Answer:
Explanation:
Magnetic field near current carrying wire
= 
i is current , r is distance from wire
B = 10⁻⁷ x 
force on second wire per unit length
B I L , I is current in second wire , L is length of wire
= 10⁻⁷ x x 33 x 1
= 3234 x 
This should balance weight of second wire per unit length
3234 x = .075
r = 
x 10⁻⁷
= .0043 m
= .43 cm .
Answer: There is only one Sun in the galaxy … that is the thing that rises in the morning and sets at night. However, there is a use of “sun” to signify any old star … nobody knows exactly there might be trillions out there
Explanation:
The component of the force in negative z-direction is -0.144 N.
The given parameters;
- <em>current in the wire, I = 2.7 A</em>
- <em>length of the wire, L = (3.2 i + 4.3j) cm</em>
- <em>magnetic filed, B = 1.24 i</em>
The force on the segment of the wire is calculated as follows;
where;
- <em>θ is the angle wire and magnetic field</em>
<em />
The force on the wire segment will be perpendicular in negative z-direction (applying right hand rule), so there won't be any x and y component of the force.
The angle between the wire and the magnetic field is calculated as follows;
The magnitude of the wire length is calculated as follows;
The component of the force in negative z-direction is calculated as;
Thus, the component of the force in negative z-direction is -0.144 N.
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Answer:
k1 + k2
Explanation:
Spring 1 has spring constant k1
Spring 2 has spring constant k2
After being applied by the same force, it is clearly mentioned that spring are extended by the same amount i.e. extension of spring 1 is equal to extension of spring 2.
x1 = x2
Since the force exerted to each spring might be different, let's assume F1 for spring 1 and F2 for spring 2. Hence the equations of spring constant for both springs are
k1 = F1/x -> F1 =k1*x
k2 = F2/x -> F2 =k2*x
While F = F1 + F2
Substitute equation of F1 and F2 into the equation of sum of forces
F = F1 + F2
F = k1*x + k2*x
= x(k1 + k2)
Note that this is applicable because both spring have the same extension of x (I repeat, EXTENTION, not length of the spring)
Considering the general equation of spring forces (Hooke's Law) F = kx,
The effective spring constant for the system is k1 + k2
Answer:
the object will travel 0.66 meters before to stop.
Explanation:
Using the energy conservation theorem:
The work done by the friction force is given by:
![W_f=F_f*d\\W_f=\µ*m*g*d\\W_f=0.35*4*9.81*d\\W_f=13.7d[J]](https://tex.z-dn.net/?f=W_f%3DF_f%2Ad%5C%5CW_f%3D%5C%C2%B5%2Am%2Ag%2Ad%5C%5CW_f%3D0.35%2A4%2A9.81%2Ad%5C%5CW_f%3D13.7d%5BJ%5D)
so:
