Because the elevator moves at a constant speed, it's in equilibrium and the net force acting on it is zero. Then the tension in the cable exactly equals the magnitude of the elevator's weight, which is
(3000 kg) (9.80 m/s²) = 29,400 N
Answer
Applying Wein's displacement

1) for sun T = 5800 K


2) for tungsten T = 2500 K


3) for heated metal T = 1500 K


4) for human skin T = 305 K


5) for cryogenically cooled metal T = 60 K


range of different spectrum
UV ----0.01-0.4
visible----0.4-0.7
infrared------0.7-100
for sun T = 5800
λ 0.01 0.4 0.7 100
λT 58 2320 4060 5.8 x 10⁵
F 0 0.125 0.491 1
fractions
for UV = 0.125
for visible = 0.441-0.125 = 0.366
for infrared = 1 -0.491 = 0.509
Answer: a) 11.76 m/s b) 7.056 m
Explanation:
The described situation is as follows:
An object is dropped from the top of a tower and when measuring the time it takes to reach the ground that turns out to be 0.02 minutes.
This situation is related to free fall, this also means we have constant acceleration, hence the equations we will use are:
(1)
(2)
Where:
Is the final velocity of the object
Is the initial velocity of the object (it was dropped)
is the acceleration due gravity
is the height of the tower
is the time it takes to the object to reach the ground
b) Begining with (1):
(3)
(4)
(5) This is the final velocity of the object
a) Substituting (5) in (2):
(6)
Clearing
:
(7)
(8) This is the height of the tower
Answer:
(a) 2.85 m
(b) 16.5 m
(c) 21.7 m
(d) 22.7 m
Explanation:
Given:
v₀ₓ = 19 cos 71° m/s
v₀ᵧ = 19 sin 71° m/s
aₓ = 0 m/s²
aᵧ = -9.8 m/s²
(a) Find Δy when t = 3.5 s.
Δy = v₀ᵧ t + ½ aᵧ t²
Δy = (19 sin 71° m/s) (3.5 s) + ½ (-9.8 m/s²) (3.5 s)²
Δy = 2.85 m
(b) Find Δy when vᵧ = 0 m/s.
vᵧ² = v₀ᵧ² + 2 aᵧ Δy
(0 m/s)² = (19 sin 71° m/s)² + 2 (-9.8 m/s²) Δy
Δy = 16.5 m
(c) Find Δx when t = 3.5 s.
Δx = v₀ₓ t + ½ aₓ t²
Δx = (19 cos 71° m/s) (3.5 s) + ½ (0 m/s²) (3.5 s)²
Δx = 21.7 m
(d) Find Δx when Δy = 0 m.
First, find t when Δy = 0 m.
Δy = v₀ᵧ t + ½ aᵧ t²
(0 m) = (19 sin 71° m/s) t + ½ (-9.8 m/s²) t²
0 = t (18.0 − 4.9 t)
t = 3.67
Next, find Δx when t = 3.67 s.
Δx = v₀ₓ t + ½ aₓ t²
Δx = (19 cos 71° m/s) (3.67 s) + ½ (0 m/s²) (3.67 s)²
Δx = 22.7 m