Answer:
The current flowing through the outer coils is
Explanation:
From the question we are told that
The number of turn of inner coil is 
The radius of inner coil is 
The current flowing through the inner coil is 
The number of turn of outer coil is 
The radius of outer coil is 
For net magnetic field at the common center of the two coils to be zero the current flowing in the outer coil must be opposite to current flowing inner coil
The magnetic field due to inner coils is mathematically represented as
The magnetic field due to inner coils is mathematically represented as
Now for magnetic field at center to be zero
So
=> 
Answer:
16.1 N
Explanation:
From the question,
F = ma.............................. Equation 1
Where F = horizontal force, m = mass of the object, a = acceleration .
Given: m = 7.0 kg, a = 2.3 m/s²
Substitute this values into equation 1
F = (7.0×2.3)
F = 16.1 N.
Hence the magnitude of the horizontal force is 16.1 N
Answer:
The puck moves a vertical height of 2.6 cm before stopping
Explanation:
As the puck is accelerated by the spring, the kinetic energy of the puck equals the elastic potential energy of the spring.
So, 1/2mv² = 1/2kx² where m = mass of puck = 39.2 g = 0.0392 g, v = velocity of puck, k = spring constant = 59 N/m and x = compression of spring = 1.3 cm = 0.013 cm.
Now, since the puck has an initial velocity, v before it slides up the inclined surface, its loss in kinetic energy equals its gain in potential energy before it stops. So
1/2mv² = mgh where h = vertical height puck moves and g = acceleration due to gravity = 9.8 m/s².
Substituting the kinetic energy of the puck for the potential energy of the spring, we have
1/2kx² = mgh
h = kx²/2mg
= 59 N/m × (0.013 m)²/(0.0392 kg × 9.8 m/s²)
= 0.009971 Nm/0.38416 N
= 0.0259 m
= 2.59 cm
≅ 2.6 cm
So the puck moves a vertical height of 2.6 cm before stopping
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