Answer: 3217.79 hours.
Explanation:
Given, A 140 lb. climber saved her potential energy as she descended from Mt. Everest (Elev. 29,029 ft) to Kathmandu (Elev. 4,600 ft).
Power = 0.4 watt
Mass of climber = 140 lb
= 140 x 0.4535 kg [∵ 1 lb= 0.4535 kg]
⇒ Mass of climber (m) = 63.50 kg
Let ![h_1=29,029\ ft= 8848.04\ m\ \ \ \ [ 1 ft=0.3048\ m ]](https://tex.z-dn.net/?f=h_1%3D29%2C029%5C%20ft%3D%208848.04%5C%20m%5C%20%5C%20%5C%20%5C%20%20%5B%201%20ft%3D0.3048%5C%20m%20%5D)
and 
Now, Energy saved =
![\text{Power}=\dfrac{\text{energy}}{\text{time}}\\\\\Rightarrow 0.4=\dfrac{4633620.91}{\text{time}}\\\\\Rightarrow\ \text{time}=\dfrac{4633620.91}{0.4}\approx11584052.28\text{ seconds}\\\\=\dfrac{11584052.28}{3600}\text{ hours}\ \ \ [\text{1 hour = 3600 seconds}]\\\\=3217.79\text{ hours}](https://tex.z-dn.net/?f=%5Ctext%7BPower%7D%3D%5Cdfrac%7B%5Ctext%7Benergy%7D%7D%7B%5Ctext%7Btime%7D%7D%5C%5C%5C%5C%5CRightarrow%200.4%3D%5Cdfrac%7B4633620.91%7D%7B%5Ctext%7Btime%7D%7D%5C%5C%5C%5C%5CRightarrow%5C%20%5Ctext%7Btime%7D%3D%5Cdfrac%7B4633620.91%7D%7B0.4%7D%5Capprox11584052.28%5Ctext%7B%20seconds%7D%5C%5C%5C%5C%3D%5Cdfrac%7B11584052.28%7D%7B3600%7D%5Ctext%7B%20hours%7D%5C%20%5C%20%5C%20%5B%5Ctext%7B1%20hour%20%3D%203600%20seconds%7D%5D%5C%5C%5C%5C%3D3217.79%5Ctext%7B%20hours%7D)
Hence, she can power her 0.4 watt flashlight for 3217.79 hours.
Ω = 2.81
A = 0.232
k = 29.8
x = A cos(ωt + Ф)
at t = 0:
x = A = A cos(ωt + Ф) = A cos(Ф)
Ф = 0
at t = 1.42, with Ф = 0:
x = A cos(ωt)
U = 1/2 k x² = 1/2 k [A cos(ωt)]²
Answer:
Oceanic crust is about 6 km (4 miles) thick. It is composed of several layers, not including the overlying sediment. The topmost layer, about 500 metres (1,650 feet) thick, includes lavas made of basalt (that is, rock material consisting largely of plagioclase [feldspar] and pyroxene).
Explanation:
it was on google hope this helps
Answer:
B) Absorbed energy results in the change in potential energy.
Explanation:
i got it wrong on usatestpreo
Answer:
Explanation:
A ) angular velocity ω = 2π / T
= 2 x 3.14 / 60
= .10467 rad / s
linear velocity v = ω R
= .10467 x 50
= 5.23 m / s
centripetal force = m v² / R
= mg v² / gR
= 834 x 5.23² / 9.8 x 50
= 46.55 N
B )
apparent weight
= mg - centripetal force
= 834 - 46.55
= 787.45 N
C ) apparent weight
= mg + centripetal force
= 834 + 46.55
= 880.55 N.
D )
For apparent weight to be zero
centripetal force = mg
mg = mv² / R
v² = gR
= 9.8 x 50
= 490
v = 22.13 m /s
time period of revolution
= 2π R /v
2 x 3.14 x 50 / 22.13
= 14.19 s
