Answer:
heya answer option b
Explanation:
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Answer:
This value is less than the maximum tension of 500 lbs, making it safe for man to go to the tip flap
Explanation:
We must work on this problem using the rotational equilibrium equations and then they compared the tension values that the cable supports.
Let's start with fixing a reference system on the hinge of the flag, we take as positive the anti-clockwise turn
They indicate the weight of the pole W₁ = 120 lb and a length of L = 9 ft, the weight of the man W₂ = 150, we assume that the cable is at the tip of the pole
-
L + W₂ L + W₁ L / 2 = 0
T_{y} = W₂ + W₁ / 2
T_{y} = 120 + 150/2
T_{y} = 195 lb
we use trigonometry to find the cable tension
sin 30 = T_{y} / T
T = T_{y} / sin 30
T = 195 / sin 30
T = 390 lb
This value is less than the maximum tension of 500 lbs, making it safe for man to go to the tip flap
T < 500 lb
Answer:
a) λ = 189.43 10⁻⁹ m b) λ = 269.19 10⁻⁹ m
Explanation:
The diffraction network is described by the expression
d sin θ= m λ
Where m corresponds to the diffraction order
Let's use trigonometry to find the breast
tan θ = y / L
The diffraction spectrum is measured at very small angles, therefore
tan θ = sin θ / cos θ = sin θ
We replace
d y / L = m λ
Let's place in the first order m = 1
Let's look for the separation of the lines (d)
d = λ L / y
d = 501 10⁻⁹ 9.95 10⁻² / 15 10⁻²
d = 332.33 10⁻⁹ m
Now we can look for the wavelength of the other line
λ = d y / L
λ = 332.33 10⁻⁹ 8.55 10⁻²/15 10⁻²
λ = 189.43 10⁻⁹ m
Part B
The compound wavelength B
λ = 332.33 10⁻⁹ 12.15 10⁻² / 15 10⁻²
λ = 269.19 10⁻⁹ m
Answer: 6.48m/s
Explanation:
First, we know that Impulse = change in momentum
Initial velocity, u = 19.8m/s
Let,
Velocity after first collision = x m/s
Velocity after second collision = y m/s
Also, we know that
Impulse = m(v - u). But then, the question said, the guard rail delivered a "resistive" impulse. Thus, our impulse would be m(u - v).
5700 = 1500(19.8 - x)
5700 = 29700 - 1500x
1500x = 29700 - 5700
1500x = 24000
x = 24000/1500
x = 16m/s
Also, at the second guard rail. impulse = ft, so that
Impulse = 79000 * 0.12
Impulse = 9480
This makes us have
Impulse = m(x - y)
9480 = 1500(16 -y)
9480 = 24000 - 1500y
1500y = 24000 - 9480
1500y = 14520
y = 14520 / 1500
y = 9.68
Then, the velocity decreases by 3.2, so that the final velocity of the car is
9.68 - 3.2 = 6.48m/s
Answer:
<em>0.45 mm</em>
Explanation:
The complete question is
a certain fuse "blows" if the current in it exceeds 1.0 A, at which instant the fuse melts with a current density of 620 A/ cm^2. What is the diameter of the wire in the fuse?
A) 0.45 mm
B) 0.63 mm
C.) 0.68 mm
D) 0.91 mm
Current in the fuse is 1.0 A
Current density of the fuse when it melts is 620 A/cm^2
Area of the wire in the fuse = I/ρ
Where I is the current through the fuse
ρ is the current density of the fuse
Area = 1/620 = 1.613 x 10^-3 cm^2
We know that 10000 cm^2 = 1 m^2, therefore,
1.613 x 10^-3 cm^2 = 1.613 x 10^-7 m^2
Recall that this area of this wire is gotten as
A = 
where d is the diameter of the wire
1.613 x 10^-7 = 
6.448 x 10^-7 = 3.142 x 
=
d = 4.5 x 10^-4 m = <em>0.45 mm</em>