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mario62 [17]
3 years ago
6

A northbound train and a southbound train meet each other on parallel tracks heading in opposite directions. The northbound trai

n travels 16 miles per hour faster than the southbound train. After 1.5 hours, they are 174 miles apart. At what speeds are the two trains traveling?
Physics
2 answers:
SSSSS [86.1K]3 years ago
4 0

Answer:

50mph,60mph

Explanation:

Let speed of southbound train=x mph

Speed of northbound train,y=x+16 mph

Distance between two trains after 1.5 hour,174 miles.

We have to find the speed of two trains.

Distance=speed\times time

Using the formula

174=1.5x+1.5(x+16)

174=1.5(x+x+16)

174=(2x+16)\times 1.5

174=3x+24

3x=174-24=150

x=\frac{150}{3}=50mph

y=x+16=50+16=66mph

marta [7]3 years ago
3 0

Answer

Given,

Let the speed of the slower train be x mph.

speed of faster train be (x+16) mph.

time, t = 1.5 hr

d₁ = 1.5 x

d₂ = (x+16)× 1.5

both train are traveling in opposite direction

d₁  + d₂ = 174

1.5 x +  (x+16)× 1.5 = 174

3 x + 24 = 174

3 x = 150

x = 50

Speed of the slower train is 50 mph

Speed of the faster train is (50+16) = 66 mph.

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Murrr4er [49]

Answer:

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Explanation:

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When the time is vo/g the projectile are in the middle of the range.

<h2>x axis:</h2>

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R=d_x(t=2*\frac{vo}{g})

R=vo*cos(A)*2\frac{vo}{g} \\\\R=\frac{(vo)^{2}*2* sin(A)cos(A)}{g} \\\\R=\frac{(vo)^{2} sin(2A)}{g}

**sin(2A)=2sin(A)cos(A)

<h2>The maximum range occurs when A=45°(because sin(90°)=1)</h2><h2>The actual range R'=(2/√3)R:</h2>

Let B the actual angle of projectile

\frac{vo^{2} }{g} =(\frac{2}{\sqrt{3} }) \frac{vo^{2} *sin(2B)}{g}\\\\1= \frac{2 }{\sqrt{3}} *sin(2B)\\\\sin(2B)=\frac{\sqrt{3}}{2}\\\\

2B=60°

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7 0
3 years ago
A ship leaves the island of Guam and sails a distance 255 km at an angle 49.0 o north of west. Part A: In which direction must i
kiruha [24]

Answer:

Explanation:

We shall represent displacement in vector form .Consider east as x axes and north as Y axes west as - ve x axes and south as - ve Y axes . 255 km can be represented by the following vector

D₁ = - 255 cos 49 i  + 255 sin49 j

= - 167.29 i + 192.45 j

Let D₂ be the further displacement which lands him 125 km east . So the resultant displacement is

D = 125 i

So

D₁ + D₂ = D

- 167.29 i + 192.45 j + D₂ = 125 i

D₂ = 125 i + 167.29 i - 192.45 j

= 292.29 i - 192.45 j

Angle of D₂ with x axes θ

tan θ = -192.45 / 292.29

= - 0.658

θ = 33.33 south of east

Magnitude of D₂

D₂² = ( 192.45)² + ( 292.29)²

D₂ = 350 km approx

Tan

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3 years ago
A ball is released from the top of a tower of height h meters. it takes T seconds to reach the ground . what is the position of
alekssr [168]

Answer:the

8/9 h

Explanation:

Height  =   1/2 a T^2     now change to T/3

 now height = 1/2 a (T/3)^2   =<u> 1/9</u>  1/2 a T^2     <===== it is 1/9 of the way down   or   8/9 h

 

3 0
2 years ago
One gallon of gasoline in an automobiles engine produces on average 9.50 kg of carbon dioxide, which is a greenhouse gas; that i
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The annual production of carbon dioxide is 124121.49×10^{6}[/tex] kg.

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4 0
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Answer:

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\mu _{k} mgd \cos 28.4 +mgd\sin 28.4  =  \frac{1}{2}mv_{i} ^{2}

d(6.60 \times 9.8 \times 0.62 \times 0.879 + 6.60 \times 9.8 \times 0.475) = \frac{1}{2} \times 6.60 \times (1.56)^{2}

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Therefore, the distance travel by block before coming to rest is 0.122 m

7 0
3 years ago
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