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bija089 [108]
1 year ago
13

A vertical wire carries a current straight up in a region where the magnetic field vector points due north. What is the directio

n of the resulting force on this current?
Physics
1 answer:
Elanso [62]1 year ago
7 0

Answer:

The direction of the resulting force on this current is due east.

Explanation:

Given;

direction of the magnetic field to be due north

Applying right hand rule which states that: to determine the direction of the magnetic force on a positive moving charge point the thumb of the right hand in the direction of velocity v, the fingers in the direction of magnetic field B, and a perpendicular to the palm points in the direction of magnetic force.

Since the magnetic force must be perpendicular to the magnetic field, and direction of the magnetic field is due north, then the magnetic force must be due East.

Therefore, the direction of the resulting force on this current is due east.

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In Paul Hewitt's book, he poses this question: "If the forces that act on a bullet and the recoiling gun from which it is fired
Sauron [17]
They have different accelerations because of their masses. According to Newton's Second Law, an objects acceleration is inversely proportional to its mass. Therefore the object with the larger mass, in this case the gun, will have a smaller acceleration. In the same way, the less massive object, being the bullet, will have a higher acceleration.

Hope this helps :)
4 0
1 year ago
A 60.0 kg object is moving east at 8.00 m/s, and then slows down to 4.00 m/s. How much work was done?
Artist 52 [7]
The answer is  -1440. That is letter A.
8 0
1 year ago
Read 2 more answers
The atomic number of beryllium (Be) is 4, and the atomic number of barium (Ba) is 56. Which comparison is best supported by this
Thepotemich [5.8K]

The answer is:

B)They are in the same group because they have similar chemical properties, but they are in different periods because they have very different atomic numbers.

The explanation:

The vertical columns on the periodic table are called groups or families because of their similar chemical behavior. All the members of a family of elements have the same number of valence electrons and similar chemical properties.

A period is a horizontal row of elements on the periodic table. For example, the elements sodium ( Na ) and magnesium ( Mg ) are both in period 3. The elements astatine ( At ) and radon ( Rn ) are both in period 6.

6 0
2 years ago
Read 2 more answers
Three point charges are arranged on a line. Charge q3 = +5.00 nC and is at the origin. Charge q2 = -2.00 nC and is at x = 5.00 c
tatuchka [14]

Answer:

q₁= +0.5nC

Explanation:

Theory of electrical forces

Because the particle q3 is close to three other electrically charged particles, it will experience two electrical forces and the solution of the problem is of a vector nature.

To solve this problem we apply Coulomb's law:

Two point charges (q1, q2) separated by a distance (d) exert a mutual force (F) whose magnitude is determined by the following formula:

o solve this problem we apply Coulomb's law:  

Two point charges (q₁, q₂) separated by a distance (d) exert a mutual force (F) whose magnitude is determined by the following formula:  

F=K*q₁*q₂/d² Formula (1)  

F: Electric force in Newtons (N)

K : Coulomb constant in N*m²/C²

q₁,q₂:Charges in Coulombs (C)  

d: distance between the charges in meters

Data:

Equivalences

1nC= 10⁻⁹ C

1cm= 10⁻² m

Data

q₃=+5.00 nC =+5* 10⁻⁹ C

q₂= -2.00 nC =-2* 10⁻⁹ C

d₂= 5.00 cm= 5*10⁻² m

d₁= 2.50 cm=  2.5*10⁻² m

k = 8.99*10⁹ N*m²/C²

Calculation of magnitude and sign of q1

Fn₃=0 : net force on q3 equals zero

F₂₃:The force F₂₃ that exerts q₂ on q₃ is attractive because the charges have opposite signs,in direction +x.

F₁₃:The force F₂₃ that exerts q₂ on q₃ must go in the -x direction so that Fn₃ is zero, therefore q₁ must be positive and F₂₃ is repulsive.

We propose the algebraic sum of the forces on q₃

F₂₃ - F₁₃=0

\frac{k*q_{2} *q_{3} }{d_{2}^{2}  } -\frac{k*q_{1} *q_{3} }{d_{1}^{2}  }=0

We eliminate k*q₃ of the equation

\frac{q_{1} }{d_{1}^{2}  } = \frac{q_{2} }{d_{2}^{2}  }

q_{1} =\frac{q_{2} *d_{1} ^{2} }{d_{2}^{2}  }

q_{1} =\frac{2*10^{-9}*2.5^{2}*10^{-4}   }{5^{2}*10^{-4}  }

q₁= +0.5*10⁻⁹ C

q₁= +0.5nC

4 0
1 year ago
The atmosphere pressure can support mercury in a tube, which the upper end is closed, up to 0.76 meter. If the mercury is replac
Leni [432]

Answer:

Maximum height the atmosphere pressure can support the

water=10.336 m

Explanation:

We know that ,

Pressure = h\cdot\rho\cdot g

Case 1 - Mercury in the tube

Density\ of\ mercury =\rho_1\\and\ height\ attained\ for\ mercury\ column = h_1

Case 2 - Water in the tube

Density\ of\ water =\rho_2\\and\ height\ attained\ for\ water\ column = h_2

Since atmospheric pressure is same

.P=h_1\cdot\rho_1\cdot g = h_2\cdot\rho_2\cdot g

or,  h_2=\frac{h_1\rho_1}{\rho_2}

Given\ h_1= 0.76\  m,\rho_1=13.6\cdot\rho_2

∴ h_2=0.76\cdot13.6=10.336\ m

Hence height of the water column =10.336 m

6 0
2 years ago
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