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Luden [163]
1 year ago
12

You drive 5 miles at 30 mi/hr and then another 3 miles at 50 mi/hr. What is your average speed for the whole8-mile trip?

Physics
1 answer:
s2008m [1.1K]1 year ago
6 0

Given data

The initial distance covered is d1 = 5 miles

The initial speed is s1 = 30 mi/hr

The final distance is d2 = 3 miles

The final speed is s2 = 50 mi/hr

The expression for the initial time taken to travel is given as:

t_1=\frac{d_1}{s_1}

The expression for the final time taken is given as:

t_2=\frac{d_2}{s_2}

The expression for the average speed for the whole trip is given as:

\begin{gathered} \text{Average sp}eed\text{ =}\frac{total\text{ distance}}{total\text{ time}} \\ S_{Avg}=\frac{d_1+d_2}{t_1+t_2} \\ S_{Avg}=\frac{d_1+d_2}{\frac{d_1}{s_1}+\frac{d_2}{s_2}_{}} \end{gathered}

Substitute the value in the above equation.

\begin{gathered} S_{Avg}=\frac{5\text{ mi+3 mi}}{\frac{5\text{ mi}}{30\text{ mi/hr}}+\frac{3\text{ mi}}{50\text{ mi/hr}}} \\ S_{Avg}=35.3\text{ mi/hr} \end{gathered}

Thus, the average speed for the whole trip is 35.3 mi/hr.

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Answer:

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Explanation:

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FTorque(3/4L)= FBcostheta(1/2L)+ FWcostheta(L)

Ftorque=(Fcostheta(1/2L)+FWcosL)/(3/4L)

Ftorque= 2/3FBcostheta+ 4/3FWcostheta

Ftorque=2/3(1350)(9.81)cos55° + 2/3(2250)(9.81)cos 55°

Ftorque= 21900N

b) components of Normal force

Efx=FNx-FTcos(90-theta)=0 static equilibrium

Fnx=21900cos(90-55)=17900N

Fy=FNy+ FTsin(90-theta)-FB-FW=0

FNy= -FTsin(90-55)+FB+FW

FNy= -21900sin(35)+(1350+2250)×9.81=22700N

The Normal force

FN=sqrt(17900^2+22700^2)

FN= 28.900N

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Answer: The correct option is (c.).

Explanation:

Mass of the cart A= 1.5 kg

Velocity of Cart A = 1.4 m/s towards right

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Velocity of Cart B = 1.4 m/s towards left

Momentum (P)= Mass × Velocity

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