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lord [1]
3 years ago
13

the record for a ski jump is 180 m set in 1989. Assume the jumper comes off the end of the ski jump horizontally and falls 90 m

vertically before contacting the ground . what was the initial horizontal speed of the jumper?
Physics
1 answer:
Mashutka [201]3 years ago
8 0
Let's write the equations of motion on both x- (horizontal) and y- (vertical) axis. On the x-axis, it's a uniform motion with constant velocity vx. On the y-axis, it is a uniformly accelerated motion with initial height h=90 m and acceleration of g=9.81 m/s^2 pointing down (so with a negative sign):
S_x(t)=v_x t
S_y(t)=h- \frac{1}{2} gt^2

First, let's find the time at which the jumper reaches the ground. This happens when Sy(t)=0:
0=h- \frac{1}{2} g t^2
and so 
t= \sqrt{ \frac{2h}{g} }= \sqrt{ \frac{2\cdot 90m}{9.81 m/s^2} }=4.28 s

Then, we can find the horizontal speed. In fact, we know that at the time t=4.28 s, when the jumper reached the ground, he covered exactly 180 m, so Sx=180 m. Using this into the law of motion in x, we find
v_x= \frac{S_x}{t} = \frac{180 m}{4.28 s} =42 m/s

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Artemon [7]

It will take the quarter 0.151 seconds to reach the ground.

<u>Given the following data:</u>

  • Height = 1.45 meters
  • Initial velocity = 10.32 m/s

We know that acceleration due to gravity (a) for an object is equal to 9.8 meter per seconds square.

To find how much time it will take the quarter to reach the ground, we  would use the second equation of motion.

Mathematically, the second equation of motion is given by the formula;

S = ut + \frac{1}{2} at^2

Where:

  • S is the height or distance covered.
  • u is the initial velocity.
  • a is the acceleration.
  • t is the time measured in seconds.

Substituting the values into the formula, we have;

1.45 = 10.32(t) + \frac{1}{2} (9.8)t^2\\\\1.45 = 10.32t + 4.9t^2\\\\4.9t^2 + 10.32t - 1.45 = 0

The standard form of a quadratic equation is:

ax^2 + bx + c = 0

a = 4.9, b = 10.32 and c = 1.45

We would solve the above quadratic equation by using the quadratic equation formula;

x = \frac{-b\; \pm \;\sqrt{b^2 - 4ac}}{2a}

Substituting the values, we have;

t = \frac{-10.32\; \pm \;\sqrt{10.32^2\; - \;4(4.9)(1.45)}}{2(4.9)}\\\\t = \frac{-10.32\; \pm \;\sqrt{106.5024\; - \;28.42}}{9.8}\\\\t = \frac{-10.32\; \pm \;\sqrt{78.0824}}{9.8}\\\\t = \frac{-10.32\; \pm \;8.84}{9.8}\\\\t = \frac{-10.32\; + \;8.84}{9.8}\\\\t = \frac{1.48}{9.8}

<em>Time, t = 0.151 seconds.</em>

Therefore, it will take the quarter 0.151 seconds to reach the ground.

Read more: brainly.com/question/8898885

3 0
2 years ago
_____Is defined as the conductors and equipment for delivering electric energy from the serving utility to the wiring system of
insens350 [35]

Answer:

Service.

Explanation:

Service equipment and service conductors runs from the service point to the service disconnecting point. Service provides the delivery of electrical energy from it's production point to the consumer where it is meant to be utilized.

3 0
3 years ago
Steam enters an adiabatic turbine steadily at 7 MPa, 5008C, and 45 m/s, and leaves at 100 kPa and 75 m/s. If the power output of
marusya05 [52]

Answer:

a) \dot m = 6.878\,\frac{kg}{s}, b) T = 104.3^{\textdegree}C, c) \dot S_{gen} = 11.8\,\frac{kW}{K}

Explanation:

a) The turbine is modelled by means of the First Principle of Thermodynamics. Changes in kinetic and potential energy are negligible.

-\dot W_{out} + \dot m \cdot (h_{in}-h_{out}) = 0

The mass flow rate is:

\dot m = \frac{\dot W_{out}}{h_{in}-h_{out}}

According to property water tables, specific enthalpies and entropies are:

State 1 - Superheated steam

P = 7000\,kPa

T = 500^{\textdegree}C

h = 3411.4\,\frac{kJ}{kg}

s = 6.8000\,\frac{kJ}{kg\cdot K}

State 2s - Liquid-Vapor Mixture

P = 100\,kPa

h = 2467.32\,\frac{kJ}{kg}

s = 6.8000\,\frac{kJ}{kg\cdot K}

x = 0.908

The isentropic efficiency is given by the following expression:

\eta_{s} = \frac{h_{1}-h_{2}}{h_{1}-h_{2s}}

The real specific enthalpy at outlet is:

h_{2} = h_{1} - \eta_{s}\cdot (h_{1}-h_{2s})

h_{2} = 3411.4\,\frac{kJ}{kg} - 0.77\cdot (3411.4\,\frac{kJ}{kg} - 2467.32\,\frac{kJ}{kg} )

h_{2} = 2684.46\,\frac{kJ}{kg}

State 2 - Superheated Vapor

P = 100\,kPa

T = 104.3^{\textdegree}C

h = 2684.46\,\frac{kJ}{kg}

s = 7.3829\,\frac{kJ}{kg\cdot K}

The mass flow rate is:

\dot m = \frac{5000\,kW}{3411.4\,\frac{kJ}{kg} -2684.46\,\frac{kJ}{kg}}

\dot m = 6.878\,\frac{kg}{s}

b) The temperature at the turbine exit is:

T = 104.3^{\textdegree}C

c) The rate of entropy generation is determined by means of the Second Law of Thermodynamics:

\dot m \cdot (s_{in}-s_{out}) + \dot S_{gen} = 0

\dot S_{gen}=\dot m \cdot (s_{out}-s_{in})

\dot S_{gen} = (6.878\,\frac{kg}{s})\cdot (7.3829\,\frac{kJ}{kg\cdot K} - 6.8000\,\frac{kJ}{kg\cdot K} )

\dot S_{gen} = 11.8\,\frac{kW}{K}

4 0
3 years ago
To form a better telescope, you can interchange eyepieces for new focal lengths. A new design lets you replace both lenses, whic
HACTEHA [7]

Answer:

option (b)

Explanation:

for a telescope, the object is placed very far from the objective lens so the focal length of objective is large and the eye piece is of short focal length.

So, long focal length objective and short focal length eyepiece.

Option (b) is correct.

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A child is thrown horizontally from a 45m tall tower with a velocity of 7m/s.
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The child is thrown for 7 seconds
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