Answer:
T₂ = 2482.34 N
Explanation:
Equations of balance of forces
Look at the force diagram in the attached graph:
∑Fx=0
T₂cos α -T₁senα = 0 Equation (1)
∑Fy=0
T₁cos α +T₂sinα-W =0 Equation(2)
Data
m=539 kg
g= 9.8 m/s²
W= m*g= 539 kg* 9.8 m/s²= 5282.2 N
T₁ = 1.88T₂
Problem development
in the equation (1)
T₂cos α-(1.88T₂)senα = 0 We divided the equation by ( T₂cos α)
1 - (1.88)tanα = 0
tanα= 1/(1.88)
tanα= 0.5319
α = 28°
T₂=(1.88T₂)tanα
in the equation (2)
(1.88T₂)cos α+ T₂sinα - 5282.2 =0 We divided the equation by cos α
1.88T₂+ T₂tanα - 5282.2/cos α =0
1.88T₂+ T₂tan(28°) - 5282.2/(cos 28°) =0
1.88T₂+ (0.53)T₂- 5982.46 =0
(2.41)T₂ = 5982.46
T₂ = 5982.46/(2.41)
T₂ = 2482.34 N
Answer:
I have a screenshot of this.
Explanation:
Answer:
80mm or 8cm
Explanation:
According to the lens formula,
1/f = 1/u+1/v
If the object distance u = 4cm = 40mm
Object height = 1.5mm
Image height = 3mm
First, we need to get the image distance (v) using the magnification formula Magnification = image distance/object distance = Image height/object height
v/40=3/1.5
1.5v = 120
v = 120/1.5
v = 80mm
The image distance is 80mm
To get the focal length, we will substitute the image distance and the object distance in the mirror formula to have;
1/f = 1/40+1/-80
Note that the image formed by the lens is an upright image (virtual), therefore the image distance will be negative.
Also the focal length of the converging lens is positive. Our formula will become;
1/f = 1/40-1/80
1/f = 2-1/80
1/f = 1/80
f = 80mm
The focal length of the lens 80mm or 8cm
Assuming our "closed tube" is closed at only one end, then
<span> v = fλ = f*4L/n
</span><span> where "n" is the harmonic number. So
</span><span> L = nv / 4f = n*346m/s / 4*256Hz = n*0.38 m
</span> <span>Since the only option in your list that is an integer multiple of 0.38 m is 1.35 m
</span><span> I'd say that we're hearing the fourth harmonic.
answer is
</span><span>A. 1.35 m</span><span>
</span>