Struggling with this, pls answer:( Brainliest to the (FIRST) right answer
1 answer:
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Answer:
the ball travelled approximately 60 m towards north before stopping
Explanation:
Given the data in the question;
First course : = 0.75 m/s²,
= 20 m,
= 10 m/s
now, form the third equation of motion;
v² = u² + 2as
we substitute
² = (10)² + (2 × 0.75 × 20)
² = 100 + 30
² = 130
= √130
= 11.4 m/s
for the Second Course:
= 11.4 m/s,
= -1.15 m/s²,
= 0
Also, form the third equation of motion;
v² = u² + 2as
we substitute
0² = (11.4)² + (2 × (-1.15) × )
0 = 129.96 - 2.3
2.3 = 129.96
= 129.96 / 2.3
= 56.5 m
so;
|d| = √( ² +
² )
we substitute
|d| = √( (20)² + (56.5)² )
|d| = √( 400 + 3192.25 )
|d| = √( 3592.25 )
|d| = 59.9 m ≈ 60 m
Therefore, the ball travelled approximately 60 m towards north before stopping
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:
E=
Explanation:
We are given that
Charge on ring= Q
Radius of ring=a
We have to find the magnitude of electric filed on the axis at distance a from the ring's center.
We know that the electric field at distance x from the center of ring of radius R is given by
Substitute x=a and R=a
Then, we get
Where K=
Hence, the magnitude of the electric filed due to charged ring on the axis of ring at distance a from the ring's center=