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Kryger [21]
3 years ago
10

The coordinates of a bird flying in the xy-plane are given by x(t)=αt and y(t)=3.0m−βt2, where α=2.4m/s and β=1.2m/s2.part a:Cal

culate the velocity vector of the bird as a function of time. Give your answer as a pair of components separated by a comma. For example, if you think the xcomponent is 3 and the y component is 4, then you should enter 3,4.
part b:Calculate the acceleration vector of the bird as a function of time. Give your answer as a pair of components separated by a comma. For example, if you think the x component is 3 and the y component is 4, then you should enter 3,4
Physics
1 answer:
8090 [49]3 years ago
3 0
Α=2.4 \frac{m}{s}

β=1.2 \frac{m}{s^2}

x(t)=at

y(t)=3-βt^2

Vx(t)=α

Vy(t)=-2βt

vectorV=[α;-2βt]

ax(t)=0

ay(t)=-2βt

vector a [0;-2βt]


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Mekhanik [1.2K]

Answer:

  • The difference in length for steel is 2.46 x 10⁻⁴ m
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Explanation:

Given;

original length of steel, L₁ = 1.00 m

original length of invar, L₁ = 1.00 m

coefficients of volume expansion for steel, \gamma_{st.} =  3.6 × 10⁻⁵ /°C

coefficients of volume expansion for invar, \gamma_{in.} =  2.7 × 10⁻⁶ /°C

temperature rise in both meter stick, θ = 20.5°C

Difference in length, can be calculated as:

L₂ = L₁ (1 + αθ)

L₂  = L₁ + L₁αθ

L₂  - L₁ = L₁αθ

ΔL = L₁αθ

Where;

ΔL is difference in length

α is linear expansivity = \frac{\gamma}{3}

Difference in length, for steel at 20.5°C:

ΔL =  L₁αθ

Given;

L₁ = 1.00 m

θ = 20.5°C

\alpha = \frac{\gamma}{3} = \frac{3.6*10^{-5}}{3} = 1.2*10^{-5} /^oC

ΔL  = 1 x 1.2 x 10⁻⁵ x 20.5 = 2.46 x 10⁻⁴ m

Difference in length, for invar at 20.5°C:

ΔL =  L₁αθ

Given;

L₁ = 1.00 m

θ = 20.5°C

\alpha = \frac{\gamma}{3} = \frac{2.7*10^{-6}}{3} = 0.9*10^{-6}/^oC

ΔL  = 1 x 0.9 x 10⁻⁶ x 20.5 = 1.845 x 10⁻⁵ m

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