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Nuetrik [128]
3 years ago
9

The plate is suspended using the three cables which exert the forces shown, express each force as a cartesian vector

Physics
1 answer:
Westkost [7]3 years ago
6 0

solution:

write the coordinates of the points A and B  

A= (0,0,14)ft

B= (5,-6,0)ft

write the position vector of a with recpect to b

R_{BA}=(0-5)i+(0-(-6))j+(14-0)k

     =-5i+6k+14k

calculation of a magnitude of the vector R_{BA}

\left | R_{BA} \right |=\sqrt{(-5)^2+6^2+14^2}

                      =\sqrt{25+36+196}

                      =16.031ft

calculation of the unit vector  

\frac{-5i+6k+14k}{16.031}

calculation of the force

F_{BA}=R_{BA}U_{BA}  

here, F_{BA} is the magnitude of the force.

substitute 350lb for F_{BA} and (\frac{-5i+6k+14k}{16.031}) for U_{BA}.

F_{BA} = 350 x (\frac{-5i+6k+14k}{16.031})

      =-109.2i+131j+305.7k lb

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Consider two point charges located on the x axis: one charge, q1= -11.0 nC, is located at x1= -1.675 m; the second charge, q2= 3
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Please refer to the figure.

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F_{13} = \frac{1}{4\pi \epsilon_0}\frac{q_1 q_3}{r_1^2} = \frac{1}{4\pi \epsilon_0}\frac{11\times 10^{-9}\times 47.5\times10^{-9}}{(-1.675 - (-1.18))^2} = 1.92\times 10^{-5}N

F_{23} = \frac{1}{4\pi\epsilon_0}\frac{q_2q_3}{r_2^2} = \frac{1}{4\pi\epsilon_0}\frac{31\times 10^{-9} \times 47.5\times 10^{-9}}{(-1.18 - 0)^2} = 9.5\times 10^{-6}

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3 years ago
A bullet of mass 0.017 kg traveling horizontally at a high speed of 290 m/s embeds itself in a block of mass 5 kg that is sittin
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Answer:

(a) vf = 0.98 m/s

(b) K₁ = 714.85 J : Total translational kinetic energy before the collision.

K₂=  2.41 J : Total translational kinetic energy after the collision.

Explanation:

Theory of collisions  

Linear momentum is a vector magnitude (same direction of the velocity) and its magnitude is calculated like this:    

p=m*v  

where  

p:Linear momentum  

m: mass  

v:velocity  

There are 3 cases of collisions : elastic, inelastic and plastic.  

For the three cases the total linear momentum quantity is conserved:  

P₀ = Pf Formula (1)  

P₀ :Initial linear momentum quantity  

Pf : Final linear momentum quantity  

Data

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m₂ = 5 kg : mass of the block

v₀₁ =  290 m/s : initial velocity of the bullet

v₀₂ = 0  : initial velocity of the block₂

(a) Speed of the block after the bullet embeds itself in the block

We appy the formula (1):

P₀ = Pf  

m₁*v₀₁ + m₂*v₀₂ = (m₁+ m₂)vf  

vf: final velocity of the block

( 0.017)*( 290) + (5)*(0) = ( 0.017 + 5 )*vf

4.93+ 0 = (  5.017 )*vf

vf = 4.93 / 5.017

vf = 0.98 m/s

b)  Total translational kinetic energy before (K₁) and after the collision(K₂).

K₁ = 1/2(m₁*v₀₁² + m₂*v₀₂²)

K₁ = 1/2(0.017*(290)² + 5*(0)²) = 714.85 J

K₂= 1/2(m₁+ m₂)*vf²

K₂= 1/2(0.017+ 5)*(0.98)²  = 2.41 J

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