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Gelneren [198K]
3 years ago
5

A segull flying horizontally at 15m/s drops a clam. The clam takes 3.0sec to hit the ground. How high was the seagull when the c

lam was dropped ?
Need formula and answer
Physics
1 answer:
lora16 [44]3 years ago
7 0

The distance a dropped object falls, with gravity and no air resistance:

Distance = (1/2) (acceleration) (falling time)²

Without air resistance, the horizontal motion has no effect on the fall.

Acceleration of Earth gravity = 9.8 m/s²

Distance = (1/2) (acceleration) (falling time)²

Distance = (1/2) (9.81 m/s²) (3.0 s)²

Distance = (0.5) x (9.81 m/s²) x (9.0 s²)

Distance = (0.5 x 9.81 x 9.0) (m-s² / s²)

Distance = 44.15 meters

We don't care how fast the bird was flying horizontally.  It doesn't change anything.  (It DOES determine how far ahead of the drop point the clam hits the ground.  Most problems like this ask for that distance.  This one didn't.)

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A skateboarder rolls off a 2.5 m high bridge into the river. If the skateboarder was originally moving at 7.0 m/s, how much time
saul85 [17]

Answer:

  t = 0.714 s and  x = 5.0 m

Explanation:

This is a projectile throwing exercise, in this case when the skater leaves the bridge he goes with horizontal speed

         vₓ = 7.0 m / s

Let's find the time it takes to get to the river

         y = y₀ + v_{oy} t - ½ g t²

the initial vertical speed is zero and when it reaches the river its height is zero

        0 = y₀ + 0 - ½ g t²

        t = \sqrt{\frac{2y_o}{g}  }

        t = ra 2 2.5 / 9.8

        t = 0.714 s

the distance traveled is

       x = vₓ t

       x = 7.0 0.714

       x = 5.0 m

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Katyanochek1 [597]

Answer:

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6 0
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dexar [7]

Answer:

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Explanation:

6 0
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At a certain location, wind is blowing steadily at 9 m/s. Determine the mechanical energy of air per unit mass and the power gen
Misha Larkins [42]

Answer:

  1. The specific mechanical energy of the air in the specific location is 40.5 J/kg.
  2. The power generation potential of the wind turbine at such place is of 2290 kW
  3. The actual electric power generation is 687 kW

Explanation:

  1. The mechanical energy of the air per unit mass is the specific kinetic energy of the air that is calculated using: \frac{1}{2} V^2 where V is the velocity of the air.
  2. The specific kinetic energy would be: \frac{1}{2}(9\frac{m}{s})^2=40.5\frac{m^2}{s^2}=40.5\frac{m^2 }{s^2}\frac{kg}{kg}=40.5\frac{N*m }{kg}=40.5\frac{J}{kg}.
  3. The power generation of the wind turbine would be obtained from the product of the mechanical energy of the air times the mass flow that moves the turbine.
  4. To calculate mass flow it is required first to calculate the volumetric flow. To calculate the volumetric flow the next expression would be: \frac{V\pi D_{blade}^2}{4} =\frac{9\frac{m}{s}\pi(80m)^2}{4} =45238.9\frac{m^3}{s}
  5. Then the mass flow is obtain from the volumetric flow times the density of the air: m_{flow}=1.25\frac{kg}{m^3}45238.9\frac{m^3}{s}=56548.7\frac{kg}{s}
  6. Then, the Power generation potential is: 40.5\frac{J}{kg} 56548.7\frac{kg}{s} =2290221W=2290.2kW
  7. The actual electric power generation is calculated using the definition of efficiency:\eta=\frac{E_P}{E_I}}, where η is the efficiency, E_P is the energy actually produced and, E_I is the energy input. Then solving for the energy produced: E_P=\eta*E_I=0.30*2290kW=687kW
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3 years ago
Question 2 of 10
tangare [24]

Explanation:

everything can be found in the picture

8 0
3 years ago
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