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Olenka [21]
2 years ago
7

tourist travels 1500 miles using two planes. The second plane averages 50 miles per hour faster than the first plane. The touris

t uses the slower plane for the first 500 and the faster plane for the next 1000 miles. The total flying time is 6.5 hours. What is the speed of the first plane?
Physics
1 answer:
melamori03 [73]2 years ago
5 0

Answer:

V₁ = 200 mi/h

Explanation:

To do this, we need to analyze the given data.

We have a total distance and time of 1500 mi and 6.5 h. We also know that the second plane goes 50 mi/h faster than the first plane, (V₂ = V₁ + 50). The firsts 500 miles were on the first plane and the remaining 1000 mi with the second plane.

Using the expression of distance, and assuming a constant speed in both planes we have:

d = V * t   (1)

Replacing this expression for both planes we have:

d₁ = V₁ * t₁

d₂ = (V₁+50) * t₂

WE are missing several data, because we do not know the times for each plane, but we can have a clue.

The total time of flight was 6.5 h, this is equivalent to just sum both times of both planes, so:

t = t₁ + t₂   (3)

From the expressions of distance for both planes, we can solve for time, and then replace in the above expression of the total time. In that way we can solve for the speed of the plane. Let's solve for both times in the planes:

t₁ = d₁ / V₁ = 500/V₁    (4)

t₂ = d₂ / V₁+50 = 1000/ V₁ + 50   (5)

Now, all we have to do is replace (4) and (5) into (3) and solve for V₁:

6.5 = [(500/V₁) + (1000/V₁ + 50)

6.5 = [500*(V₁+50) + 1000V₁ / V₁(V₁+50)]

6.5 = [500V₁ + 25000 + 1000V₁ / V₁² + 50V₁

6.5(V₁² + 50V₁) = 1500V₁ + 25000

6.5V₁² + 325V₁ - 1500V₁ - 25000 = 0

6.5V₁² - 1175V₁ - 25000 = 0

From here, we use the quadratic equation to solve for V₁:

X = 1175 ± √(1175)² + 4 * 6.5 * 25000 / 2 * 6.5

X = 1175 ± √1380625 + 650000 / 13

X = 1175 ± 1425 / 13

X₁ = 1175 + 1425 / 13 = 200

X₂ = 1175 - 1425 / 13 = -19.23

So, it's very clear that the speed cannot be negative, so we stick with the first value. Therefore the speed pf the first plane is:

<h2>V₁ = 200 mi/h</h2><h2>V₂ = 250 mi/h</h2>

Hope this helps

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L=(0.35)(9.0)(3.0)=9.5 kg m^2/s

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Section 2.2
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A child is riding in a child-restraint chair, securely fastened to the seat of a car. Assume the car has speed 47 km/h when it h
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Answer: F = 1235 N

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Acceleration is the change of velocity in a period of time: a=\frac{\Delta v}{\Delta t}

Velocity of the car is in km/h. Transforming it in m/s:

v=\frac{47.10^{3}}{36.10^{2}}

v = 13 m/s

At the moment the car decelerates, acceleration is

a=\frac{13}{0.2}

a = 65 m/s²

Then, force will be

F_{net}=19(65)

F_{net} = 1235 N

The horizontal net force the straps of the restraint chair exerted on the child to hold her is 1235 newtons.

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2 years ago
A proton that has a mass m and is moving at 270 m/s in the i hat direction undergoes a head-on elastic collision with a stationa
Nataly_w [17]

Answer:

V_p = 267.258 m/s

V_n = 38.375 m/s      

Explanation:

using the law of the conservation of the linear momentum:

P_i = P_f

where P_i is the inicial momemtum and P_f is the final momentum

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:

36450 =  \frac{1}{2}V_p^2 +\frac{1}{2}(14)V_n^2  (eq. 2)

from this last equation we solve for V_n as:

V_n = \sqrt{\frac{36450-\frac{1}{2}V_p^2 }{\frac{1}{2} } }

and replace in the other equation as:

270 = V_p + 14\sqrt{\frac{36450-\frac{1}{2}V_p^2 }{\frac{1}{2} } }

so,

V_p = -267.258 m/s

Vp is negative because the proton go in the -i hat direction.

Finally, replacing this value on eq. 1 we get:

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

V_n = 38.375 m/s  

3 0
3 years ago
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