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tamaranim1 [39]
2 years ago
6

An ice rink is 200 feet long and 85 feet wide. The four corners of the rink are rounded to a radius of 24 feet. What is the surf

ace area of the rink? Round the answer to the nearest square foot.
Physics
1 answer:
Gennadij [26K]2 years ago
4 0

The surface area of the ice rink is equal to 15190.44ft^{2}

Why?

To calculate the area of the ice rink, we need to remember that the surface area of the four coners (with radius of 24 feet) is equal to the area of a single circle with a radius of 24 feet.

So, calculating the area we have:

SurfaceArea=TotalArea-CircleArea\\\\SurfaceArea=200ft*85ft-\pi *radius^{2}\\\\SurfaceArea=17000ft^{2}-\pi *(24ft)^{2}=17000ft^{2}-1809.56ft^{2}=15190.44ft^{2}

Hence, we have that the surface area of the ice rink is equal to 15190.44ft^{2}

Have a nice day!

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natta225 [31]

Answer:

799.54 ft

Explanation:

Linear thermal expansion is:

ΔL = α L₀ ΔT

where ΔL is the change in length,

α is the linear thermal expansion coefficient,

L₀ is the original length,

and ΔT is the change in temperature.

Given:

α = 1.2×10⁻⁵ / °C

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ΔT = -17°C − 31°C = -48°C

Find: ΔL

ΔL = (1.2×10⁻⁵ / °C) (800 ft) (-48°C)

ΔL = -0.4608

Rounded to two significant figures, the change in length is -0.46 ft.

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Answer:

After 1 sec = 4.9 m

After 2 sec = 19.6 m

After 3 sec = 44.1 m

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After 5 sec = 122.5 m

Explanation:

After 1 sec:

<em>u=0m/s   t=1 s  a=9.8m/s²</em>

s = ut + (1/2)at²

=0(1) + (1/2)(9.8)(1²) = 4.9m

After 2 sec:

<em>u=0m/s   t=2 s  a=9.8m/s²</em>

s = ut + (1/2)at²

=0(2) + (1/2)(9.8)(2²) = 19.6m

After 3 sec:

<em>u=0m/s   t=3 s  a=9.8m/s²</em>

s = ut + (1/2)at²

=0(3) + (1/2)(9.8)(3²) = 44.1m

After 4 sec:

<em>u=0m/s   t=4 s  a=9.8m/s²</em>

s = ut + (1/2)at²

=0(4) + (1/2)(9.8)(4²) = 78.4m

After 5 sec:

<em>u=0m/s   t=5 s  a=9.8m/s²</em>

s = ut + (1/2)at²

=0(5) + (1/2)(9.8)(5²) = 122.5m

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Explanation:

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