Answer:
hai điện tích điểm bằng nhau có độ lớn 6.10^-8C đặt cách nhau một khoảng là r trong dầu hỏa có hằng số điện môi E=2,1. Lực tương tác tĩnh điện giữa chúng laf3,85.10^-4. tìm r
The block has maximum kinetic energy at the bottom of the curved incline. Since its radius is 3.0 m, this is also the block's starting height. Find the block's potential energy <em>PE</em> :
<em>PE</em> = <em>m g h</em>
<em>PE</em> = (2.0 kg) (9.8 m/s²) (3.0 m)
<em>PE</em> = 58.8 J
Energy is conserved throughout the block's descent, so that <em>PE</em> at the top of the curve is equal to kinetic energy <em>KE</em> at the bottom. Solve for the velocity <em>v</em> :
<em>PE</em> = <em>KE</em>
58.8 J = 1/2 <em>m v</em> ²
117.6 J = (2.0 kg) <em>v</em> ²
<em>v</em> = √((117.6 J) / (2.0 kg))
<em>v</em> ≈ 7.668 m/s ≈ 7.7 m/s
A front is a narrow region between two air masses of different densities.
Answer:
The angle of banked curve that makes the reliance on friction unnecessary is

Explanation:
In order the car to stay on the curve without friction, the net force in the direction of radius should be equal or smaller than the centripetal force. Otherwise the car could slide off the curve.
The only force in the direction of radius is the sine component of the weight of the car

The cosine component is equivalent to the normal force, which we will not be using since friction is unnecessary.
Newton’s Second Law states that

Also, the car is making a circular motion:

Combining the equations:

Finally the angle is
