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ale4655 [162]
3 years ago
8

The blue curve is the plot of the data. The straight orange line is tangent to the blue curve at t = 40 s. A plot has the concen

tration of A in molar on the y axis and time in seconds on the x axis. A curve contains the points (0, 0.52), (20, 0.43), (40, 0.35), (60, 0.29), (80, 0.24), and (100, 0.20). A line touches the curve at (40, 0.35) and has a y intercept of (0, 0.48). Approximate the instantaneous rate of this reaction at time t = 40 s.

Physics
2 answers:
victus00 [196]3 years ago
7 0

Answer:

.00325

Explanation:

YOu have to look at the red graph

Sever21 [200]3 years ago
5 0

Answer:

  0.00325 moles/liter/second

Explanation:

The tangent line has a slope of (y2 -y1)/(x2 -x1) = (0.35-0.48)/(40-0) = -0.00325.

The rate of the reaction is about 0.00325 moles/liter/second.

_____

This is the rate of decrease of the concentration of A.

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Answer:

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Explanation:

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Work output = 240 N

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To find the mechanical efficiency of a machine;

Efficiency = \frac {Out-put \; work}{In-put \; work} * 100

Substituting into the equation, we have;

Efficiency = \frac {240}{300} * 100

Efficiency = 0.8 * 100

Efficiency = 80%

Therefore, the mechanical efficiency of the machine is 80 percent.

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3 years ago
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When gravity increases what happens to weight?
Fynjy0 [20]
Weight increases but mass stays the same
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A car with a mass of 1. 2 × 103 kilograms starts from rest and attains a speed of 20 meters/second in 5 seconds. What net force
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Answer:

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Explanation:

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2 years ago
A projectile is thrown with velocity v at an angle θ with horizontal. When the projectile is at a height equal to half of the ma
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  • Let, the maximum height covered by projectile be \sf{H_m}

\purple{ \longrightarrow  \bf{h_m =  \dfrac{ {v}^{2} \: {sin}^{2} \theta  }{2g} }}

  • Projectile is thrown with a velocity = v
  • Angle of projection = θ

  • Velocity of projectile at a height half of the maximum height covered be \sf{v_0}

\qquad______________________________

Then –

\qquad \pink{  \longrightarrow \bf{ \dfrac{h_m}{2}  = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta  }{2g} }}

\qquad \longrightarrow \sf{ \dfrac{ {v}^{2}  \: {sin}^{2} \theta  }{2g} \times  \dfrac{1}{2}  =  \dfrac{ {v_0}^{2} \: {sin}^{2} \theta  }{2g} }

\qquad\longrightarrow  \sf{ \dfrac{ {v}^{2}  \: {sin}^{2} \theta  }{4g}  =  \dfrac{ {v_0}^{2} \: {sin}^{2} \theta  }{2g} }

\qquad\longrightarrow  \sf{ \dfrac{ {v}^{2}  \: {sin}^{2} \theta  }{2}  =   {v_0}^{2} \: {sin}^{2} \theta }

\qquad\longrightarrow  \sf{ \dfrac{ {v}^{2} }{2}  =   {v_0}^{2} }

\qquad\longrightarrow \bf{v_0 =   \sqrt{ \dfrac{ {v}^{2} }{2} } =  \dfrac{v}{ \sqrt{2} }  }

  • Now, the vertical component of velocity of projectile at the height half of \sf{h_m} will be –

\qquad \longrightarrow   \bf{v_{(y)}=v_0 \: sin \theta }

\qquad \longrightarrow \bf{v_{(y)} = \dfrac{v}{ \sqrt{2} }  \: sin \theta =  \dfrac{v \: sin \: \theta}{ \sqrt{2} }  }

Therefore, the vertical component of velocity of projectile at this height will be–

☀️\qquad\pink {\bf{ \dfrac{v \: sin \:  \theta}{ \sqrt{2} }} }

6 0
3 years ago
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