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Leona [35]
3 years ago
10

A stunt man jumps from the top of a building and lands 10 meters below his initial height. In case A, the stunt man lands on a s

tack of foam mattresses. In case B, the stunt man lands on a stack of concrete blocks. In which case will the person be more likely to sustain serious injury?
Physics
1 answer:
krek1111 [17]3 years ago
8 0

Answer:

Explanation:

The stunt will likely sustain serious injury in case of concrete blocks because the average force acting on the person will be more because concrete blocks do not squeeze to provide more time for the force to act on the body instead it acts for a small amount of interval.

Impulse=F_{avg}\times \Delta T

As impulse is constant so time requires to act force on the body is more as compared to concrete block and thus average force in mattress case is less.  

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D

Explanation:

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What method is used to find the number of neutrons in an atom?
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B

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In a women's 100-m race, accelerating uniformly, Laura takes 1.82 s and Healan 3.07 s to attain
Sav [38]

Answer:

Laura is ahead and for a distance of 3.22 m

Explanation:

To solve this problem of one-dimensional kinematics, we have to find the acceleration and the final speed of each runner. Let's start with Laura

Lura data is acceleration time 1.82s, total run time 10.4 s and total distance 100m.  In all the races the  rest starts, so the initial speed is zero (Vo = 0)

   Vf1= Vo + a1 t1    

   Vf1 = x/t                

   XT  = X1 + X2

   X1 = Vo t1 + ½ a1 t1²  

   X1 = ½ a1 t1²  

   X2 = Vf1 (t-t1)

This is the remaining time of the race after the acceleration is over.

    XT = ½ a1 t1² + Vf1 (t-t1)

We remplace the expression of Vf1

     XT = ½ a1 t1² + a1 t1 (t-t1)

Laura's aceleration (a1) is

   a1= XT / [ ½  t1² + t1 (t-t1)]

   a1= 100/ [ ½ 1.82²+ 1.82 (10.4 -1.82)]

   a1=  5.79m/s2  

We repeat the same calculation for the other Healan runner, whose data are: total distance 100m, acceleration time 3.07 s and total time 10.4 s

Vf2= Vo + a2 t2    

Vf2 = x/t                

XT  = X3 + X4

X3 = Vo t2 + ½ a2 t2²  

X3 = ½ a2 t2²  

X4 = Vf2 (t-t2)

XT = ½ a2 t2² + Vf2 (t- t2)

XT = ½ a2 t2² + a2 t2 (t-t2)

The aceleration of Healan (a2)

a2 = XT / [½ t2² + t2 (t-t2)]

a2 = 100 / [½ 3,07²+ 3.07 (10.4 -3.07)]

a2 = 3.67 m / s2

We also need the final speeds of each runner

Laura Vf1 = Vo + a1 t1

          Vf1 = 0 + 5.79 1.82

          Vf1 = 10.54 m / s

Healan Vf2 = Vo + a2 t2

            Vf2 = 0 + 3.67 3.07

            Vf2 = 11.27 m / s

Having the acceleration and speed of each runner, you can start answering the questions

a) For t3 = 6.15s

Laura

The time to stop with constant speed is what remains after accelerating

XL= ½ a1 t1² + Vf1 (t3-t1)

XL= ½ 5.79 1.82² + 10.54 (6.15 – 1.82)    

XL= 55.23 m

Healan  

XH= ½ a2 t2² + Vf2 (t3-t2)

             XH= ½  3.67 3.07² + 11.27 (6.15-3.07)

             XH= 52.01 m

             (XL -XH)= 55.23- 52.01

             (XH -XL)=  3.22 m

It is appreciated from these results that Laura is ahead and for a distance of 3.22 m

b) If we analyze the acceleration values ​​of each runner, knowing that they leave the rest and that Healan at the end has a speed greater than Laura, the point of maximum distance difference is when Laura stops accelerating t = 1.82 s

      XL= ½  a1 t12  

      XL= ½ 5.79 1.822

      XL= 9.59 m

      XH = ½ a2 t12

      XH= ½ 3.67 1.822

      XH= 6.08 m

The maximum distance difference is 3.51 m

c) Already analyzed in the previous part 1.82 s, since the Laura stop accelerating and Heala continue with acceleration will travel greater distances in equal time units

6 0
3 years ago
B) A non-inductive load takes a current of 15 A at 125 V. An inductor is then connected
levacccp [35]

Answer:

i. 43.5 mH ii.  16 Ω. In phasor form Z = (8.33 + j13.66) Ω iii 58.64°

Explanation:

i. The resistance , R of the non-inductive load R = 125 V/15 A = 8.33 Ω

The reactance X of the inductor is X = 2πfL where f = frequency = 50 Hz.

So, x = 2π(50)L = 100πL Ω = 314.16L Ω

Since the current is the same when the 240 V supply is applied, then

the impedance Z = √(R² + X²) = 240 V/15 A

√(R² + X²) = 16 Ω

8.33² + X² = 16²

69.3889 + X² = 256

X² = 256 - 69.3889

X² = 186.6111

X = √186.6111

X = 13.66 Ω

Since X = 314.16L = 13.66 Ω

L = 13.66/314.16

= 0.0435 H

= 43.5 mH

ii. Since the same current is supplied in both circuits, the impedance Z of the circuit is Z = 240 V/15 A = 16 Ω.

So in phasor form Z = (8.33 + j13.66) Ω

iii. The phase difference θ between the current and voltage is  

θ = tan⁻¹X/R

= tan⁻¹(314.16L/R)

= tan⁻¹(314.16 × 0.0435 H/8.33 Ω)

= tan⁻¹(13.66/8.33)

= tan⁻¹(1.6406)

= 58.64°

3 0
3 years ago
Please help meh its due
Fed [463]
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