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zavuch27 [327]
3 years ago
7

A freight train leaves the train station 4hrs before a passenger train. The two trains are traveling in the same direction on pa

rallel tracks. If the rate of the passenger train is 40 mph faster than the freight train, how fast is each train traveling if the passenger train passes the freight train in 6hrs?
Physics
1 answer:
enot [183]3 years ago
8 0
It can be said that most of the required information's are already given in the question. 
Let us assume the speed of the passenger train = x
Speed of the freight train = y
x = y + 40
The second equation will be 
6 * x = 10 * y
6x = 10y
Dividing both sides by 2, we get
3x = 5y
x = 5y/3
Putting the value of "x" in the first equation, we get
x = y + 40
5y/3 = y + 40
5y = 3y + 120
5y - 3y = 120
2y = 120
y = 60 mph
Putting the value of y in the first equation, we get
x = y + 40
   = 60 + 40
   = 100 mph
From the above deduction, we can conclude that the passenger train is traveling at 100 mph and the freight train is traveling at 60 mph.
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A cannon, positioned on a hill, shoots a cannonball horizontally at 23 m/s. The cannonball hits the stone wall 1.96 m below the
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Answer: 14. 49 m

Explanation:

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x=V_{o} cos \theta t (1)

y-y_{o}=V_{o} sin \theta t-\frac{1}{2}gt^{2} (2)

Where:

x is the horizontal distance between the cannon and the ball

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t is the time

y=0 is the final height of the cannonball

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g=9.8 m/s^{2} is the acceleration due gravity

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3 years ago
what is the dot product and cross product of of two vectors if the angle is between them is 90 degree?​
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\sf{\pink{\underline{\underline{\blue{GIVEN:-}}}}}

  • The angle between the two vectors is 90° .

\sf{\pink{\underline{\underline{\blue{TO\: FIND:-}}}}}

  1. The dot product of two vectors .
  2. The cross product of two vectors .

\sf{\pink{\underline{\underline{\blue{SOLUTION:-}}}}}

⚡ Let \rm{\vec{a}} and \rm{\vec{b}} are the two vectors .

✍️ We have know that,

\orange\bigstar\:\rm{\pink{\boxed{\green{\vec{a}\:.\:\vec{b}\:=\:ab\cos{\theta}\:}}}}

Where,

  • θ = 90°

\rm{\implies\:\vec{a}\:.\:\vec{b}\:=\:ab\cos{90^{\degree}}\:}

  • cos 90° = <u>0</u>

\rm{\implies\:\vec{a}\:.\:\vec{b}\:=\:ab\times{0}\:}

\rm{\implies\:\vec{a}\:.\:\vec{b}\:=\:0\:}

\rm{\red{\therefore}} [1] The dot product of two vectors is “ <u>0</u> ” .

✍️ We have know that,

\orange\bigstar\:\rm{\pink{\boxed{\green{\vec{a}\:\times\:\vec{b}\:=\:ab\sin{\theta}\:}}}}

Where,

  • θ = 90°

\rm{\implies\:\vec{a}\:\times\:\vec{b}\:=\:ab\sin{90^{\degree}}\:}

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\rm{\implies\:\vec{a}\:\times\:\vec{b}\:=\:ab\times{1}\:}

\rm{\implies\:\vec{a}\:\times\:\vec{b}\:=\:ab\:}

\rm{\red{\therefore}} [2] The cross product of two vectors is “ <u>ab</u> ” .

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