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LiRa [457]
2 years ago
15

Suppose that Salim hits a 100 g fastball moving towards him at a speed of 42 m/s with his bat as shown in the image.the impulse

of the fastball is 8 kg.m/s in the antiparallel direction of the initial velocity. Then the final momentum of the fastball is --------------( Suppose that the direction of the fastball toward the bat is negative and away from the bat is positive)
1.8 kg m/s
0.6 kg m/s
3.8 kg m/s
2.4 kg m/s
Physics
1 answer:
FinnZ [79.3K]2 years ago
3 0

The final momentum of the ball is 3.8 kgm/s.

<h3>Change in momentum of the ball</h3>

The impulse received by the ball is equal to change in momentum of the ball.

J = ΔP

where;

  • J is the impulse
  • ΔP is change in momentum

ΔP = P₂ - P₁

P₂ = ΔP + P₁

<h3>Final momentum of the ball</h3>

The final momentum of the ball is calculated as follows;

P₂ = 8 + (- 0.1 x 42)

P₂ = 8 - 4.2

P₂ = 3.8 kgm/s

Learn more about change in momentum here: brainly.com/question/7538238

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Consider two less-than-desirable options.
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The force would be the same in both cases - option C.

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I = F(average)*ΔT. Where F(average) is the average force and ΔT is the time interval.

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Air flows through an adiabatic turbine that is in steady operation. The air enters at 150 psia, 900oF, and 350 ft/s and leaves a
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1486.5\frac{Btu}{s}

Explanation:

The inlet specific volume of air is given by:

v_1=\frac{RT_1}{P_1}\\\\v_1=\frac{(0.3704\frac{psia.ft^3}{lbm.R})(1360R)}{150psia}\\\\v_1=3.358\frac{ft^3}{lbm} \ \ \ \  \ \  \ \ \...i

The mass flow rates is expressed as:

\dot m=\frac{1}{v_1}A_1V_1\\\\\dot m=\frac{1}{3.358ft^3/psia}(0.1ft^2)(350ft/s)\\\\\dot m=10.42\frac{lbm}{s}

The energy balance for the system can the be expresses in the rate form as:

E_{in}-E_{out}=\bigtriangleup \dot E=0\\\\E_{in}=E_{out}\\\\\dot m(h_1+0.5V_1^2)=\dot W_{out}+\dot m(h_2+0.5V_2^2)+Q_{out}\\\\\dot W_{out}=\dot m(h_2-h_1+0.5(V_2^2-V_1^2))=-m({cp(T_2-t_1)+0.5(V_2^2-V_1^2)})\\\\\\\dot W_{out}=-(10.42lbm/s)[(0.25\frac{Btu}{lbm.\textdegree F})(300-900)\textdegree F+0.5((700ft/s)^2-(350ft/s)^2)(\frac{1\frac{Btu}{lbm}}{25037ft^2/s^2})]\\\\\\\\=1486.5\frac{Btu}{s}

Hence, the mass flow rate of the air is 1486.5Btu/s

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