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Vanyuwa [196]
3 years ago
13

1.a bag is dropped from a hovering helicopter. the bag has fallen for 2 s. what is the ball's velocity at the instant its hittin

g the ground? what was the height of the helicopter? if the bag was dropped from a helicopter descending at the speed of 2 m/s what is the time of the bag fall and its velocity near the ground?
Physics
1 answer:
omeli [17]3 years ago
4 0

1. The bag's velocity immediately before hitting the ground.

Recall this kinematics equation:

Vf = Vi + aΔt

Vf is the final velocity, Vi is the initial velocity, a is the acceleration, and Δt is the time elapsed.

Given values:

Vi = 0m/s (you assume this because the bag is dropped, so it falls starting from rest)

a is 9.81m/s² (this is the near-constant acceleration of objects near the surface of the earth)

Δt = 2s

Plug in the values and solve for Vf:

Vf = 0 + 9.81×2

Vf = 19.62m/s

2. The height of the helicopter.

Recall this other kinematics equation:

d = ViΔt + 0.5aΔt²

d is the distance traveled by the object, Vi is the initial velocity, a is the acceleration, and Δt is the time elapsed.

Given values:

Vi = 0m/s (bag is dropped starting from rest)

a = 9.81m/s² (acceleration due to gravity of the earth)

Δt = 2s

Plug in the values and solve for d:

d = 0×2 + 0.5×9.81×2²

d = 19.62m

3. Time of the bag's fall and its velocity immediately before hitting the ground... if it started falling at 2m/s

Reuse the equation from question 2:

d = ViΔt + 0.5aΔt²

Given values:

d = 19.6m (height of the helicopter obtained from question 2)

Vi = 2m/s

a = 9.81m/s² (acceleration due to earth's gravity)

Plug in the values and solve for Δt:

19.6 = 2Δt + 0.5×9.81Δt²

4.91Δt² + 2Δt - 19.6 = 0

Use the quadratic formula to get values of Δt (a quick Google search will give you the formula and how to use it to solve for unknown values):

Δt = 1.8s, Δt = −2.2s

The formula gives us 2 possible answers for Δt but within the situation of our problem, only the positive value makes sense. Reject the negative value.

Δt = 1.8s

Now we can use this new value of Δt to get the velocity before hitting the ground:

Vf = Vi + aΔt

Given values:

Vi = 2m/s

a = 9.81m/s²

Δt = 1.8s (result from previous question)

Plug in the values and solve for Vf:

Vf = 2 + 9.81×1.8

Vf = 19.66m/s

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frosja888 [35]

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Question:

A 1.60 m cylindrical rod of diameter 0.450 cm is connected to a power supply that maintains a constant potential difference of 12.0 V across its ends, while an ammeter measures the current through it. You observe that at room temperature (20.0 ∘C) the ammeter reads 18.4 A , while at 92.0 ∘C it reads 17.4 A . You can ignore any thermal expansion of the rod.

a) Find the resistivity and for the material of the rod at 20° C .

b) Find the temperature coefficient of resistivity at 20° C for the material of the rod.

Given Information:

Room temperature = T₀ = 20° C

Temperature = T = 92° C

Current at 20° C = I₀ = 18.4 A

Current at 92° C = I = 17.4 A

Voltage = V = 12 V

Length = L = 1.60 m

Diameter = d = 0.450 cm = 0.0045 m

Required Information:

Resistivity of the material at 20° C = ρ = ?

Temperature coefficient of resistivity at 20° C = α = ?

Answer:

Resistivity of the material at 20° C = 2.062x10⁻⁶ Ω.m

Temperature coefficient of resistivity at 20° C = 7.986x10⁻⁴ per °C

Explanation:

a) We want to find out the resistivity of the material at 20° C

The resistivity of any material can be found using,

ρ = R₀A/L

Where R₀ is the resistance of the rod at 20° C, A is the area of rod and L is the length of the cylindrical rod.

We also know that area is given by

A = πr²

where r = d/2 = 0.0045/2 = 0.00225 m

A = π(0.00225)²  

A = 5.062⁻⁶ m²

We know that resistance of the material is given by

R₀ = V/I₀

R₀ = 12/18.4

R₀ = 0.6521 Ω

Therefore, the resistivity of the material is

ρ = R₀A/L

ρ = (0.6521*5.062⁻⁶)/1.60

ρ = 2.062x10⁻⁶ Ω.m

b) We want to find out the temperature coefficient of resistivity of the rod at 20° C

The temperature coefficient of resistivity is given by

α = R/R₀ - 1/(T - T₀)

Where R is the resistance of the rod at 90° C

R = V/I

R = 12/17.4

R = 0.6896 Ω

α = R/R₀ - 1/(T - T₀)

α = (0.6896/0.6521) - 1/(92° - 20°)

α = 0.0575/72°

α = 0.000798 per °C

α = 7.986x10⁻⁴ per °C

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