Answer:
For your kind information, read the question thoroughly!
The bottom is the container part and not the liquid.
I hope it will be useful.
Answer:
The magnitude of the magnetic field is 1.83 x
T.
Explanation:
The flow of an electric current in a straight wire induces magnetic field around the wire. When current is flowing through two wires in the same direction, a force of attraction exists between the wires. But if the current flows in opposite directions, the force of repulsion is felt by the wires.
In the given question, the direction of flow of current through the wires is opposite, thus both wires applies the same field on each other. The result to repulsion between them.
The magnetic field (B) between the given wires can be determined by:
B = 
where: I is the current, r is the distance between the wires and
is the magnetic field constant.
But, I = 11 A, r = 0.12 m and
= 4
x
Tm/A
So that;
B = 
= 1.8333 x 
B = 1.83 x
T
Answer:
y maximum 3.54 m, value X 2.35 m
Explanation:
We have a projectile launch problem, let's calculate the maximum height of the projectile, where the vertical speed must be zero
Vyf² = Vyo² - 2g (Y-Yo)
Where Yo is the initial height of the ramp 1.5 m
0 = Vyo² -2g (Y-Yo)
Y-Yo = Voy² / 2g
Y = Yo + Voy² / 2g
Let's calculate the velocity components using trigonometry
Voy = vo without T
Vox = Vo cost
Voy = 7.3 sin 60
Vox = 7.3 cos 60
Voy = 6.32 m / s
Vox = 3.65 m / s
Let's calculate the maximum height
Y = 1.5 +6.32²/2 9.8
Y = 3.54 m
This is the maximum height from the ground
b) They ask us for the position of this point horizontally, we can calculate it looking for the time it took for the skateboarder to reach the highest point
Vfy = Voy - gt
0 = Voy - gt
t = Voy / g
t = 6.32 / 9.8
t = 0.645 s
Since there is no acceleration on the x-axis, we have a uniform movement, we can calculate the distance for this time
X = Vox t
X = 3.65 0.645
X= 2.35 m
Answer:
C
Explanation:
That is where the most heat and light is showing on this diagram.
Answer: 0.5 m
Explanation:
Given
Mass of the person is 
Trampoline launches the person into the air up to height of 
Force experience by springs is 
Here, the work done on displacing the springs is equivalent to the Potential energy acquired by the person i.e.
![\Rightarrow F\cdot x=mgh\quad [\text{x=displacement of the trampoline}]\\\\\text{Insert the values}\\\\\Rightarrow x=\dfrac{50\times 9.8\times 2}{1960}\\\\\Rightarrow x=\dfrac{980}{1960}\\\\\Rightarrow x=0.5\ m](https://tex.z-dn.net/?f=%5CRightarrow%20F%5Ccdot%20x%3Dmgh%5Cquad%20%5B%5Ctext%7Bx%3Ddisplacement%20of%20the%20trampoline%7D%5D%5C%5C%5C%5C%5Ctext%7BInsert%20the%20values%7D%5C%5C%5C%5C%5CRightarrow%20x%3D%5Cdfrac%7B50%5Ctimes%209.8%5Ctimes%202%7D%7B1960%7D%5C%5C%5C%5C%5CRightarrow%20x%3D%5Cdfrac%7B980%7D%7B1960%7D%5C%5C%5C%5C%5CRightarrow%20x%3D0.5%5C%20m)