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JulijaS [17]
2 years ago
7

A soccer field is viewed from above, while a ball is kicked eastward with an initial speed of 10.0 m/s. The ball experiences a c

onstant eastward acceleration of 0.75 m/s 2 over a distance of 20.0 m. Immediately after this 20.0 m distance, the ball is kicked northward in a manner that does not change its speed at that point, just its direction. The ball is allowed to come to a complete stop. How far north does the ball travel before stopping if the acceleration after this kick is 1.15 m/s 2
Physics
1 answer:
satela [25.4K]2 years ago
7 0

Answer:

the ball travelled approximately 60 m towards north before stopping

Explanation:

 Given the data in the question;

First course : a_{x} = 0.75 m/s², d_{x} = 20 m, u_{x} = 10 m/s

now, form the third equation of motion;

v² = u² + 2as

we substitute

v_{x}² = (10)² + (2 × 0.75 × 20)

v_{x}² = 100 + 30

v_{x}² = 130

v_{x} = √130

v_{x} = 11.4 m/s

for the Second Course:

u_{y} =  11.4 m/s,  a_{y} = -1.15 m/s²,  v_{y} = 0

Also, form the third equation of motion;

v² = u² + 2as

we substitute

0² = (11.4)² + (2 × (-1.15) ×  d_{y} )

0 = 129.96 - 2.3d_{y}

2.3d_{y}  = 129.96

d_{y} = 129.96 / 2.3

d_{y} = 56.5 m

so;

|d| = √( d_{x}² + d_{y}² )

we substitute

|d| = √( (20)² + (56.5)² )

|d| = √( 400 + 3192.25 )

|d| = √( 3592.25 )

|d| = 59.9 m ≈ 60 m

Therefore, the ball travelled approximately 60 m towards north before stopping

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Answer:

a.  9.99625 cm b. 68 °C

Explanation:

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Before we find the length of the rod, we need to find the coefficient of linear expansion, α = (L - L₀)/[L₀(T - T₀)] where L₀ = length of rod at temperature T₀ = 10.0 cm, T₀ = 20 °C, L = length of rod at temperature T = 10.015 cm and T = 100 °C

Substituting the values of the variables into the equation, we have

α = (L - L₀)/[L₀(T - T₀)]

α = (10.015 cm - 10.0 cm)/[10.0 cm(100 °C - 20 °C)]

α = 0.015 cm/[10.0 cm × 80 °C]

α = 0.015 cm/[800.0 cm °C]

α = 0.00001875 /°C

We now find the length L₁ at T₁ = 0 °C from

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L₁ = L₀(1 + α(T₁ - T₀))

L₁ = 10.0 cm[1 +  0.00001875 /°C(0° C - 20 °C)]

L₁ = 10.0 cm[1 +  0.00001875 /°C × -20° C]

L₁ = 10.0 cm[1 - 0.000375]

L₁ = 10.0 cm[0.999625]

L₁ = 9.99625 cm

(b) What is the temperature if the length of the rod is 10.009 cm?

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L₃ = L₀(1 + α(T₃ - T₀))

making T₃ subject of the formula, we have

L₃/L₀ = 1 + α(T₃ - T₀)

L₃/L₀ - 1 = α(T₃ - T₀)

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T₃ = T₀ + (L₃/L₀ - 1)/α

substituting the values of the variables into the equation, we have

T₃ = 20 °C + (10.009 cm/10.0 cm - 1)/0.00001875 /°C

T₃ = 20 °C + (1.0009 - 1)/0.00001875 /°C

T₃ = 20 °C + 0.0009/0.00001875 /°C

T₃ = 20 °C + 48 °C

T₃ = 68 °C

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The total distance traveled would be the sum of the distances.

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