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FromTheMoon [43]

2 years ago

9

An engineer in a locomotive sees a car stuck on the track at a railroad crossing in front of the train. When the engineer first

sees the car, the locomotive is 260 m from the crossing and its speed is 26 m/s. If the engineer's reaction time is 0.51 s, what should be the magnitude of the minimum deceleration to avoid an accident? Answer in units of m/s 2

1 answer:

GarryVolchara [31]2 years ago

3 0

Answer:

The right answer is "1.369 m/s²".

Explanation:

The given values are:

Distance (s)

= 260 m

Initial speed (u)

= 26 m/s

Reaction time (t')

= 0.51 s

During reaction time, the distance travelled by locomotive will be:

⇒ $s'=ut'$

$=26\times 0.51$

$=13.26 \ m$

Remained distance between locomotive and car:

⇒ $x=s-s'$

$=260-13.26$

$=246.74 \ m$

Now,

The final velocity to avoid collection is, V = 0 m/s

From third equation of motion:

⇒ $V^2=u^2+2ax$

On putting the estimated values, we get

⇒ $0=(26)^2+2\times a\times 246.74$

⇒ $0=676+493.48a$

⇒ $493.48a=-676$

⇒ $a=-\frac{676}{493.48}$

⇒ $a=1.369 \ m/s^2$

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Dominik [7]

Answer:

a. All the laboratory equipment given are very basic equipment used in all the laboratories. Name of each equipment is as follows:

V refers to the tape measure.
W refers to dropper.
X refers to an inoculating loop.
Y refers to a stopwatch.
Z refers to the microscope.

b. Use of each laboratory equipment identified is:

Tape measures (V) is used to measure the length of objects or distance in a laboratory.
Dropper (W) is used to measure unit of drop required to dispensed as one drop or several drops in any experiment.
Inoculating loop (X) is used by microbiologists to cultivate microbes on plates and retrieving an inoculum from a culture of microorganisms.
Stopwatch (Y) is used to measure the time of any experiment.
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3 0

2 years ago

A car travels 90 meters north in 15 seconds. Then turns around and travels at 40 meters south in 5.0 seconds. what is the averag

Viefleur [7K]

Average velocity = (displacement) / (total time)

Displacement = distance and direction between
the starting point and the end point

The car travels 90 meters north, then 40 meters south.
So it ends up 50 meters north of where it began.

Displacement = 50 meters north
Total time = (15 + 5) = 20 seconds

Average velocity = (50 meters north) / (20 seconds)

= 2.5 m/s north

4 0

2 years ago

A -0.00325 C charge q1 is placed 5.62 m from a second charge q2. The first charge is repelled with a 48900 N force. What is the

blagie [28]

Answer: q2 = -0.05286

Explanation:

Given that

Charge q1 = - 0.00325C

Electric force F = 48900N

The electric field strength experienced by the charge will be force per unit charge. That is

E = F/q

Substitute F and q into the formula

E = 48900/0.00325

E = 15046153.85 N/C

The value of the repelled second charge will be achieved by using the formula

E = kq/d^2

Where the value of constant

k = 8.99×10^9Nm^2/C^2

d = 5.62m

Substitutes E, d and k into the formula

15046153.85 = 8.99×10^9q/5.62^2

15046153.85 = 284634186.5q

Make q the subject of formula

q2 = 15046153.85/ 28463416.5

q2 = 0.05286

Since they repelled each other, q2 will be negative. Therefore,

q2 = -0.05286

6 0

2 years ago

xConsider the following reduction potentials: Cu2+ + 2e– Cu E° = 0.339 V Pb2+ + 2e– Pb E° = –0.130 V For a galvanic cell employi

slega [8]

Answer:

Approximately $\rm 90\; kJ$ .

Explanation:

Cathode is where reduction takes place and anode is where oxidation takes place. The potential of a electrochemical reaction ( $E^{\circ}(\text{cell})$ ) is equal to

$E^{\circ}(\text{cell}) = E^{\circ}(\text{cathode}) - E^{\circ}(\text{anode})$ .

There are two half-reactions in this question. $\rm Cu^{2+} + 2\,e^{-} \rightleftharpoons Cu$ and $\rm Pb^{2+} + 2\,e^{-} \rightleftharpoons Pb$ . Either could be the cathode (while the other acts as the anode.) However, for the reaction to be spontaneous, the value of $E^{\circ}(\text{cell})$ should be positive.

In this case, $E^{\circ}(\text{cell})$ is positive only if $\rm Cu^{2+} + 2\,e^{-} \rightleftharpoons Cu$ is the reaction takes place at the cathode. The net reaction would be

$\rm Cu^{2+} + Pb \to Cu + Pb^{2+}$ .

Its cell potential would be equal to $0.339 - (-0.130) = \rm 0.469\; V$ .

The maximum amount of electrical energy possible (under standard conditions) is equal to the free energy of this reaction:

$\Delta G^{\circ} = n \cdot F \cdot E^{\circ} (\text{cell})$ ,

where

$n$ is the number moles of electrons transferred for each mole of the reaction. In this case the value of $n$ is $2$ as in the half-reactions.
$F$ is Faraday's Constant (approximately $96485.33212\; \rm C \cdot mol^{-1}$ .)

$\begin{aligned}\Delta G^{\circ} &= n \cdot F \cdot E^{\circ} (\text{cell})\cr &= 2\times 96485.33212 \times (0.339 - (-0.130)) \cr &\approx 9.0 \times 10^{4} \; \rm J \cr &= 90\; \rm kJ\end{aligned}$ .

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