Explanation:
<em>an </em><em>eclipse </em><em>happens </em><em>when </em><em>one </em>
<em>astronomical </em><em>body </em><em>block </em><em>light </em>
<em>from </em><em>or </em><em>to </em><em>another</em><em>,</em><em> </em><em>the </em><em>moon </em><em>moves </em><em>into</em>
<em>the </em><em>shadow </em><em>of </em><em>earth </em><em>cast </em><em>by </em><em>sun</em><em>.</em><em>.</em><em>.</em><em> </em><em>In </em><em>a </em><em>solar </em>
<em>eclipse</em><em>,</em><em> </em><em>the </em><em>moon </em><em>passes </em><em>between </em><em>Earth </em>
<em>and </em><em>the </em><em>sun </em><em>stops </em><em>some </em><em>all </em><em>of </em><em>the </em><em>sun's </em><em>light</em>
<em>from </em><em>reaching </em><em>Earth </em>
- Initial velocity=u=30m/s
- Final velocity=v=10m/s
- Time=5s




Now
For the second question
Using 2nd equation of motion





Answer:
b. increasing the number of turns per unit length on the solenoid
e. increasing the current in the solenoid
Explanation:
As we know that energy density depends on the strength of the magnetic field. The magnetic field strength depends on the no of turns of the solenoid and the current passing through it. The greater the number of turns per unit length, greater the current passing through it, more stronger the magnetic field is. As
B = μ₀nI
n = no of turns
I = current through the wire
So the right options are
b. increasing the number of turns per unit length on the solenoid
e. increasing the current in the solenoid
Answer:
D. Perfectly inelastic
Explanation:
Kinetic energy is lost so the two bodies stick together.
Answer:
A. The period of an oscillation does not depend upon amplitude.
Explanation:
The period of a spring-mass system is:
T = 1/f = 2π√(m/k)
where f is the frequency, m is the mass, and k is the spring constant.
The answer isn't B. There are no frictionless systems in the real world.
The answer isn't C or D. As shown, the frequency is a function of both the mass and the spring constant.
The answer isn't E. Turning motion into heat is not an advantage for a clock.
The correct answer is A. The period of the system does not depend on the amplitude.