Let
V the speed of the girl per step
v the speed of the escalator per step
L the length between ground and floor in number of steps
V*60+v*60=L => V+v=L/60
V*90-v*90=L => V-v=L/90
By adding the equations V+v=L/60 and V-v=L/90 we get:
2V=L(3+2)/180
V=5L/360
V = L/72
Time to climb (in steps) with v=0 (escalator standing still) is:
V*t=L
t = L/V
t = L/(L/72)
t =72 steps
The girl will count 72 steps.
Answer:
Displacement
Explanation:
The quantity 45m north is a typical example of displacement.
Displacement is the distance traveled by a body in a specific direction. Displacement is a vector quantity with both magnitude and direction.
- When we are specifying the displacement of a body, the direction must be indicated accurately.
- Therefore, the quantity given is displacement
Answer:
1.Free edge
2.Nail Root
3.Nail matrix
4.lunula
Note: The numbering coincides respectively with the dashes in the questions.
Explanation:
Free edge is the distal white nail ending.
Nail Root is the proximal part of the nail embedded in the skin.
Nail matrix is the actively growing part of the nail were the nail root thickens
Lunula the whitish semilunar area of the proximal end of the nail body it appears whitish because stratum basala obscures. the underlying blood vessels.
Answer:
60 kg
80 kg
Explanation:
Work is equal to the change in energy.
W = ΔE = E − E₀
Let's start with block B. The work done by the tension force is equal to the change in energy. Initially, the block has potential energy. Finally, the block has kinetic energy.
W = ΔE
FΔy = ½ mv² − mgh
T (-2.0 m) = ½ m (6.00 m/s)² − m (10 m/s²) (2.0 m)
T (-2.0 m) = m (-2 m²/s²)
T = m (1 m/s²)
Now let's look at block A. The work done by tension and against friction is equal to the change in energy. Initially, the block has no energy. Finally, it has both kinetic and potential energy.
W = ΔE
Fd = ½ mv² + mgh − 0
(T − Nμ) (2.0 m) = ½ (4.00 kg) (6.00 m/s)² + (4.00 kg) (10 m/s²) (⅗ × 2.0 m)
(T − Nμ) (2.0 m) = 120 J
T − Nμ = 60 N
Draw a free body diagram of block A and sum the forces in the perpendicular direction to find the normal force N.
N = mg cos θ
N = (4.00 kg) (10 m/s²) (⅘)
N = 32 N
Substitute:
T − 32μ = 60 N
If μ = 0, then T = 60 N and m = 60 kg.
If μ = ⅝, then T = 80 N and m = 80 kg.
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