Step-#1:
Ignore the wire on the right.
Find the strength and direction of the magnetic field at P,
caused by the wire on the left, 0.04m away, carrying 5.0A
of current upward.
Write it down.
Step #2:
Now, ignore the wire on the left.
Find the strength and direction of the magnetic field at P,
caused by the wire on the right, 0.04m away, carrying 8.0A
of current downward.
Write it down.
Step #3:
Take the two sets of magnitude and direction that you wrote down
and ADD them.
The total magnetic field at P is the SUM of (the field due to the left wire)
PLUS (the field due to the right wire).
So just calculate them separately, then addum up.
The vertical weight carried by the builder at the rear end is F = 308.1 N
<h3>Calculations and Parameters</h3>
Given that:
The weight is carried up along the plane in rotational equilibrium condition
The torque equilibrium condition can be used to solve
We can note that the torque due to the force of the rear person about the position of the front person = Torque due to the weight of the block about the position of the front person
This would lead to:
F(W*cosθ) = mgsinθ(L/2) + mgcosθ(W/2)
F(1cos20)= 197/2(3.10sin20 + 2 cos 20)
Fcos20= 289.55
F= 308.1N
Read more about vertical weight here:
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Answer:
Part a)
![h' = \frac{10}{14} h](https://tex.z-dn.net/?f=h%27%20%3D%20%5Cfrac%7B10%7D%7B14%7D%20h)
Part b)
if both sides are rough then it will reach the same height on the other side because the energy is being conserved.
Part c)
Since marble will go to same height when it is rough while when it is smooth then it will go to the height
![h' = \frac{10}{14} h](https://tex.z-dn.net/?f=h%27%20%3D%20%5Cfrac%7B10%7D%7B14%7D%20h)
so on smooth it will go to lower height
Explanation:
As we know by energy conservation the total energy at the bottom of the bowl is given as
![\frac{1}{2} mv^2 + \frac{1}{2}I\omega^2 = mgh](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%20mv%5E2%20%2B%20%5Cfrac%7B1%7D%7B2%7DI%5Comega%5E2%20%3D%20mgh)
here we know that on the left side the ball is rolling due to which it is having rotational and transnational both kinetic energy
now on the right side of the bowl there is no friction
so its rotational kinetic energy will not change and remains the same
so it will have
![\frac{1}{2}mv^2 = mgh'](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7Dmv%5E2%20%3D%20mgh%27)
now we know that
![I = \frac{2}{5}mr^2](https://tex.z-dn.net/?f=I%20%3D%20%5Cfrac%7B2%7D%7B5%7Dmr%5E2)
![\omega = \frac{v}{r}](https://tex.z-dn.net/?f=%5Comega%20%3D%20%5Cfrac%7Bv%7D%7Br%7D)
so we have
![\frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mr^2)(\frac{v}{r})^2 = mgh](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7Dmv%5E2%20%2B%20%5Cfrac%7B1%7D%7B2%7D%28%5Cfrac%7B2%7D%7B5%7Dmr%5E2%29%28%5Cfrac%7Bv%7D%7Br%7D%29%5E2%20%3D%20mgh)
![\frac{1}{2}mv^2 + \frac{1}{5}mv^2 = mgh](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7Dmv%5E2%20%2B%20%5Cfrac%7B1%7D%7B5%7Dmv%5E2%20%3D%20mgh)
![\frac{7}{10}mv^2 = mgh](https://tex.z-dn.net/?f=%5Cfrac%7B7%7D%7B10%7Dmv%5E2%20%3D%20mgh)
![\frac{1}{2}mv^2 = \frac{10}{14}mgh](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7Dmv%5E2%20%3D%20%5Cfrac%7B10%7D%7B14%7Dmgh)
so the height on the smooth side is given as
![h' = \frac{10}{14} h](https://tex.z-dn.net/?f=h%27%20%3D%20%5Cfrac%7B10%7D%7B14%7D%20h)
Part b)
if both sides are rough then it will reach the same height on the other side because the energy is being conserved.
Part c)
Since marble will go to same height when it is rough while when it is smooth then it will go to the height
![h' = \frac{10}{14} h](https://tex.z-dn.net/?f=h%27%20%3D%20%5Cfrac%7B10%7D%7B14%7D%20h)
so on smooth it will go to lower height
A joule is a unit of energy.
The work done to transport an electron from the positive to the negative terminal is 1.92×10⁻¹⁹ J.
Given:
Potential difference, V = 1.2 V
Charge on an electron, e = 1.6 × 10⁻¹⁹ C
Calculation:
We know that the work done to transport an electron from the positive to the negative terminal is given as:
W.D = (Charge on electron)×(Potential difference)
= e × V
= (1.6 × 10⁻¹⁹ C)×(1.2 V)
= 1.92 × 10⁻¹⁹ J
Therefore, the work done in bringing the charge from the positive terminal to the negative terminal is 1.92 × 10⁻¹⁹ J.
Learn more about work done on a charge here:
<u>brainly.com/question/13946889</u>
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