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vlabodo [156]
3 years ago
12

The 1994 Winter Olympics included the aerials competition in skiing. In this event skiers speed down a ramp that slopes sharply

upward at the end. The sharp upward slope launches them into the air, where they perform acrobatic maneuvers. In the women's competition, the end of a typical launch ramp is directed 61° above the horizontal. With this launch angle, a skier attains a height of 16 m above the end of the ramp. What is the skier's launch speed?

Physics
1 answer:
kykrilka [37]3 years ago
4 0

Answer: Got It!

<em>Explanation:  </em>let s = speed at launch

v = 0 at top = s sin 63 - g t

so at top

t = s sin 63/g = .0909 s

h = 13.6 = s sin 63 t - 4.9 t^2

13.6 = .081s^2 - .0405 s^2

s^2 = 336

s = 18.3 m/s

0  0

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At 1 AM, a handsome astronomy instructor has just completed a late night dinner with Jennifer Garner and observes a nearly full
oksano4ka [1.4K]

Answer:

The correct answer is: waxing gibbous, 3 days

Explanation:

Waning quarter moon: hair removal time and bangs cuts.

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We must take advantage of the influence of the lunar cycle in our favor according to the action we are going to take. If you have trouble growing your hair, try to go to the hairdresser in a crescent moon: it will grow faster. It is no nonsense. Since I cut my bangs to the Cleopatra, the touch-ups last me for another 1-1.5 weeks. As I reviewed the bangs in a growing room, in just a couple of weeks I was returning to the hairdresser.

That affects hair removal. There are many people who take appointments to the beautician to shave by consulting the lunar calendar. The hair removal done as soon as the dwindling is the best because it lasts longer, lasts for another week until the next appointment.

6 0
3 years ago
An object of irregular shape has a characteristic length of L = 0.5 m and is maintained at a uniform surface temperature of Ts =
goblinko [34]

Answer:

The value of the average convection coefficient is 20 W/Km².

Explanation:

Given that,

For first object,

Characteristic length = 0.5 m

Surface temperature = 400 K

Atmospheric temperature = 300 K

Velocity = 25 m/s

Air velocity = 5 m/s

Characteristic length of second object = 2.5 m

We have same shape and density of both objects so the reynold number will be same,

We need to calculate the value of the average convection coefficient

Using formula of  reynold number for both objects

R_{1}=R_{2}

\dfrac{u_{1}L_{1}}{\eta_{1}}=\dfrac{u_{2}L_{2}}{\eta_{2}}

\dfrac{h_{1}L_{1}}{k_{1}}=\dfrac{h_{2}L_{2}}{k_{2}}

Here, k_{1}=k_{2}

h_{2}=h_{1}\times\dfrac{L_{1}}{L_{2}}

h_{2}=\dfrac{q}{T_{2}-T_{1}}\times\dfrac{L_{1}}{L_{2}}

Put the value into the formula

h_{2}=\dfrac{10000}{400-300}\times\dfrac{0.5}{2.5}

h_{2}=20\ W/Km^2

Hence, The value of the average convection coefficient is 20 W/Km².

7 0
3 years ago
A car starts from rest and accelerates uniformly for a five seconds along a straight road. If speed obtained by the car is 72 km
Step2247 [10]

Answer:

50 meters

Explanation:

Let's start by converting to m/s. There are 3600 seconds in an hour and 1000 meters in a kilometer, meaning that 72km/h is 20m/s.

v_f=v_o+at

Since the car starts at rest, you can write the following equation:

20=0+a(5) \\\\a=20\div 5=4 m/s^2

Now that you have the acceleration, you can do this:

d=v_o+\dfrac{1}{2}at^2

Once again, there is no initial velocity:

d=\dfrac{1}{2}(4)(5)^2=2 \cdot 25=50m

Hope this helps!

8 0
3 years ago
Except when necessary for takeoff and landing, what is the minimum safe altitude required for a pilot to operate an aircraft ove
MAVERICK [17]
Except when necessary for takeoff and landing, <span>the minimum safe altitude required for a pilot to operate an aircraft over other than congested area is an altitude of 1000 ft above the highest obstacle within a 2000 ft horizontal radius of the aircraft.

It is also good to know that apart from taking off and landing, the aircraft must not operate at a distance less than 500 ft from any person, vessel, structure or vehicle.</span>
5 0
3 years ago
A speedboat moving at 28 m/s approaches a no-wake buoy marker 91 m ahead. The pilot slows the boat with a constant acceleration
Tpy6a [65]

Answer:

7.5 m/s

Explanation:

We can find its velocity when it reaches the buoy by applying one of Newton's equations of motion:

v^2 = u^2 + 2as

where v = final velocity

u = initial velocity

a = acceleration

s = distance traveled

From the question:

u = 28 m/s

a = -4 m/s^2

s = 91 m

Therefore:

v^2 = 28^2 + 2 * (-4) * 91\\\\v^2 = 784 + -728 = 56\\\\v = \sqrt{56}\\ \\v = 7.5 m/s

The velocity of the boat when it reaches the buoy is 7.5 m/s.

6 0
3 years ago
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