Solution :
a). B at the center :

Here, one of the current is in the clockwise direction and therefore, the other current must be in the clockwise direction in order to cancel out the effect of the magnetic field that is produced by the other.
Therefore, the answer is ANTICLOCKWISE or COUNTERCLOCKWISE
b). Also, the sum of the fields must be zero.
Therefore,

So,


A
Therefore, the current in the outer wire is 24.38 ampere.
Answer:
The actual angle is 30°
Explanation:
<h2>Equation of projectile:</h2><h2>y axis:</h2>

the velocity is Zero when the projectile reach in the maximum altitude:

When the time is vo/g the projectile are in the middle of the range.
<h2>x axis:</h2>

R=Range


**sin(2A)=2sin(A)cos(A)
<h2>The maximum range occurs when A=45°
(because sin(90°)=1)</h2><h2>The actual range R'=(2/√3)R:</h2>
Let B the actual angle of projectile

2B=60°
B=30°
Answer:
The possible range of wavelengths in air produced by the instrument is 7.62 m and 0.914 m respectively.
Explanation:
Given that,
The notes produced by a tuba range in frequency from approximately 45 Hz to 375 Hz.
The speed of sound in air is 343 m/s.
To find,
The wavelength range for the corresponding frequency.
Solution,
The speed of sound is given by the following relation as :

Wavelength for f = 45 Hz is,


Wavelength for f = 375 Hz is,


So, the possible range of wavelengths in air produced by the instrument is 7.62 m and 0.914 m respectively.
Answer:0.61 m
Explanation:
Given
Initial velocity (u)=7.40 m/s
launch angle 
height of fence=0.7 m
Trajectory of Projectile is given by
For x=2.1 m
At x=2.1 m Y=1.31
therefore Fox clear fence by a margin =1.31-0.7=0.61 m
Why do teachers give such hard questions? anyways hopefully your question will get answered! (btw i also need the points but yeah gl!!!)