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3 years ago

A ball has 2 kg*m/s of momentum when thrown with a velocity of 8 m/s outwards. Find the mass of the ball.

Physics

1 answer:

worty [1.4K]3 years ago

5 0

Answer:

As, we know momentum = mass × velocity that is p = mv

we have momentum (p) = 2 kg m/s

and velocity (v) = 8 m/s

so substitute values in p=mv

2 kg m/s = m × 8 m / s

m = 2 / 8 kg = 0.25 kg

Therefore, mass of substance is 0.25 kg.

2/8 in fractional form.....

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3 years ago

A big lump of meat of mass 5Kg is hung from a spring balance in an elevator. Find the reading of the balance of (I) the elevator

Mrrafil [7]

The reading of the balance if ,

I ) If the elevator is moving with a steady speed = 50 N

II ) If the elevator is moving upwards with acceleration of 0.2 m / s² = 51 N

T = m g + m a

T = Force

m = Mass

g = Acceleration due to gravity

a = Acceleration

m = 5 kg

g = 10 m / s²

I ) If the elevator is moving with a steady speed,

At steady speed, a = 0

T = ( 5 * 10 ) + ( 5 * 0 )

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II ) If the elevator is moving upwards with acceleration of 0.2 m / s²,

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T = ( 5 * 10 ) + ( 5 * 0.2 )

T = 50 + 1

T = 51 N

Therefore, the reading of the balance if ,

I ) If the elevator is moving with a steady speed = 50 N

II ) If the elevator is moving upwards with acceleration of 0.2 m / s² = 51 N

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1 year 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,

Sedbober [7]

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 : $a_r_3=4.553 \ and\ h_3=1909.62kJ/kg$