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MAXImum [283]
3 years ago
15

A passenger train takes 2 hours less than a freight train takes to travel from norman to enid. the passenger train averages 96 k

m/hr, while the freight train averages 64 km/hr. what is the distance from norman to enid?
Physics
1 answer:
valentina_108 [34]3 years ago
6 0

Passenger

r = 96 km/hr

t = t - 2

Freight

r = 64

t = t    You have to solve for t before you can solve for d

The distance is the same for both trains.

Formula

r_passenger_train * time_passenger = r_freight_train * time_freight

Solve

96 km/hr * (t - 2) = 64*t         the easiest way to go on is to divide by 96

(t - 2) = 64t/96

t - 2 = 0.667t                         Subtract 0.667t from both sides.

t - 0.667t - 2 = 0                   Add 2 to both sides. combine the two times

0.333t = 2                             Divide by  0.333

t = 2/0.333 = 6 hours

Answer

Now solve for the distance.

d_ passenger = r_passenger * t_passenger

d_passenger = ???

r_ passenger = 96

t = 6 - 2 = 4 hours.

d = 96 * 4

d = 384 km.

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On a cold winter day, a penny (mass 2.50 g) and a nickel (mass 5.00 g) are lying on the smooth (frictionless) surface of a froze
sp2606 [1]

Answer:

0.78333 m/s in the opposite direction

1.566 m/s in the same direction

Explanation:

m_1 = Mass of penny = 0.0025 kg

m_2 = Mass of nickel = 0.005 kg

u_1 = Initial Velocity of penny = 2.35 m/s

u_2 = Initial Velocity of nickel = 0 m/s

v_1 = Final Velocity of penny

v_2 = Final Velocity of nickel

As momentum and Energy is conserved

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}

{\tfrac {1}{2}}m_{1}u_{1}^{2}+{\tfrac {1}{2}}m_{2}u_{2}^{2}={\tfrac {1}{2}}m_{1}v_{1}^{2}+{\tfrac {1}{2}}m_{2}v_{2}^{2}

From the two equations we get

v_{1}=\frac{m_1-m_2}{m_1+m_2}u_{1}+\frac{2m_2}{m_1+m_2}u_2\\\Rightarrow v_1=\frac{0.0025-0.005}{0.0025+0.005}\times 2.35+\frac{2\times 0.5}{0.4005+0.5}\times 0\\\Rightarrow v_1=-0.78333\ m/s

The final velocity of the penny is 0.78333 m/s in the opposite direction

v_{2}=\frac{2m_1}{m_1+m_2}u_{1}+\frac{m_2-m_1}{m_1+m_2}u_2\\\Rightarrow v_2=\frac{2\times 0.0025}{0.0025+0.005}\times 2.35+\frac{0.005-0.0025}{0.005+0.0025}\times 0\\\Rightarrow v_2=1.566\ m/s

The final velocity of the nickel is 1.566 m/s in the same direction

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