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DochEvi [55]
3 years ago
13

It is well known that bullets fired at Superman simply bounce off his chest. Suppose that a gangster sprays Superman's chest wit

h bullets of mass m = 1.7 g at a rate of R = 165 bullets/min. The speed of each bullet is v = 380 m/s. Suppose the bullets rebound with no change in speed. What is the average force exerted by the stream of bullets on Superman's chest?
Physics
1 answer:
Zanzabum3 years ago
6 0

Answer:

1.77 N

Explanation:

Mass of bullet m = 1.7 x 10⁻³ kg

velocity v = 380 m /s .

Momentum of one bullet

= 1.7 x 10⁻³ x 380

= 646 x 10⁻³ kg m/s

momentum of 165 bullets

= 165 x 646 x 10⁻³ kg m/s

= 106.59 kg m/s

Final momentum after bouncing

= 0

change in momentum

= 106.59  - 0

= 106.59 kg m/s

This change occurs in one minute

so rate of change in momentum

= 106.59 / 60 kg m/s per second

= 1.77 kg m/s per second

rate of change in momentum = force

This is force on superman's chest

Force required = 1.77 N .

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If a fish is trying to capture an insect hovering above the surface of the water – how will it jump to catch it? Will it aim abo
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2 years ago
If an object is projected upward from ground level with an initial velocity of 160 ft per​ sec, then its height in feet after t
sesenic [268]

Answer:

Time will be 5 sec

And maximum height will be 1200 feet    

Explanation:

We have given that initial velocity of the object u = 160 ft/sec'

Function of height s(t)=-16t^2+160t

Take derivative of the distance

We know that first derivative of distance is velocity

So v(t)=-32t+160

At maximum height we know that velocity is zero

So -32t+160=0

t = 5 sec

So it will take 5 sec to reach maximum height

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3 years ago
Consider a composite cube made of epoxy with fibers aligned along one axis of the cube (the fibers are parallel to four of the t
coldgirl [10]

The question is incomplete. The complete question is :

Consider a composite cube made of epoxy with fibers aligned along one axis of the cube (the fibers are parallel to four of the twelve cube edges). If the cube can only be loaded in axial tension such that the force is uniformly applied over - and is normal to - a cube face, what is the lowest possible positive length change the cube can experience under this tension? The applied tensile force is 102 KN. The unloaded cube edge length is 56 mm. The glass fibers have an elastic modulus of 200 GPa. The epoxy has an elastic modulus of 38 GPa. The cube is comprised of 18 vol% epoxy (the balancing vol % is glass fiber). Hint: The loading axis is intentionally unspecified. Answer Format: Lowest possible length increase (change of length) under tension.

Solution :

Given :

$E_{glass fibre}$ = 200 GPa

$V_{glass fibre} = 82\%$

$E_{epoxy}$ = 38 GPa

$V_{epoxy} = 82\%$

Edge length = 56 mm

Cube is loaded in axial tension such that the force is uniformly applied over a cube face.

$E_{\text{composite}}=\frac{E_{glass fibre} \times E_{epoxy}}{(E_{glass fibre .E_{epoxy}})+(E_{fibre}.V_{glass fibre})}$

$E_{composite} = \frac{200 \times 38}{(200 \times 0.18)+(38\times 0.82)}$

               $=113.16 $  GPa

Applied stress $=\frac{\text{applied load}}{\text{area}}$

                    $\sigma=\frac{102 \times 10^3 \ N}{56 \times 56 \times 10^{-6} \ m^2}$

                       = 32.5 MPa

By Hooke's law

$\sigma = E . \epsilon$

$\sigma = E. \frac{\Delta l}{l}$

$\Delta l = \frac{\sigma}{E}\times l$

Length change, $\Delta l =\frac{32.5 \times 10^6 \ Pa}{113.16 \times 10^9 \ Pa}\times 56 \times 10^{-2} \ m$

$\Delta l = \frac{32.5 \times 56}{113.16} \times 10^{-3} \ mm$

   = 0.016 mm

7 0
3 years ago
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