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oksano4ka [1.4K]
1 year ago
14

A cheetah is known to be the fastest mammal on Earth, at least for short runs. Cheetahs have been observed running a distance of

5.50 × 102 m with an average speed of 1.00 × 102 km/h.
How long would it take a cheetah to cover this distance at this speed?
Physics
1 answer:
Arlecino [84]1 year ago
4 0

The cheetah needs 197.8 seconds to complete the 5.50 *100 m distance.

The distance traveled divided by the time taken is the most popular formula for calculating average speed. The average speed is determined by dividing the whole distance by the total time required to complete the distance. The other calculation, provided you have the initial and final speeds, adds them together and divides by 2.20. calculating the time needed for a cheetah to travel 5.50 102 meters.

10 km/h average speed is equal to 10 / 3.6, or 2.78 m/s.

Dist. = 5.50 102

Time =?

average speed = time / distance

time=5.50 x 100/ 2.78

Time = 197.8 s

Therefore, it takes a cheetah 197.8 seconds to travel 5.50 102 meters.

Learn more about average speed here-

brainly.com/question/12322912

#SPJ9

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d = 1.55 * 10⁻⁶ m

Explanation:

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The wavelength of the green laser pointer, \lambda = 532 nm = 532 * 10⁻⁹ m

Distance between the central maximum and the first order maximum = 1.1 m

Distance between the screen and the CD = 3 m

A_{m} = Angle between the incident light and the diffracted light

From the setup shown in the attachment, it is a right angled triangle in which

sin(A_{m}) = \frac{opposite}{Hypotenuse} \\sin(A_{m}) =\frac{1.1}{\sqrt{1.1^{2}+3^{2}}}

sin(A_{m} ) = 0.344\\A_{m} = sin^{-1} 0.344\\A_{m} = 20.14^{0}

Putting all appropriate values into equation (1)

d = \frac{1* 532*10^{-9} }{0.344 }\\d = 0.00000155 m\\d = 1.55 * 10^{-6} m

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The initial height of the water in a sealed container of diameter 100.0 cm is 5.00 m. The air pressure inside the container is 0
katen-ka-za [31]

Answer:

a)  F = 2.66 10⁴ N, b)   h = 1.55 m

Explanation:

For this fluid exercise we use that the pressure at the tap point is

Exterior

          P₂ = P₀ = 1.01 105 Pa

inside

         P₁ = P₀ + ρ g h

the liquid is water with a density of ρ=1000 km / m³

         P₁ = 0.85   1.01 10⁵ + 1000   9.8  5

         P₁ = 85850 + 49000

         P₁ = 1.3485 10⁵ Pa

the net force is

         ΔP = P₁- P₂

         Δp = 1.3485 10⁵ - 1.01 10⁵

         ΔP = 3.385 10⁴ Pa

Let's use the definition of pressure

         P = Fe / A

         F = P A

the area of ​​a circle is

         A = pi r² = [i d ^ 2/4

let's reduce the units to the SI system

         d = 100 cm (1 m / 100 cm) = 1 m

         F = 3.385 104 pi / 4 (1) ²

         F = 2.66 10⁴ N

b) the height for which the pressures are in equilibrium is

        P₁ = P₂

        0.85 P₀ + ρ g h = P₀

        h = \frac{P_o ( 1-0.850)}{\rho \ g}

        h = \frac{1.01 \ 10^5 ( 1 -0.85)}{1000 \ 9.8}

        h = 1.55 m

4 0
3 years ago
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