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Alexus [3.1K]
3 years ago
5

Small-plane pilots regularly compete in "message drop" competitions, dropping heavy weights (for which air resistance can be ign

ored) from their low-flying planes and scoring points for having the weights land close to a target. A plane 75m above the ground is flying directly toward a target at 40m/s .
A) At what distance from the target should the pilot drop the weight?
B) The pilot looks down at the weight after she drops it. Where is the plane located at the instant the weight hits the ground?
not yet over the target
past the target
directly over the target
not enough information to determine
Physics
1 answer:
MrMuchimi3 years ago
7 0

Answer:

a. 156.41m

b. directly over the target

Explanation:

a) At the point of release, the weight will have:

Horizontal:

velocity of 40m/s (same as the plane)

Acceleration of zero since it is not under the influence of any force.

Vertical:

initial Velocity - zero at this instant

Acceleration - Towards the ground with g (9.81m/s^2)

from newton's equation of motion, since its falling under gravity, the horizontal component wont be involved

s = ut + (1/2)at^2

u = initial velocity

a = acceleration

s = displacement

t = time taken

75 = 0 + (1/2)(9.81)t^2

t^2=15.29

t=3.91

t = 3.91s

Applying the time of flight to the horizontal components:

Distance = speed x time

= 40 x 3.91

= 156.41m

B) The pilot looks down at the weight after she drops it. Where is the plane located at the instant the weight hits the ground?

not yet over the target

past the target

directly over the target

not enough information to determine

c. directly over the target. will be the correct answer

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Explanation:

As the given spheres are connected by a thin wire so, the potential on the spheres are the same.

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Using the above two equations, the final equation will be as follows.

          q_{2} = \frac{Qr_{2}}{r_{1} + r_{2}}

and,    q_{1} = \frac{Qr_{1}}{r_{1} + r_{2}}

Hence, we will calculate the charge on sphere B after the equilibrium is reached as follows.

          q_{2} = \frac{Qr_{2}}{r_{1} + r_{2}}

                     = \frac{-95.5 \times 74.4 cm}{(11.5 + 74.4) cm}

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At 20 C, a steel rod of length 40.000 cm and a brass rod
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Answer:

a. stress in steel rod is -9.6 X 10⁷pa while stress in brass rod is -7.2 X 10⁷pa

b. new junction from steel rod is 40.0192cm while new junction from brass rod is 30.024cm

Explanation:

<u>Step 1:</u> <u>identify the given parameters and standard parameters</u>

⇒Steel rod: length of rod = 40.000 cm

                    Young modulus(Υ) = 20 X 10¹⁰ pa

                    coefficient of linear expansion(α) = 1.2 X 10⁻⁵ K⁻¹

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 ⇒Brass rod: length of rod = 30.000 cm

                    Young modulus(Υ) = 9 X 10¹⁰ pa

                    coefficient of linear expansion(α) = 2.0 X 10⁻⁵ K⁻¹

                     stress in the rod =?

--------------------------------------------------------------------------------------------

<u>Step 2:</u> <u>calculate the stress in each rod</u>

⇒Steel rod: stress in the rod = -Y*α*ΔT

                                                = (-20 X 10¹⁰ pa) (1.2 X 10⁻⁵ K⁻¹)(60-20)K

                                                = (-20 X 10¹⁰ pa) (1.2 X 10⁻⁵ K⁻¹)(40)K

                                                =  -9.6 X 10⁷ pa

--------------------------------------------------------------------------------------------

⇒Brass rod: stress in the rod = -Y*α*ΔT

                                                = (-9 X 10¹⁰ pa) (2.0 X 10⁻⁵ K⁻¹)(60-20)K

                                                = (-9 X 10¹⁰ pa) (2.0 X 10⁻⁵ K⁻¹)(40)K

                                                =  -7.2 X 10⁷ pa

--------------------------------------------------------------------------------------------

∴ stress in steel rod is -9.6 X 10⁷pa while stress in brass rod is -7.2 X 10⁷pa

--------------------------------------------------------------------------------------------

<u>Step 3:</u> <u>calculate the new length of each rod</u>

⇒Steel rod: New length = ΔL + L₀

                                   ΔL = α*L₀*ΔT

                                  ΔL = (1.2 X 10⁻⁵ K⁻¹)(40cm)(40)K

                                   ΔL = 1920 X 10⁻⁵ cm = 0.0192cm

                    New length = ΔL + L₀ = 0.0192cm + 40cm

                    New length = 40.0192cm

--------------------------------------------------------------------------------------------

⇒Brass rod: New length = ΔL + L₀

                                    ΔL = α*L₀*ΔT

                                   ΔL = (2.0 X 10⁻⁵ K⁻¹)(30cm)(40)K

                                   ΔL = 2400 X 10⁻⁵ cm = 0.024cm

                    New length = ΔL + L₀ = 0.024cm + 30cm

                    New length = 30.024cm

--------------------------------------------------------------------------------------------

Therefore, new junction from steel rod is 40.0192cm while new junction from brass rod is 30.024cm

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