The torque value is 560 Nm, so the result does not depend on the value of Ф.
Given:
Mass=80.0 g
Radius=20.0 cm
Magnetic field=0.35T
In a magnetic field, a loop of current-carrying wire will experience the following torque:
τ=μ×Β
Here, is the μ magnetic moment and Β is the magnetic field.
τ= 1600 × 0.35 T
τ= 560 Nm
<h3>
What is torque and its SI unit?</h3>
The rotating equivalent of force is torque. It is also known as the moment, moment of force, rotating force, or turning effect depending on the field of research. It serves as an example of how a force can alter a body's rotational motion.
The Newton-metre, or kgm2sec-2, is the SI unit for torque. How did we get to this point? Considering the equation Torque = Force X Distance Torque is measured in newton-meters, whereas distance is measured in meters and force is measured in newtons.
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Answer:
The kinetic energy of the ejected electrons increases.
Explanation:
As we know that electrons are only ejected from a metal surface if the frequency of the incident light increases the work function of the metal. If the frequency of the incident light is less than the work function of the metal no matter how intense the beam the electrons will not be ejected from the surface.
Using conservation of energy principle we have
If we increase the intensity of incident light the term on the LHS of the above equation increases this increase appears in the kinetic energy term in RHS of the equation since remains constant.
Heat must be added to raise the temperature from 20.0 •C to 82 •C is
Q= (4200 x 1.8 + 880 x 1.3) x (82 - 20) = 539648J
The true statement according to Newton's laws of motion is; "there is no net force on the block during the 2nd second as zero net force keeps the motion constant."
According to the Newton first law of motion, an object will continue in its state of rest or uniform motion unless it is acted upon by a net force. The action of a net force leads to an acceleration.
In the 2nd second of the journey, there is no net force on the block hence the block continues to move at constant velocity.
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To solve this exercise it is necessary to apply the equations concerning Work, both by general definition and by conservation of energy.
In other words, the work done by an object due to gravity is the equivalent to that defined by the potential energy equations, that is
Where,
m=mass
g=gravitational acceleration
Change in height
On the other hand we have that the work done by tension is defined by the conservation of kinetic and potential energy, that is to say
Where,
Change in Kinetic Energy
Change in Potential Energy
PART A) As defined by the work done by gravity would be given by,
Therefore the work done by gravity is 25.725kJ
PART B) The work done by the tension applies the energy conservation equation, that is to say
Replacing with our values,
Therefore the work done by tension is 25.9kJ