Answer:
The strength of the magnetic field that the line produces is 
.
Explanation:
From Biot-Savart law, the equation to determine the strength of the magnetic field for any straight wire can be deduced:
(1)
Where 
is the permiability constant, I is the current and r is the distance from the wire.
Notice that it is necessary to express the current, I, from kiloampere to ampere.
⇒ 
Finally, equation 1 can be used:
Hence, the strength of the magnetic field that the line produces is .
Answer:
Friction force on the bullet is 58.7 N opposite to its velocity
Explanation:
As we know that initial speed of the bullet is 55 m/s
after travelling into the sand bag by distance d = 1.34 m it comes to rest
so final speed
now we can use kinematics top find the acceleration of the bullet
so we have
now by Newton's II law we know that
so we have
Answer:
a) 
= 0.25 m / s b) u = 0.25 m / s
Explanation:
a) To solve this problem let's start with the conservation of the moment, for this we define a system formed by the ball plus the dog, in this case all the forces are internal and the moment is conserved
We will write the data
m₁ = 0.40 kg
v₁₀ = 9.0 m / s
m₂ = 14 kg
v₂₀ = 0
Initial
po = m₁ v₁₀
Final
= (m₁ + m₂) vf
po = pf
m₁ v₁₀ = (m₁ + m₂)
= v₁₀ m₁ / (m₁ + m₂)
= 9.0 (0.40 / (0.40 +14)
= 0.25 m / s
b) This is the reference frame of the center of mass of the system in this case the speed of this frame is the speed of the center of mass
u = 0.25 m / s
In the direction of movement of the ball
c) Let's calculate the kinetic energy in both moments
Initial
K₀ = ½ m₁ v₁₀² +0
K₀ = ½ 0.40 9 2
K₀ = 16.2 J
Final
= ½ (m₁ + m₂) 2
= ½ (0.4 +14) 0.25 2
= 0.45 J
ΔK = K₀ -
ΔK = 16.2-0.445
ΔK = 1575 J
These will transform internal system energy
d) In order to find the kinetic energy, we must first find the velocities of the individual in this reference system.
v₁₀’= v₁₀ -u
v₁₀’= 9 -.025
v₁₀‘= 8.75 m / s
v₂₀ ‘= v₂₀ -u
v₂₀‘= - 0.25 m / s
‘= - u
= 0
Initial
K₀ = ½ m₁ v₁₀‘² + ½ m₂ v₂₀‘²
Ko = ½ 0.4 8.75² + ½ 14.0 0.25²
Ko = 15.31 + 0.4375
K o = 15.75 J
Final
= ½ (m₁ + m₂) vf’²
= 0
All initial kinetic energy is transformed into internal energy in this reference system
That's unaccelerated motion,
and constant velocity.
