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serious [3.7K]
2 years ago
15

Trained dolphins are capable of a vertical leap of 7.0m straight up from the surface of the water-an impressive feat.Suppose you

could train a dolphin to launch itself out of the water at this same speed but at an angle.What maximum horizontal range could the dolphin achieve?
Physics
1 answer:
anzhelika [568]2 years ago
5 0

Answer:

   θ = 45º

Explanation:

To find the solution, let's use the projectile launch equation

    R = vo² sin 2θ / g

Where vo is the initial speed of the dolphin, T is the angle of the jump and g the gravity acceleration.

To obtain a maximum range, the sine T function must be 1,

     sin 2θ = 1

     2θ = sin⁻¹ 1

     2θ = 90

     θ = 45º

Therefore, the dolphin should jump at an angle of 45º from the horizontal

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For this problem, we use the conservation of momentum as a solution. Since momentum is mass times velocity, then,

m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'
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v₁' and v₂' are final velocities of cart A and B, respectively
m₁ and m₂ are masses of cart A and B, respectively

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7 0
2 years ago
pitot tube on an airplane flying at a standard sea level reads 1.07 x 105 N/m2. What is the velocity of the airplane?
Allushta [10]

Answer:

V_infinty=98.772 m/s

Explanation:

complete question is:

The following problem assume an inviscid, incompressible flow. Also, standard sea level density and pressure are 1.23kg/m3(0.002377slug/ft3) and 1.01imes105N/m2(2116lb/ft2), respectively. A Pitot tube on an airplane flying at standard sea level reads 1.07imes105N/m2. What is the velocity of the airplane?

<u>solution:</u>

<u>given:</u>

<em>p_o=1.07*10^5 N/m^2</em>

<em>ρ_infinity=1.23 kg/m^2</em>

<em>p_infinity=1.01*10^5 N/m^2</em>

p_o=p_infinity+(1/2)*(ρ_infinity)*V_infinty^2

V_infinty^2=9756.097

V_infinty=98.772 m/s

8 0
2 years ago
When Lucy saw a shark, a limbic system structure known as the _____ became activated and enabled her to rapidly respond to the t
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Answer:

i think it's the paleomammalian cortex

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Gold and silicon are mutually insoluble in the solid state and form a eutectic system with a eutectic temperature of 636 k and a
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Yupp its c because my dad farted 
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2 years ago
A cubical box, 5.00 cm on each side, is immersed in a fluid. The gauge pressure at the top surface of the box is 594 Pa and the
Zolol [24]

Answer:

The density of the fluid is 1100 kg/m³.

Explanation:

Given that,

Height = 5.00 cm

Pressure at top =594 Pa

Pressure at bottom = 1133 Pa

We need to calculate the change in pressure

Using formula of change in pressure

\Delta P=P_{b}-P_{t}

Where, P_{b} = Pressure at bottom

P_{t} = Pressure at top

put the value into the formula

\Delta P=1133-594

\Delta P=539\ Pa

Using formula of pressure for density

\Delta P = \rho g h

\rho =\dfrac{\Delta P}{gh}

Where, \rho = density

P = pressure

h = height

Put the value in to the formula

\rho =\dfrac{539}{5.00\times10^{-2}\times9.8}

\rho =1100\ kg/m^3

Hence, The density of the fluid is 1100 kg/m³.

4 0
2 years ago
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