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yan [13]
3 years ago
14

Comic-book superheroes are sometimes able to punch holes through steel walls. (a) If the ultimate shear strength of steel is tak

en to be 2.50 x 108 Pa, what force is required to punch through a steel plate 2.00 cm thick? Assume the superhero’s fist has cross-sectional area of 1.00 x 102 cm2 and is approximately circular. (b) Qualitatively, what would happen to the superhero on delivery of the punch? What physical law applies?
Physics
1 answer:
Katena32 [7]3 years ago
7 0
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions here.

shear stress is the measure of the material's resistance to cause "sliding" motion of its layers. 
<span>Since the hero has to take a cylindrical piece of metal out, the area which will cause this force is 2*pi*r*l (put r and l in SI units to get answer in Newtons) </span>
<span>so the force required is shear strength into this area (2.60109 * 2 * pi * r * l) </span>
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A projectile is launched vertically at 100 m/s. If air resistance can be ignored, at what speed will it return to its initial le
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<h2>Speed with which it return to its initial level is 100 m/s</h2>

Explanation:

We have equation of motion v² = u² + 2as

Initial velocity, u = 100 m/s  

Acceleration, a = -9.81 m/s²  

Final velocity, v = ?

Displacement, s = 0 m  

Substituting  

v² = u² + 2as

v² = 100² + 2 x -9.81 x 0

v² = 100²

v = ±100 m/s

+100 m/s is initial velocity and -100 m/s is final velocity.

Speed with which it return to its initial level is 100 m/s

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2 years ago
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3 years ago
The membrane that surrounds a certain type of living cell has a surface area of 4.7 x 10-9 m2 and a thickness of 1.3 x 10-8 m. A
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Answer:

Q = 1.2*10⁻¹² C

Explanation:

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       C = \frac{Q}{V}  (1)

  • For the special case of a parallel plate capacitor, just by application of Gauss' law to a rectangular surface half out of the outer surface, and half inside it, it can be showed that the value of the capacitance C is a parameter defined only by geometric constants, as follows:

       C = \frac{\epsilon_{0}*\epsilon _{r} * A}{d}  (2)

  • So, due to the left sides in (1) and (2) are equal each other, right sides must be equal too.
  • Replacing ε₀, εr (dielectric constant), A, d and V by their values, we can solve for Q, as follows:

       Q =\frac{\epsilon_{0} * \epsilon_{r} *A* V}{d} = \frac{(8.85*(4.7)^{2}*79.5)e-24 (F/m*m2*V)}{1.3e-8m} =  1.2e-12 C = 1.2 pC  (3)

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Actually what the problem meant about the westward component of the ball’s displacement is the horizontal component of the displacement. To help us better understand the problem, I attached a figure of the situation.

We can see from the figure that to solve for the value of the horizontal component, we have to make use of the sin function. That is:

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sin 42 = x / 40 m

x = (40 m) sin 42

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Therefore the ball has a westward displacement of about 26.77 m

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