Answer:
2940.1 joules would you burn in climbing stairs all day.
Explanation:
Work = W = F
d
going up stairs would be against force of gravity
W = mgh
where h is the height
the question is not complete because we need speed or distance
h = v t
so assuming 1 step per second
h = 86,400 steps 7inchs/step 0.0254 m/inch
h = 15362 m
so from this
W = 800 N 15362
= 12289600 J
that means YOU need 12289600 J to walk 1 step per second all day
divide that by 4180 J /Kcal
Kcal = 
= 
= 2940.1 Kcal
if you ran faster you would use more energy 2 steps per second would mean 5880 Kcal.
Explanation:
The angle of the handle relative to the horizontal is 35°. The angle of the ramp to the horizontal is 7°. So the angle of the handle relative to the ramp is 28°.
cos 28° = 50 / F
F = 50 / cos 28°
F = 56.6 lbs
Answer:
b) q large and m small
Explanation:
q is large and m is small
We'll express it as :
q > m
As we know the formula:
F = Eq
And we also know that :
F = Bqv
F = 
Bqv =
or Eq =
Assume that you want a velocity selector that will allow particles of velocity v⃗ to pass straight through without deflection while also providing the best possible velocity resolution. You set the electric and magnetic fields to select the velocity v⃗ . To obtain the best possible velocity resolution (the narrowest distribution of velocities of the transmitted particles) you would want to use particles with q large and m small.
Answer:
hellooooo :) ur ans is 33.5 m/s
At time t, the displacement is h/2:
Δy = v₀ t + ½ at²
h/2 = 0 + ½ gt²
h = gt²
At time t+1, the displacement is h.
Δy = v₀ t + ½ at²
h = 0 + ½ g (t + 1)²
h = ½ g (t + 1)²
Set equal and solve for t:
gt² = ½ g (t + 1)²
2t² = (t + 1)²
2t² = t² + 2t + 1
t² − 2t = 1
t² − 2t + 1 = 2
(t − 1)² = 2
t − 1 = ±√2
t = 1 ± √2
Since t > 0, t = 1 + √2. So t+1 = 2 + √2.
At that time, the speed is:
v = at + v₀
v = g (2 + √2) + 0
v = g (2 + √2)
If g = 9.8 m/s², v = 33.5 m/s.
