1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
harkovskaia [24]
3 years ago
10

Two trains traveling towards one another on a straight track are 300m apart when the engineers on both trains become aware of th

e impending Collision and hit their brakes. The eastbound train, initially moving at 97.0 km/h Slows down at 3.50ms^2. The westbound train, initially moving at 127 km/h slows down at 4.20 m/s^2. What is the distance between them once they stop (express your answer to the two significant figures)
Physics
1 answer:
sattari [20]3 years ago
5 0

Answer:

48 m

Explanation:

Two trains traveling towards one another on a straight track are 300m apart when the engineers on both trains become aware of the impending Collision and hit their brakes. The eastbound train, initially moving at 97.0 km/h Slows down at 3.50ms^2. The westbound train, initially moving at 127 km/h slows down at 4.20 m/s^2.

The eastbound train

First convert km/h to m/s

(97 × 1000)/3600

97000/3600

26.944444 m/s

As the train is decelerating, final velocity V = 0 and acceleration a will be negative. Using third equation of motion

V^2 = U^2 - 2as

O = 26.944^2 - 2 × 3.5 S

726 = 7S

S = 726/7

S1 = 103.7 m

The westbound train

Convert km/h to m/s

(127×1000)/3600

127000/3600

35.2778 m/s

Using third equation of motion

V^2 = U^2 - 2as

0 = 35.2778^2 - 2 × 4.2 × S

1244.52 = 8.4S

S = 1244.52/8.4

S2 = 148.2 m

S1 + S2 = 103.7 + 148.2 = 251.86

The distance between them once they stop will be

300 - 251.86 = 48.14 m

Therefore, the distance between them once they stop is 48 metres approximately.

You might be interested in
Compared to the charge on a proton, the amount of charge on an electron is,
lbvjy [14]
Compared to the charge on a proton, the amount of charge on an electron is same and has the opposite sign
3 0
4 years ago
Two teams of nine members each engage in tug-of-war. Each of the first team's members has an average mass of 68 kg and exerts an
diamong [38]

Answer:

(a) Acceleration  = 0.1063 m/s^2      (Second team wins)

(b) Tension in rope = 65.106 N

Explanation:

Total mass of first team = 68 * 9 = 612 kg

Total force of first team = 1350 * 9 = 12150 N

Total mass of second team = 73 * 9 = 657 kg

Total force of seconds team = 1365 * 9 = 12285 N

Difference in force = 12285 - 12150 = 135 N   (towards the second team as it has more force)

(a) For acceleration we get:

F = m * a

135 = (mass of both teams) * a

a = 135 / (612 + 657)

acceleration  = 0.1063 m/s^2      (Second team wins)

(b) Since we know the acceleration of the first team (pulling being pulled towards the second team at an acceleration of 0.1063 m/s^2) , we can find out the force required to move them:

Force required for first team = mass of first team * acceleration

Force required = 612 * 0.1063

Force required = 65.106 N

This is the force exerted on the first team through the rope, so the tension in the rope will also be 65.106 N.

7 0
4 years ago
1.)how is climate different from weather
Zolol [24]
(sorry I can only answer no. 1)
the climate is the weather over a long period of time (could be days, weeks, months, or years) while weather is only how the day has been on one day (e.g. sunny)
6 0
3 years ago
Three charged particles are positioned in the xy plane: a 50-nC charge at y = 6 m on the y axis, a −80-nC charge at x = −4 m on
disa [49]

Answer:

V = 48 Volts

Explanation:

Since we know that electric potential is a scalar quantity

So here total potential of a point is sum of potential due to each charge

It is given as

V = V_1 + V_2 + V_3

here we have potential due to 50 nC placed at y = 6 m

V_1 = \frac{kQ}{r}

V_1 = \frac{(9\times 10^9)(50 \times 10^{-9})}{\sqrt{6^2 + 8^2}}

V_1 = 45 Volts

Now potential due to -80 nC charge placed at x = -4

V_2 = \frac{kQ}{r}

V_2 = \frac{(9\times 10^9)(-80 \times 10^{-9})}{12}

V_2 = -60 Volts

Now potential due to 70 nC placed at y = -6 m

V_3 = \frac{kQ}{r}

V_3 = \frac{(9\times 10^9)(70 \times 10^{-9})}{\sqrt{6^2 + 8^2}}

V_3 = 63 Volts

Now total potential at this point is given as

V = 45 - 60 + 63 = 48 Volts

3 0
3 years ago
I sure do need help with this one
Neko [114]
Am not sure, but i think its c.

8 0
3 years ago
Other questions:
  • The following table lists the work functions of a few commonmetals, measured in electron volts.
    14·1 answer
  • Which of the following is true about light waves and sound waves? A. Sound waves can move at different speeds, but light waves a
    8·1 answer
  • The particles ejected from the sun during a coronal mass ejection is called
    12·1 answer
  • What kind of quantity is displacement?
    14·1 answer
  • De la 2surse coerente aflate la distanta d=2 metri una de alta se propaga intrun mediu elastic unde cu viteza =340 m/s.la distan
    5·1 answer
  • What #of protons ,electrons ,and neutrons are in zinc
    13·1 answer
  • How many atoms are there
    14·1 answer
  • A frog jumps up at time t = 0.
    15·2 answers
  • The frequency of a wave is 250hz. What is the period
    5·1 answer
  • At what angle should the roadway on a curve with a 45 m radius be banked to allow cars to negotiate the curve at 12 m/s even if
    11·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!