The height, h to which the package of mass m bounces to depends on its initial velocity, v and the acceleration due to gravity, g and is given below:
<h3>What are perfectly elastic collision?</h3>
Perfectly elastic collisions are collisions in which the momentum as well as the energy of the colliding bodies is conserved.
In perfectly elastic collisions, the sum of momentum before collision is equal to the momentum after collision.
Also, the sum of kinetic energy before collision is equal to the sum of kinetic energy after collision.
Since some of the Kinetic energy is converted to potential energy of the body;
Therefore, the height to which the package m bounces to depends on its initial velocity and the acceleration due to gravity.
Learn more about elastic collisions at: brainly.com/question/7694106
The correct answer is C) towards the center of the circle.
Although the object is moving at a constant speed it is constantly accelerating due to the constant change in direction as it describes the circular path. This causes a constant change in velocity as velocity is a vector quantity.
For the object to maintain the circular path there has to be centripetal force acting on the object and this centripetal force is directed towards the center of the circle.
Answer:
a) B = 1.99 x 10⁻⁴ Tesla
b) B = 0.88 x 10⁻⁴ Tesla
Explanation:
According to Biot - Savart Law, the magnetic field due to a currnt carrying straight wire is given as:
B = μ₀ I L/4πr²
where,
μ₀ = permebility of free space = 1.25 x 10⁻⁶ H m⁻¹
I = current = 2 A
L = Length of wire = 40 cm = 0.4 m
a)
r = radius of magnetic field = 2 cm = 0.02 m
Therefore,
B = (1.25 x 10⁻⁶ H m⁻¹)(2 A)(0.4 m)/4π(0.02 m)²
<u>B = 1.99 x 10⁻⁴ Tesla</u>
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b)
r = radius of magnetic field = 3 cm = 0.03 m
Therefore,
B = (1.25 x 10⁻⁶ H m⁻¹)(2 A)(0.4 m)/4π(0.03 m)²
<u>B = 0.88 x 10⁻⁴ Tesla</u>
Answer:
The current in the primary is 0.026 A
Explanation:
Using the formula
I1 = (V1/V2)*I2
we have
I1 = (6.4/120)*0.500
I1 = 0.026 A
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