30mi/6hrs is a speed of 5 mph, which converts to a pace of 12 min/mi.
Answer:
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Answer:
statement - 'The work done by friction is equal to the sum of the work done by the gravity and the initial push' is correct.
Explanation:
The statement ''The work done by friction is equal to the sum of the work done by the gravity and the initial push" is correct.
The above statement is correct because, the initial push will tend to slide down the block thus the work done by the initial push will be in the downward direction. Also, the gravity always acts in the downward direction. thus, the work done done by the gravity will also be in the downward direction
here, the downward direction signifies the downward motion parallel to the inclined plane.
Now we know that the work done by the friction is against the direction of motion. Thus, the friction force will tend to move the block up parallel to the inclined plane.
Hence, for the block to stop sliding the the above statement should be true.
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;
![F = ILBsin(\theta)](https://tex.z-dn.net/?f=F%20%3D%20ILBsin%28%5Ctheta%29)
where;
- <em>θ is the angle wire and magnetic field</em>
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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;
![\theta = tan^{-1} (\frac{y}{x} )\\\\\theta = tan^{-1} (\frac{4.3}{3.2} )\\\\\theta = 53.3 \ ^0](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20tan%5E%7B-1%7D%20%28%5Cfrac%7By%7D%7Bx%7D%20%29%5C%5C%5C%5C%5Ctheta%20%3D%20tan%5E%7B-1%7D%20%28%5Cfrac%7B4.3%7D%7B3.2%7D%20%29%5C%5C%5C%5C%5Ctheta%20%3D%2053.3%20%5C%20%5E0)
The magnitude of the wire length is calculated as follows;
![|l | = \sqrt{3.2^2 + 4.3^2} = 5.36 \ cm = 0.0536 \ m](https://tex.z-dn.net/?f=%7Cl%20%7C%20%3D%20%5Csqrt%7B3.2%5E2%20%2B%204.3%5E2%7D%20%3D%205.36%20%5C%20cm%20%3D%200.0536%20%5C%20m)
The component of the force in negative z-direction is calculated as;
![F_z = -ILB sin(\theta)\\\\F_z = -2.7 \times 0.0536 \times 1.24 \times sin(53.3)\\\\F_z = -0.144 \ N](https://tex.z-dn.net/?f=F_z%20%3D%20-ILB%20sin%28%5Ctheta%29%5C%5C%5C%5CF_z%20%3D%20-2.7%20%5Ctimes%200.0536%20%5Ctimes%201.24%20%5Ctimes%20%20sin%2853.3%29%5C%5C%5C%5CF_z%20%3D%20-0.144%20%5C%20N)
Thus, the component of the force in negative z-direction is -0.144 N.
Learn more here:brainly.com/question/22719779
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