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3 years ago

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A 3.52 kg steel ball is tossed upward from a height of 6.93 meters above the floor with a vertical velocity of 2.99 m/s. What is

the velocity (m/s) of the ball just before it strikes the floor?

1 answer:

Dafna1 [17]3 years ago

7 0

Answer : The final velocity of the ball is, 12.03 m/s

Explanation :

By the 3rd equation of motion,

$v^2-u^2=2as$

where,

s = distance covered by the object = 6.93 m

u = initial velocity = 2.99 m/s

v = final velocity = ?

a = acceleration = $9.8m/s^2$

Now put all the given values in the above equation, we get the final velocity of the ball.

$v^2-(2.99m/s)^2=2\times (9.8m/s^2)\times (6.93m)$

$v=12.03m/s$

Thus, the final velocity of the ball is, 12.03 m/s

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Air enters a 200 mm diameter adiabatic nozzle at 195 deg C, 500 kPa and 100 m/s. It exits at 85 kPa. If the exit diameter is 158

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Answer:

$v_2 = 160.23 m/s$

$T_2 = 475.797 k$

Explanation:

given data:

Diameter = $d_1 = 200mm$

$t_1 =195 degree$

$p_1 =500 kPa$

$v_1 = 100m/s$

$p_2 = 85kPa$

$d_2 = 158mm$

from continuity equation

$A_1v_1 = A_2v_2$

$v_2 = \frac{\frac{\pi}{4}d_1^2 v_1^2}{\frac{\pi}{4}d_2^2}$

$v_2 = \frac{d_2v_1}{d_2^2}$

$v_2 = [\frac{d_1}{d_2}]^2 v_1$

$= [\frac{0.200}{0.158}]^2 \times 100$

$v_2 = 160.23 m/s$

by energy flow equation

$h_1 + \frac{v_1^2}{2} +gz_1 +q =h_2 + \frac{v_2^2}{2} +gz_2 +w$

$z_1 =z_2$ and q =0, w =0 for nozzle

therefore we have

$h_1 -h_2 =\frac{v_1^2}{2} -\frac{v_2^2}{2}$

$dh = \frac{1}{2} (v_1^2 -v_2^2)$

but we know dh = Cp dt

hence our equation become

$Cp(T_2 -T_1) = \frac{1}{2} (v_1^2 -v_2^2)$

$Cp (T_2 -T_1) = 7836.94$

$(T_2 -T_1) = \frac{7836.94}{1.005*10^3}$

$(T_2 -T_1) = 7.797$

$T_2 = 7.797 +468 = 475.797 k$

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The two shafts of a Hooke’s coupling have their axes inclined at 20°.The shaft A revolves at a uniform speed of 1000 rpm. The sh

lapo4ka [179]

Answer:

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Explanation:

Given :

Inclination angle of two shaft, α = 20°

Speed of shaft A, $N_{A}$ = 1000 rpm

Mass of flywheel, m = 30 kg

Radius of Gyration, k =100 mm

= 0.1 m

Now we know that for maximum velocity,

$\frac{N_{B}}{N_{A}} = \frac{cos\alpha }{1 - sin^{2}\alpha }$

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$N_{B}$ = 1064.1 rpm

Now we know

Mass of flywheel, m = 30 kg

Radius of Gyration, k =100 mm

= 0.1 m

Therefore moment of inertia of flywheel, I = m. $k^{2}$

=30 X $0.1^{2}$

= 0.3 kg- $m^{2}$

Now torque on the output shaft

T₂ = I x ω

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Nitrogen gas is compressed at steady state from a pressure of 14.2 psi and a temperature 60o F to a pressure of 120 psi and a te

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Answer:

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b) 262.15 ft3/min.

Explanation:

Given data:

P1 = 14.2 psi

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T2 = 500°F = 960° R

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attached below is the detailed solution

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