Answer:
[tex]B=5\times 10^{-7}\ T[/tex]
Explanation:
Given that,
Current in a wire, I = 5 A
We need to find the magnetic field at 2m from a straight wire carrying a current of 5 A.
The magnetic field due to a wire is given by :
[tex]B=\dfrac{\mu_0I}{2\pi r}\\\\B=\dfrac{4\pi \times 10^{-7}\times 5}{2\pi \times 2}\\\\B=5\times 10^{-7}\ T[/tex]
So, the required magnetic field is [tex]5\times 10^{-7}\ T[/tex].
The skater lowers her arms as shown in the adjacent
figure decreasing her radius to 0.15 m. Find her new speed.
Answer:
is there more?
Explanation:
On a sunny day, a rooftop solar panel delivers 60 W of power to the house at an emf of 17 V. How much current flows through the panel
Answer:
3.53 amps
Explanation:
Given data
Power= 60W
Voltage= 17V
The expression relating current, power, and voltage is
P= IV
substitute
60= I*17
I= 60/17
I= 3.53 amps
Hence the current that flows is 3.53 amps
The water side of the wall of a 60-m-long dam is a quarter-circle with a radius of 7 m. Determine the hydrostatic force on the dam and its line of action when the dam is filled to the rim. Take the density of water to be 1000 kg/m3.
Answer:
[tex]26852726.19\ \text{N}[/tex]
[tex]57.52^{\circ}[/tex]
Explanation:
r = Radius of circle = 7 m
w = Width of dam = 60 m
h = Height of the dam will be half the radius = [tex]\dfrac{r}{2}[/tex]
A = Area = [tex]rw[/tex]
V = Volume = [tex]w\dfrac{\pi r^2}{4}[/tex]
Horizontal force is given by
[tex]F_x=\rho ghA\\\Rightarrow F_x=1000\times 9.81\times \dfrac{7}{2}\times 7\times 60\\\Rightarrow F_x=14420700\ \text{N}[/tex]
Vertical force is given by
[tex]F_y=\rho gV\\\Rightarrow F_y=1000\times 9.81\times 60\times \dfrac{\pi 7^2}{4}\\\Rightarrow F_y=22651982.59\ \text{N}[/tex]
Resultant force is
[tex]F=\sqrt{F_x^2+F_y^2}\\\Rightarrow F=\sqrt{14420700^2+22651982.59^2}\\\Rightarrow F=26852726.19\ \text{N}[/tex]
The hydrostatic force on the dam is [tex]26852726.19\ \text{N}[/tex].
The direction is given by
[tex]\theta=\tan^{-1}\dfrac{F_y}{F_x}\\\Rightarrow \theta=\tan^{-1}\dfrac{22651982.59}{14420700}\\\Rightarrow \theta=57.52^{\circ}[/tex]
The line of action is [tex]57.52^{\circ}[/tex].