1. Find potential of point A

A water skier leaves the end of an 8 foot tall ski ramp with a speed of 20 mi/hr and at an angle of 250. He lets go of the tow r

klemol [59]

Answer:

At highest point:

y1 = 10.4 ft

v1 = (26.5*i + 0*j) ft/s

When he lands:

x2 = 31.5 ft (distance he travels)

t2 = 1.19 s

V2 = (26.5*i - 25.9*j) ft/s

a2 = -44.3°

Explanation:

Since he let go of the tow rope upon leaving the ramp he is in free fall from that moment on. In free fall he is affected only by the acceleration of gravity. Gravity has a vertical component only, so the movement will be at constant acceleration in the vertical component and at constant speed in the horizontal component.

20 mi / h = 29.3 ft/s

If the ramp has an angle of 25 degrees, the speed is

v0 = (29.3 * cos(25) * i + 29.3 * sin(25) * j) ft/s

v0 = (26.5*i + 12.4*j) ft/s

I set up the coordinate system with the origin at the base of the ramp under its end, so:

R0 = (0*i + 8*j) ft

The equation for the horizontal position is:

X(t) = X0 + Vx0 * t

The equation for horizontal speed is:

Vx(t) = Vx0

The equation for vertical position is:

Y(t) = Y0 + Vy0 * t + 1/2 * a * t^2

The equation for vertical speed is:

Vy(t) = Vy0 + a * t

In this frame of reference a is the acceleration of gravity and its values is -32.2 ft/s^2.

In the heighest point of the trajectory the vertical speed will be zero because that is the point where it transitions form going upwards (positive vertical speed) to going down (negative vertical speed), and it crosses zero.

0 = Vy0 + a * t1

a * t1 = -Vy0

t1 = -Vy0 / a

t1 = -12.4 / -32.2 = 0.38 s

y1 = y(0.38) = 8 + 12.4 * 0.38 + 1/2 * (-32.2) * (0.38)^2 = 10.4 ft

The velocity at that moment will be:

v1 = (26.5*i + 0*j) ft/s

When he lands in the water his height is zero.

0 = 8 + 12.4 * t2 + 1/2 * (-32.2) * t2^2

-16.1 * t2^2 + 12.4 * t2 + 8 = 0

Solving this equation electronically:

t2 = 1.19 s

Replacing this time on the position equation:

X(1.19) = 26.5 * 1.19 = 31.5 ft

The speed is:

Vx2 = 26.5 ft/s

Vy2 = 12.4 - 32.2 * 1.19 = -25.9 ft/s

V2 = (26.5*i - 25.9*j) ft/s

a2 = arctg(-25.9 / 26.5) = -44.3

3 0

3 years ago

In addition to bandages and gauze, it's recommended that a first aid kit also contain ___________. A) A finger splintB) Portable

valina [46]

Answer: (D: Tweezers)

A basic first aid kit is always good to have in case of an emergency. The typical items found in a first aid kit are pain reliever, tweezers, alcohol wipes, gloves, antiseptic, medical tape, sterile gauze, insect bite swaps, triple-antibiotic ointment, hydrogen peroxide, bandage scissors, instant cold compresses, and of course bandages. There are various types of bandages that can be included; elastic bandages, adhesive bandages, and triangular bandages for starters.

Explanation:

4 0

3 years ago

12.28 LAB: Output values in a list below a user defined amount - functions Write a program that first gets a list of integers fr

Anastaziya [24]

Answer:

def output_ints_less_than_or_equal_to_threshold(user_values, upper_threshold):

for value in user_values:

if value < upper_threshold:

print(value)

def get_user_values():

n = int(input())

lst = []

for i in range(n):

lst.append(int(input()))

return lst

if __name__ == '__main__':

userValues = get_user_values()

upperThreshold = int(input())

output_ints_less_than_or_equal_to_threshold(userValues, upperThreshold)

8 0

3 years ago

on the same scale for stress, the tensile true stress-true strain curve is higher than the engineeringstress-engineering strain

Bess [88]

Answer:

The condition does not hold for a compression test

Explanation:

For a compression test the engineering stress - strain curve is higher than the actual stress-strain curve and this is because the force needed in compression is higher than the force needed during Tension. The higher the force in compression leads to increase in the area therefore for the same scale of stress the there is more stress on the Engineering curve making it higher than the actual curve.

Hence the condition of : on the same scale for stress, the tensile true stress-true strain curve is higher than the engineering stress-engineering strain curve. does not hold for compression test

5 0

3 years ago

1). Mention any four operations that requires airlines. 2). Explain how airflow is applicable to the above mentioned operations.

vladimir2022 [97]

Answer:

Following are the answer to this question:

Explanation:

1)

Following the four operations in the airlines:

Landside operations:

In Airlines, the airports are divided into areas on the countryside and on the airside, in which landside region is available to the public, although strictly controlled access to the airside zone. Its area covers all areas of the airport across the aircraft, including parts of the buildings which can only be reached by customers and employees.

Airside Operations:

It's also committed to ensuring which air operations military exercises Ballarat airfields are safe and secure. It includes the provision of parking and flight escort services to itinerant and automates. Organizing operational response to incidents, accidents, or emergencies at the airport.

Billing and invoicing Operations:

This requires several steps, each of which must be performed with absolute accuracy to ensure that perhaps the airport operator is adequately paid for supplying passengers with all the services and infrastructure. After this, the receipts want to be produced and sent to customers on the airline.

Information management:

In this, it collects all the data about the customers and employees and it also helps in finding new routes and after collecting the data it processes on them.

2)

It involves many activities in the airline, including dispatch, flight preparation, flight watch, weather information source, activities control, ground-to-air communications, and staff coordination, scheduling, and maintenance planning. Computing and expert programs are constantly being used to handle unpredictable activities.

4 0

3 years ago