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Elis [28]
2 years ago
12

A baseball (A, weight 0.33 lb) moves horizontally at 20 ft/s when it strikes a stationary block (B, weight 10 lb), supported by

a frictionless surface. If the impact occurs over a time interval of 0.01 sec, with a coefficient of restitution of e=0.75, find the average normal force between A and B during impact.
Physics
1 answer:
Allushta [10]2 years ago
5 0

Answer:

<u>F = 154.85 N </u>

Explanation:

<u>Applying consevation of linear momentum for ball A and block B:</u>

<em>ma₁ × Va₁ + m b₁ × Vb₁ = ma₂ × Va₂ + mb₂ × Vb₂ </em>  

Block B is at rest → Vb₁ = 0  

0.33 × 20 + 10 (0) = 0.33 (Va₂) + 10 (Vb₂)        

6.6 = 0.33 Va₂ + 10 Vb₂    (1)

<u>Using the ecuation of the coefficient of restitution</u>:

<em>e = (Vb₂ -  Va₂) / (Va₁ - Vb₁) </em>

0.75 = (Vb₂ - Va₂) / (20 - 0)  

0.75 × 20 = Vb₂ - Va₂

15 = Vb₂ - Va₂     (2)

<u>From (1), (2):</u>

6.6 = 0.33 ×  (Vb₂ - 15) + 10Vb₂

6.6 = 0.33Vb₂ - 4.95 + 10Vb₂  

Vb₂ = 11.55 / 10.33  

Vb₂ = 1.12 ft/s    

<u>Into (2):</u>

15 = 1.12 - Va₂

Va₂ = -13.88 ft/s

<u>Using the ecuation:</u>

<em>F = Δp / Δt,  F=Normal force between A y B</em>

F =  [(mb₂ × Vb₂) - (m b₁ × Vb₁)] / Δt    

F = [(10 × 1.12) - (10 × 0)] / 0.01  

F =  1120 lb ft/s²  = 154.85 N

Have a nice day!

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W=(F)(d) (1)  

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3 years ago
A square coil ℓ = 2cm on a side with 30 turns rotates in a uniform magnetic field, B~ = B0zˆ = 0.1Tˆz, such that the normal of t
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Answer:

a) 1.2*10^{-3}cos(1.25t)

b) 0.49mV

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\Phi_B(t)=NABcos(\omega t)\\\\\omega=\frac{2\pi}{T}=1.25\frac{rad}{s}\\\\A=l^2=(0.02m)^2=4*10^{-4}m^2\\\\B=0.1T\\\\\Phi_B(t)=1.2*10^{-3}cos(1.25 t) W

where we have used the values given by the information of the problem for N B and A.

b)

the emf is given by:

emf=-\frac{d\Phi_B}{dt}=-NBA\omega sin(\omega t)\\\\emf(t=12.5s)=-(30)(0.1T)(4*10^{-4})(1.25\frac{rad}{s})sin(1.25*12.5)=1.49*10^{-4}V=0.49mV

hope this helps!!

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3 years ago
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