The force of Earth's gravity pulls down on a stone after someone tosses it off a cliff. What is the "equal and opposite force" d
1 answer:
The equal and opposite force is the force of the stone's gravity pulling up on thy Earth.
<u>Explanation</u>:
- When someone tosses a stone off a cliff, the force of gravity of Earth pulls down and the force of the stone's gravity pulls up on the Earth due to the equal and opposite reaction exerted between the Earth and the stone.
- In Newton's third law, two important features are present. One is the force exerted on two bodies will have the same magnitude but opposite direction. The other feature is force is acting on two different objects.
- This exerted force is equal in magnitude but opposite in direction whereas the gravity exerts a downward force on a stone and upward force on an Earth which is opposite in direction.
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Answer:
hello your question is incomplete attached below is missing part of the question
answer:
1 ) Magnetic field due to long current carrying wire :
Therefore the net magnetic field due the both wires ; B = B + B
. when we adjust the current I
= I
then the Netfield (B ) = zero
2) The distance between the field lines are not equally spaced and this is because the separation between field lines increases with the increase in the distance between the wires
3) Increase in current through the wire will lead to increase in force and this can be explained via this equation
Explanation:
1 ) Magnetic field due to long current carrying wire :
Therefore the net magnetic field due the both wires ; B = B + B
. when we adjust the current I
= I
then the Netfield (B ) = zero
2) The distance between the field lines are not equally spaced and this is because the separation between field lines increases with the increase in the distance between the wires
3) Increase in current through the wire will lead to increase in force and this can be explained via this equation
Explanation:
Given that,
Mass, m = 0.08 kg
Radius of the path, r = 2.7 cm = 0.027 m
The linear acceleration of a yo-yo, a = 5.7 m/s²
We need to find the tension magnitude in the string and the angular acceleration magnitude of the yo‑yo.
(a) Tension :
The net force acting on the string is :
ma=mg-T
T=m(g-a)
Putting all the values,
T = 0.08(9.8-5.7)
= 0.328 N
(b) Angular acceleration,
The relation between the angular and linear acceleration is given by :
(c) Moment of inertia :
The net torque acting on it is, , I is the moment of inertia
Also,
So,
Hence, this is the required solution.