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artcher [175]
3 years ago
10

Two planes are about to drop an empty fuel tank. At the moment of release each plane has the same speed of 195 m/s, and each tan

k is at the same height of around 1.10 km above the ground. Although the speeds are the same, the velocities are different at the instant of release, because one plane is flying at an angle of 15° above the horizontal (Plane A) and the other is flying at an angle of 15° below the horizontal (Plane B).
1. Find the:

a. magnitude.
b. direction of the velocity with which the fuel tank hits the ground if it is from plane A.

2. Find the:

a. magnitude
b. direction of the velocity with which the fuel tank hits the ground if it is from plane B.

In each part, give the direction as a positive angle with respect to the horizontal.
Physics
1 answer:
ratelena [41]3 years ago
5 0

Answer:

376.71

Explanation:

1.a

To find the angle between them we add 15 +15=30°

Since they aren't perpendicular we use cosine law

\alpha  = 180 - 30

\sqrt{ {a}^{2}  +  {b}^{2}  - 2ab \cos( \alpha ) }

\sqrt{ {195}^{2}  +  {195}^{2} - 2 \times 195 \times 195 \times  \cos(150)  }

=376.711

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Having difficulty finding the PE and KE for these values no mass is given. Does anyone know to go solve these?
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11) 1.04\cdot 10^7 J

12) 1.04\cdot 10^7 J

13) 50.0 m/s

14) 41.6 m/s

Explanation:

11)

The potential energy of an object is the energy possessed by the object due to its position relative to the ground. It is given by

PE=mgh

where

m is the mass of the object

g is the acceleration due to gravity

h is the height relative to the ground

Here in this problem, when the train is at the top, we have:

m = 8325 kg (mass of the train + riders)

g=9.8 m/s^2 (acceleration due to gravity)

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PE=(8325)(9.8)(127)=1.04\cdot 10^7 J

12)

According to the law of conservation of energy, the total mechanical energy of the train must be conserved (in absence of friction). So we can write:

KE_t + PE_t = KE_b + PE_b

where

KE_t is the kinetic energy at the top

PE_t is the potential energy at the top

KE_b is the kinetic energy at the bottom

PE_b is the potential energy at the bottom

The kinetic energy is the energy due to motion; since the train is at rest at the top, we have

KE_t=0

Also, at the bottom the height is zero, so the potential energy is zero

PE_b=0

Therefore, we find:

KE_b=PE_t=1.04\cdot 10^7 J

13)

The kinetic energy of an object is the energy of the object due to its motion. Mathematically, it is given by

KE=\frac{1}{2}mv^2

where

m is the mass of the object

v is the speed of the object

From question 12), we know that the kinetic energy of the train at the bottom is

KE=1.04\cdot 10^7 J

We also know that the mass is

m = 8325 kg

Therefore, we can calculate the speed of the train at the bottom:

v=\sqrt{\frac{2KE}{m}}=\sqrt{\frac{2(1.04\cdot 10^7)}{8325}}=50.0 m/s

14)

At the top of the second hill, the total mechanical energy of the train is still conserved.

Therefore, we can write again:

KE_1 + PE_1 = KE_2 + PE_2

where

KE_1 is the kinetic energy at the top of the 1st hill

PE_1 is the potential energy at the top of the 1st hill

KE_2 is the kinetic energy at the top of the 2nd hill

PE_2 is the potential energy at the top of the 2nd hill

From the previous questions, we know that

KE_1=0

and

PE_1=1.04\cdot 10^7 J

The height of the second hill is

h = 39 m

So we can also find the potential energy at the second hill:

PE_2=mgh=(8325)(9.8)(39)=3.2\cdot 10^6 J

So, the kinetic energy at the second hill is

KE_2=PE_1-PE_2=1.04\cdot 10^7 - 3.2\cdot 10^6 =7.2\cdot 10^6 J

And so, the speed is

v=\sqrt{\frac{2KE_2}{m}}=\sqrt{\frac{2(7.2\cdot 10^6)}{8325}}=41.6 m/s

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Answer:

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