A point charge is placed 3 m from a 4 uC charge. What is the strength of the electric field on the point charge at this
2 answers:
Answer:
Explanation:
the information we have is:
distance:
charge:
since the prefix is equal to
the charge is:
------------------
To calculate the strength of the electric field we use the following formula:
where is the charge,
is the distance and
is Coulomb's constant
substituting all the known values into the formula:
The strength of the electric field on the point charge at the distance of 3m is 4,000 N/C.
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Answer:
a) the magnitude of r is 184.62
b) the direction is 37.74° south of the negative x-axis
Explanation:
Given the data in the question;
as illustrated in the image blow;
To find the the magnitude of r, we will use the Pythagoras theorem
r² = y² + x²
r = √( y² + x²)
we substitute
r = √((-113)² + (-146)²)
r = √(12769 + 21316 )
r = √(34085 )
r = 184.62
Therefore, the magnitude of r is 184.62
To find its direction, we need to find ∅
from SOH CAH TOA
tan = opposite / adjacent
tan∅ = -113 / -146
tan∅ = 0.77397
∅ = tan⁻¹( 0.77397 )
∅ = 37.74°
Therefore, the direction is 37.74° south of the negative x-axis
Answer:
The bullet's initial speed is 243.21 m/s.
Explanation:
Given that,
Mass of the bullet,
Mass of the pendulum,
The center of mass of the pendulum rises a vertical distance of 10 cm.
We need to find the bullet's initial speed if it is assumed that the bullet remains embedded in the pendulum. Let it is v. In this case, the energy of the system remains conserved. The kinetic energy of the bullet gets converted to potential energy for the whole system. So,
V is the speed of the bullet and pendulum at the time of collision
Now using conservation of momentum as :
Put the value of V from equation (1) in above equation as :
So, the bullet's initial speed is 243.21 m/s.