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diamong [38]
3 years ago
13

"Angel Falls in southeastern Venezuela is the highest uninterrupted waterfall in the world, dropping 979 m (3212 ft). Ignoring a

ir resistance, it would take 14 s for the water to fall from the lip of the falls to the river below. If the water lands 50 m from the base of the verti-cal cliff, what was its horizontal speed at the top"

Physics
1 answer:
Furkat [3]3 years ago
4 0

Complete Question

The complete question is shown on the first uploaded image

Answer:

The value  is v  = 3.57 \  m/s

Explanation:

From the question we are told that

  The depth  is  d =  979 \  m

   The time taken for water to fall to below the river is  t =  14 \ s

   The  horizontal distance from the base of the vertical cliff is  h  =  50 \  m

     Generally  the horizontal speed at the top is mathematically represented as

         v  =  \frac{h}{t}

=>      v  =  \frac{  50}{ 14 }

=>     v  = 3.57 \  m/s

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a pool ball leaves a 0.60-meter high table with an initial high table with an initial horizontal velocity of 2.4m/s. what is the
inysia [295]

Answer:

0.84 m

Explanation:

Given in the y direction:

Δy = 0.60 m

v₀ = 0 m/s

a = 9.8 m/s²

Find: t

Δy = v₀ t + ½ at²

0.60 m = (0 m/s) t + ½ (9.8 m/s²) t²

t = 0.35 s

Given in the x direction:

v₀ = 2.4 m/s

a = 0 m/s²

t = 0.35 s

Find: Δx

Δx = v₀ t + ½ at²

Δx = (2.4 m/s) (0.35 s) + ½ (0 m/s²) (0.35 s)²

Δx = 0.84 m

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3 years ago
What is Nuclear energy
Yuliya22 [10]

Answer:

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3 years ago
The pendulum consists of two slender rods AB and OC which have a mass of 3 kg/m. The thin plate has a mass of 12 kg/m2 . a) Dete
jeka57 [31]

Answer:

The answer is below

Explanation:

a) The location ӯ of the center of mass G of the pendulum is given as:

y=\frac{0+(\pi*(0.3\ m) ^2*12kg/m^2*1.8\ m-\pi*(0.1\ m) ^2*12kg/m^2*1.8\ m)+0.75\ m*1.5\ m *3\ kg/m}{(\pi*(0.3\ m) ^2*12kg/m^2-\pi*(0.1\ m) ^2*12kg/m^2)+3\ kg/m^2*0.8\ m+3\ kg/m^2*1.5\ m} \\\\y=0.88\ m

b)  the mass moment of inertia about z axis passing the rotation center O is:

I_G=\frac{1}{12}*3(0.8)(0.8)^2+ 3(0.8)(0.888)^2-\frac{1}{2}*(12)(\pi)(0.1)^2(0.1)^2 -(12)(\pi)(0.1)^2(1.8-\\0.888)^2+\frac{1}{2}*(12)(\pi)(0.3)^2(0.3)^2 +(12)(\pi)(0.3)^2(1.8-0.888)^2+\frac{1}{12}*3(1.5)(1.5)^2+\\3(1.5)(0.888-0.75)^2\\\\I_G=13.4\ kgm^2

c) The mass moment of inertia about z axis passing the rotation center O is:

I_o=\frac{1}{12}*3(0.8)(0.8)^2+ \frac{1}{3}* 3(1.5)(1.5)^2+\frac{1}{2}*(12)(\pi)(0.3)^2(0.3)^2 +(12)(\pi)(0.3)^2(1.8)^2-\\\frac{1}{2}*(12)(\pi)(0.1)^2(0.1)^2 -(12)(\pi)(0.1)^2(1.8)^2\\\\I_o=13.4\ kgm^2

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I believe the answer is C - 78.4 m/s.
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The answer is 100 my nilla. 
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