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2 years ago

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Two gliders on an air track collide in a perfectly elastic collision. Glider A has a mass of 1.1 kg and is initially travelling

at a velocity of 2.7 m/s [E]. It collides head-on with glider B with a mass of 2.4 kg, travelling at a velocity of 1.9 m/s [W]. Determine the final velocity of glider A.

1 answer:

Eva8 [605]2 years ago

6 0

m1= mass 1 = 1.1 kg

Vi1 = initial velocity 1 = 2.7 m/s

m2= 2.4 kg

V2i = -1.9 m/s

We assume east as positive and west as negative.

Apply the formulas:

Vf1 = ?

$vf1=(\frac{m1-m2}{m1+m2})Vi1+(\frac{2m2}{m1+m2})Vi2$

Replacing:

$Vf1=\frac{(1.1-2.4)}{(1.1+2.4)}2.7+\frac{(2\times2.4)}{(1.1+2.4)}-1.9$ $Vf1=(\frac{-1.3}{3.5})2.7+(\frac{4.8}{3.5})-1.9$ $Vf1=-1-2.6=-3.6\text{ m/s}$

Answer: 3.6 m/s west

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3 years ago

98 POINTS, 5 simple questions!! HELP

lozanna [386]

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3 years ago

Read 2 more answers

Carry out the following arithmetic operations. (Enter your answers to the correct number of significant figures.) My Notes Ask Y

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Answer :

(a) The answer will be $\Rightarrow 562$

(b) The answer will be $\Rightarrow 2.0$

(c) The answer will be $\Rightarrow 49.32$

Explanation :

Significant figures : The figures in a number which express the value -the magnitude of a quantity to a specific degree of accuracy is known as significant digits.

The rule apply for the multiplication and division is :

The least number of significant figures in any number of the problem determines the number of significant figures in the answer.

The rule apply for the addition and subtraction is :

The least precise number present after the decimal point determines the number of significant figures in the answer.

(a) The sum of the measured values 521, 31.4, 0.85, and 9.0

$521+31.4+0.85+9.0$

$\Rightarrow 562.25$

In the given expression, 521 has 3 significant figures, 31.4 has 3 significant figures, 0.85 has 2 significant figures and 9.0 has 2 significant figures. From this we conclude that least precise number present after the decimal point is 0.

Thus, the answer will be $\Rightarrow 562$

(b) The product 0.0052 × 386.2

$(0.0052)\times (386.2)$

$\Rightarrow 2.00824$

In the given expression, 0.0052 has 2 significant figures and 386.2 has 4 significant figures. From this we conclude that 2 is the least significant figures in this problem. So, the answer should be in 2 significant figures.

Thus, the answer will be $\Rightarrow 2.0$

(c) The product $15.70\times \pi$

The value of $\pi$ = 3.141414 = $3.\bar {14}$

$\pi$ is an irrational number.

$(15.70)\times (3.1414)$

$\Rightarrow 49.31998$

In the given expression, 15.70 has 4 significant figures and $3.\bar {14}$ has infinite significant figures. From this we conclude that 4 is the least significant figures in this problem. So, the answer should be in 4 significant figures.

Thus, the answer will be $\Rightarrow 49.32$

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Answer:

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Similarly, time spent on the second half of the trip:

$\displaystyle t_2 = \frac{s_2}{v_2} = \frac{25}{80} = \frac{5}{16}\; \rm hours$ .

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$\displaystyle \frac{5}{12} + \frac{5}{16} = \frac{35}{48} \; \rm hours$ .

Average speed:

$\begin{aligned} \text{Average speed} &= \frac{\text{Total Distance}}{\text{Total Time}}\\ &= 50 \left/\frac{35}{48}\right.\\ &= 50 \cdot \frac{48}{35} \\&= \frac{480}{7}\approx 68.6\; \rm mph \end{aligned}$ .

This value turned out to be slightly different from the average of the speed during the two halves of the journey. The reason is that the object traveled at each speed for a different amount of time. It spent more time at the slower speed, which gives that speed a greater weight in the average. That explains why the average speed is closer to $\rm 60\; mph$ rather than $\rm 80\; mph$ .

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