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erica [24]
3 years ago
6

Julie drives 100 mi to Grandmother's house. On the way to Grandmother's, Julie drives half the distance at 30.0 mph and half the

distance at 70.0 mph . On her return trip, she drives half the time at 30.0 mph and half the time at 70.0 mph . What is Julie's average speed on the way to Grandmother's house?
What is her average speed on the return trip?
Physics
1 answer:
salantis [7]3 years ago
6 0

Answer:

Explanation:

Given

Distance to grandmother's house=100 mi

it is given that during return trip Julie spend equal time driving with speed 30 mph and 70 mph

Let Julie travel x mi with 30 mph and 100-x with 70 mph

\frac{x}{30}=\frac{100-x}{70}

x=30 mi

Therefore

Julie's Average speed on the way to Grandmother's house=\frac{100}{\frac{50}{30}+\frac{50}{70}}

=42 mph

On return trip

=\frac{100}{2\frac{30}{30}}=50 mph

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Answer:

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the linear momentum is calculated by the next equation

P = MV

where M is the mass and V is the velocity.

so:

P_i = m(270 m/s)

P_f = mV_P + M_nV_n

where m is the mass of the proton and V_p is the velocity of the proton after the collision, M_n is the mass of the nucleus and V_n is the velocity of the nucleus after the collision.

therefore, we can formulate the following equation:

m(270 m/s) = mV_p + 14mV_n

then, m is cancelated and we have:

270 = V_p + 14V_n

This is a elastic collision, so the kinetic energy K is conservated. Then:

K_i = \frac{1}{2}MV^2 = \frac{1}{2}m(270)^2

and

Kf = \frac{1}{2}mV_p^2 +\frac{1}{2}(14m)V_n^2

then,

\frac{1}{2}m(270)^2 =  \frac{1}{2}mV_p^2 +\frac{1}{2}(14m)V_n^2

here we can cancel the m and get:

\frac{1}{2}(270)^2 =  \frac{1}{2}V_p^2 +\frac{1}{2}(14)V_n^2

now, we have two equations and two incognites:

270 = V_p + 14V_n  (eq. 1)

\frac{1}{2}(270)^2 =  \frac{1}{2}V_p^2 +\frac{1}{2}(14)V_n^2

in the second equation, we have:

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from this last equation we solve for V_n as:

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and replace in the other equation as:

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V_n = \frac{270+267.258}{14}

V_n = 38.375 m/s  

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