Answer:
The force is the same
Explanation:
The force per meter exerted between two wires carrying a current is given by the formula
![\frac{F}{L}=\frac{\mu_0 I_1 I_2}{2\pi r}](https://tex.z-dn.net/?f=%5Cfrac%7BF%7D%7BL%7D%3D%5Cfrac%7B%5Cmu_0%20I_1%20I_2%7D%7B2%5Cpi%20r%7D)
where
is the vacuum permeability
is the current in the 1st wire
is the current in the 2nd wire
r is the separation between the wires
In this problem
![I_1=2.79 A\\I_2=4.36 A\\r = 9.15 cm = 0.0915 m](https://tex.z-dn.net/?f=I_1%3D2.79%20A%5C%5CI_2%3D4.36%20A%5C%5Cr%20%3D%209.15%20cm%20%3D%200.0915%20m)
Substituting, we find the force per unit length on the two wires:
![\frac{F}{L}=\frac{(4\pi \cdot 10^{-7})(2.79)(4.36)}{2\pi (0.0915)}=2.66\cdot 10^{-5}N](https://tex.z-dn.net/?f=%5Cfrac%7BF%7D%7BL%7D%3D%5Cfrac%7B%284%5Cpi%20%5Ccdot%2010%5E%7B-7%7D%29%282.79%29%284.36%29%7D%7B2%5Cpi%20%280.0915%29%7D%3D2.66%5Ccdot%2010%5E%7B-5%7DN)
However, the formula is the same for the two wires: this means that the force per meter exerted on the two wires is the same.
The same conclusion comes out from Newton's third law of motion, which states that when an object A exerts a force on an object B, then object B exerts an equal and opposite force on object A (action-reaction). If we apply the law to this situation, we see that the force exerted by wire 1 on wire 2 is the same as the force exerted by wire 2 on wire 1 (however the direction is opposite).
Answer:
<em>The force of friction acting on the block has a magnitude of 15 N and acts opposite to the applied force.</em>
Explanation:
<u>Net Force
</u>
The Second Newton's law states that an object acquires acceleration when an unbalanced net force is applied to it.
The acceleration is proportional to the net force and inversely proportional to the mass of the object.
If the object has zero net force, it won't get accelerated and its velocity will remain constant.
The m=2 kg block is being pulled across a horizontal surface by a force of F=15 N and we are told the block moves at a constant velocity. This means the acceleration is zero and therefore the net force is also zero.
Since there is an external force applied to the box, it must have been balanced by the force of friction, thus the force of friction has the same magnitude acting opposite to the applied force.
The force of friction acting on the block has a magnitude of 15 N opposite to the applied force.
The question is incomplete. Here is the complete question.
A floating ice block is pushed through a displacement vector d = (15m)i - (12m)j along a straight embankment by rushing water, which exerts a force vector F = (210N)i - (150N)j on the block. How much work does the force do on the block during displacement?
Answer: W = 4950J
Explanation: <u>Work</u> (W), in physics, is done when a force acts on an object that has a displacement form a place to another:
W = F · d
As the formula shows, Work is a scalar product, i.e, it results in a number, so, Work only has magnitude.
Force and displacement for the ice block are in 2 dimensions, then work will be:
W = (210)i - (150)j · (15)i - (12)j
W = (210*15) + (150*12)
W = 3150 + 1800
W = 4950J
During the displacement, the ice block has a work of 4950J
Answer:
I think my friend got it all wrong, as coolness can not be transferred but heat was actually transferred between my hand and the glass windows
Explanation:
In thermodynamics, coolness can not be transferred, only heat can be transferred
Here is how the mechanism of why i felt cold works, my body gave out heat, hence there was heat transfer from a region of high to a low heat region, equilibrium was reached and I started feeling the coolness in my hands.
Answer:
c) It increases by a factor of 8
Explanation:
According to Faraday's law (and Lenz' law), the induced EMF is given as the rate of change of magnetic flux.
Mathematically:
V = -dФ/dt
Magnetic flux, Ф, is given as:
Ф = BA
where B = magnetic field strength and A = Area of object
Hence, induced EMF becomes:
V = -d(BA)/dt or -BA/t
If the magnetic field is increased by a factor of 4, (
) and the time required for the rod to move is decreased by a factor of 2 (
), the induced EMF becomes:
![V_n = -(B_nA)/t_n](https://tex.z-dn.net/?f=V_n%20%3D%20-%28B_nA%29%2Ft_n)
![V_n = \frac{-4BA}{(t/2)}\\\\V_n = \frac{-8BA}{t} \\\\V_n = 8V\\](https://tex.z-dn.net/?f=V_n%20%3D%20%5Cfrac%7B-4BA%7D%7B%28t%2F2%29%7D%5C%5C%5C%5CV_n%20%3D%20%5Cfrac%7B-8BA%7D%7Bt%7D%20%5C%5C%5C%5CV_n%20%3D%208V%5C%5C)
The EMF has increased by a factor of 8.