of the masses (m_1 + m_2)

     Also at equilibrium the moment about the point where the string holding the bar (where (m_1 +m_2)  and  m_3 are hanged ) is attached is  zero

   So basically

          (m_1 + m_2 ) * 20  = m_3 * 30

          (0.42 + 0.47)  * 20 = 30 * m_3

 Making m_3 subject

          m_3 = \frac{(0.42 + 0.47) * 20 }{30 }

                m_3 = 0.59 kg

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An air-standard Diesel cycle has a compression ratio of 16 and a cutoff ratio of 2. At the beginning of the compression process,
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Answer:

a.T_3=1723.8kPa\\b.n=0.563\\c.MEP=674.95kPa

Explanation:

a. Internal energy and the relative specific volume at s_1 are determined  from A-17:u_1=214.07kJ/kg, \ \alpha_r_1=621.2.

The relative specific volume at s_2 is calculated from the compression ratio:

\alpha_r_2=\frac{\alpha_r_1}{r}\\=\frac{621.2}{16}\\=38.825

#from this, the temperature and enthalpy at state 2,s_2 can be determined using interpolations T_2=862K and h_2=890.9kJ/kg. The specific volume at s_1 can then be determined as:

\alpha_1=\frac{RT_1}{P_1}\\\\=\frac{0.287\times 300}{95} m^3/kg\\0.906316m^3/kg

Specific volume,s_2:

\alpha_2=\frac{\alpha_1}{r}\\=\frac{0.906316}{16}m^3/kg\\=0.05664m^3/kg

The pressures at s_2 \ and\  s_3 is:

P_2=P_3=\frac{RT_2}{\alpha_2}\\\\=\frac{0.287\times862}{0.05664}\\=4367.06kPa

.The thermal efficiency=> maximum temperature at s_3 can be obtained from the expansion work at constant pressure during s_2-s_3

\bigtriangleup \omega_2_-_3=P(\alpha_3-\alpha_2)\\R(T_3-T_2)=P\alpha(r_c-1)\\T_3=T_2+\frac{P\alpha_2}{R}(r_c-1)\\\\=(862+\frac{4367\times 0.05664}{0.287}(2-1))K\\=1723.84K

b.Relative SV and enthalpy  at s_3 are obtained for the given temperature with interpolation with data from A-17 :a_r_3=4.553 \ and\  h_3=1909.62kJ/kg

Relative SV at s_4 is

a_r_4=\frac{r}{r_c}\alpha _r_3

==\frac{16}{2}\times4.533\\=36.424

Thermal efficiency occurs when the heat loss is equal to the internal energy decrease and heat gain equal to enthalpy increase;

n=1-\frac{q_o}{q_i}\\=1-\frac{u_4-u_1}{h_3-h_2}\\=1-\frac{65903-214.07}{1909.62-890.9}\\=0.563

Hence, the thermal efficiency is 0.563

c. The mean relative pressure is calculated from its standard definition:

MEP=\frac{\omega}{\alpa_1-\alpa_2}\\=\frac{q_i-q_o}{\alpha_1(1-1/r)}\\=\frac{1909.62-890.9-(65903-214.7)}{0.90632(1-1/16)}\\=674.95kPa

Hence, the mean effective relative pressure is 674.95kPa

3 0
3 years ago
How do you solve 0.004 dm + 0.12508 dm?
Effectus [21]
0.004 of something added to 0.12508 of the same thing
adds up to 0.12908 of it.  

The thing could be a glass of water, a sheet of paper,
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7 0
2 years ago
A total resistance of 3.03 Ω is to be produced by connecting an unknown resistance to a 12.18 Ω resistance. (a) What must be the
insens350 [35]

Answer:

(a) 4.0334Ω

(b)parallel

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for resistors connected in parallel;

\frac{1}{R_{eq} } =\frac{1}{R1}+\frac{1}{R2}

Req =3.03Ω , R1 =12.18Ω

\frac{1}{3.03 } =\frac{1}{12.18}+\frac{1}{R2}

\frac{1}{R2}=\frac{1}{3.03 }-\frac{1}{12.18}

\frac{1}{R2}=0.2479

R2=1/0.2479

R2=4.0334Ω

(b)parallel connection is suitable for the desired total resistance. series connection can not be used to achieve a lower resistance as the equation for series connection is.

Req = R1+R2

3 0
3 years ago
A student hangs a weight on a newtonmeter. The energy currently stored in the spring in the newton meter is 0.045N. The student
castortr0y [4]

Answer:

5x10^-3

Explanation:

Hooke's Law states that the force needed to compress or extend a spring is directly proportional to the distance you stretch it.

Hooke's Law can be represented as

<h3> F = kx, </h3>

<em>where F is the force </em>

<em>            k is the spring constant</em>

<em>            x is the extension of the material </em>

<em />

Plug values in the equation

Step 1 find the original extension

0.045 = (400)x

x = 1.125x 10^-4 m d

Step 2 find the new extension

0.045+2 = 400(x)

2.045 = 400x

x = 5.1125x10^-3

Step 3 subtract the new extension with original

Total extension of the spring =  5.1125x10^-3 - 1.125x 10^-4 m = 5x10^-3

8 0
3 years ago
A ball is thrown vertically upward, which is the positive direction. A little later, it returns to its point of release. The bal
Aleks [24]

Answer:

The initial velocity of the ball is <u>39.2 m/s in the upward direction.</u>

Explanation:

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Upward direction is positive. So, downward direction is negative.

Tota time the ball remains in air (t) = 8.0 s

Net displacement of the ball (S) = Final position - Initial position = 0 m

Acceleration of the ball is due to gravity. So, a=g=-9.8\ m/s^2(Acting down)

Now, let the initial velocity be 'u' m/s.

From Newton's equation of motion, we have:

S=ut+\frac{1}{2}at^2

Plug in the given values and solve for 'u'. This gives,

0=8u-0.5\times 9.8\times 8^2\\\\8u=4.9\times 64\\\\u=\frac{4.9\times 64}{8}\\\\u=4.9\times 8=39.2\ m/s

Therefore, the initial velocity of the ball is 39.2 m/s in the upward direction.

3 0
3 years ago
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