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juin [17]
3 years ago
13

A rectangular metal bar has length of 20cm, width of 5cm. Mass of the block is 2kg and its density is 5000kg/m3. Find the height

of the metal bar.
Physics
1 answer:
Black_prince [1.1K]3 years ago
8 0

Answer: 0.04m

Explanation:

Mass of the block (m) = 2kg

density (p) = 5000kg/m3

Volume (v) = ?

Recall that density is obtained by dividing mass by volume. i.e density = mass / volume

p = m / v

Make v the subject formula

v = m / p

So, v = 2kg / 5000kg/m3

v = 0.0004 m3

Now given that the volume of the rectangular metal bar is 0.0004m3, find its height:

length of metal bar (l) = 20cm

convert length in centimeters to meters

(If 100cm = 1m

20cm = 20/100 = 0.2m)

width of bar (w) = 5cm

convert width in centimeters to meters

(If 100cm = 1m

5cm = 5/100 = 0.05m)

height (h) = ?

Volume of metal bar = l x w x h

0.0004 m3 = 0.2m x 0.05m x h

0.0004 m3 = 0.2m x 0.05m x h

0.0004 m3 = 0.01m² x h

h = 0.0004 m3 / 0.01m²

h = 0.04m

Thus, the height of the rectangular metal bar is 0.04 meters

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Question:<em> </em><em>Find, separately, them mass of the balloon and the basket (incidentally, most of the balloon's mass is air)</em>

Answer:

The mass of the balloon is 2295 kg, and the mass of the basket is 301 kg.

Explanation:

Let us call the mass of the balloon m_1 and the mass of the basket m_2, then according to newton's second law:

(1). \:F = (m_1+m_2)a,

where a =0.265m/s^2 is the upward acceleration, and F = 688N is the net propelling force (counts the gravitational force).

Also, the tension T in the rope is 79.8 N more than the basket's weight; therefore,

(2). \:T = m_2g+79.8

and this tension must equal

T -m_2g =m_2a

(3). \:T = m_2g +m_2a

Combining equations (2) and (3) we get:

m_2a = 79.8

since a =0.265m/s^2, we have

\boxed{m_2 = 301.13kg}

Putting this into equation (1) and substituting the numerical values of F and a, we get:

688N = (m_1+301.13kg)(0.265m/s^2)

\boxed{m_1 = 2295 kg}

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8 0
3 years ago
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:

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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

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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

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t = \frac{80 m}{11.62m/s } = 6.88 \ s

Thus, the time delay between green lights on successive blocks to keep the traffic moving continuously is 6.88 s


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