Lake Tahoe Traveler! » Ski Squaw

Squaw Valley Ski School

admin — Thu, 25 Mar 2010 17:28:13 +0000

http://www.patrickmcintosh.com IS where this video is housed. The fun atmosphere runs alongside the professional program that changes lives. St. Patrick’s Day was never so shred-filled and fun.

Duration : 0:5:25

Technorati Tags: NextFit, Ski Tahoe, squaw valley, Tahoe

Squaw Valley Ski

admin — Sun, 21 Mar 2010 17:34:10 +0000

Training flight turned scenic flight after fresh snow. Cruising around Squaw Valley Ski Resort around Lake Tahoe California in the cessna 172.

Duration : 0:3:24

Technorati Tags: airplane, bush flight, cockpit, homewood ski, mountains, pilot, ski, sled, snow, snowboard, squaw, truckee, valley, wing

Physics Question! Difficult!?

admin — Sun, 21 Mar 2010 01:27:09 +0000

Squaw Valley ski area in California claims that irs lifts can move 47,000 people per hour. If the average lift carries people about 200m (vertically) higher, estimate the power needed.

1. Assume average skier plus gear, checks in at 80kg

2. Power = Enegy/time
P= ?m/?t * gh

P = 47,000 people/hr * 1 hour/3600 sec *80 kg /person * 9.81 m/s * 200m
P = 2.05 x10^6 Watts

Squaw Valley Ski Team, Boreal Slalom Dec 12 2009

admin — Fri, 19 Mar 2010 19:42:52 +0000

The first Far West Open race of the season, in a major snow storm.

Duration : 0:1:53

Technorati Tags: ski, squaw, team, valley

physics help please!??

admin — Thu, 18 Mar 2010 02:32:32 +0000

Squaw Valley ski area in California claims that its lifts can move 47,000 people per hour. If the average lift carries people about 200 m (vertically) higher, estimate maximum total power needed.

any help??

The total work done per hour is

mgh = ( m )( 47,000 )( 200 m )

The power is then work divided by the time in seconds

P = mgh / t = ( m )( 47,000 )( 200 m ) / ( 3600 s )

If your problem also gives you the average mass per person, plug it in and there’s your answer. If not, you’ll have to estimate. 60 kg would be pretty close to the weight of the average ski-er.

SQUAW VALLEY 1960 (Ski Jumping / Skispringen)

admin — Wed, 17 Mar 2010 17:34:14 +0000

SQUAW VALLEY 1960

Duration : 0:3:35

Technorati Tags: 1964, 1976, 2010, alpine, Ammann, Anders, Bardal, Biathlon, Bobsleigh, Combined, Cross-Country, Curling, Figure, Franz, Freestyle, Gregor, Harri, Hautamäki, Hilde, Hockey, ice, Innsbruck, Jacobsen, Jumping, Kalle, Keituri, Klammer, Larinto, Ljøkelsøy, Loitzl, Luge, Martin, Matti, Morgenstern, nordic, Olli, Olympiade, olympics, Olympische, planica, Roar, Sailer, Schlierenzauer, Schmitt, Short, simon, skating, Skeleton, ski, skiing, snowboard, Speed, Thomas, Tirol, Tom, Toni, Track, Vancouver, Ville, Winterspiele, Wolfgang, Zakopane, Österreich

Skullcandy Ski Team Throws Down at Squaw!

admin — Mon, 15 Mar 2010 20:32:40 +0000

Freeskier Mag came out to do a special shoot on the Skullcandy team being that we have the best team in the world! Check it out as the team members get crazy!

Duration : 0:1:18

Technorati Tags: graig coker, mike Riddle, peter olenick, Tanner Rainville, tim russel, Tom Wallisch

Physics Question. Please Help me.?

admin — Mon, 15 Mar 2010 05:38:39 +0000

Squaw Valley ski area in California claims that its lifts can move 49500 people per hour. If the average lift carries people about 210 m (vertically) higher, estimate the maximum total power needed. (Assume an average mass per person of 70 kg.)

Answer in W.

49500 people per hour x 70 kg/person x 1 hour/3600 sec = 963 kg/sec

PE = mgh
power = energy/time = mgh/time = 963 kg/sec x 9.8 m/s² x 210m = 1980000 watts

Squaw Valley Ski Team, SG Training 2-20-2010

admin — Sat, 13 Mar 2010 23:31:03 +0000

Imagine going 40mph and hucking a big jump. That is SuperG training Squaw Valley style. Very brave athletes.

Duration : 0:1:0

Technorati Tags: J3, ski, squaw, team, valley

physics help please thanks?

admin — Fri, 12 Mar 2010 07:29:37 +0000

Squaw Valley ski area in California claims that its lifts can move 46400 people per hour. If the average lift carries people about 200 m (vertically) higher, estimate the maximum total power needed. (Assume an average mass per person of 70 kg.)

The energy to move vertically 200 m one 70 kg person is mgh = 70Kg*10m/sec^2*200m = 140000 Joules.

The minimum power to move vertically 200 m 464000 people is

=140000 Joules * 46400/sec (answer is in Watts)

Note this is the minimum power assuming 100 % efficiency. In reality, you almost never achieve 100 % efficiency. Hence, the maximum power is impossible to calculate without an assumed efficiency.

Ron