The mechanical advantage of a wheel and axle is the radius of the wheel divided by the radius of the axle. On wheel and axle A,
2 answers:
Answer:
Wheel and axle B
Explanation:
It is given that the mechanical advantage of a wheel and axle is the radius of the wheel divided by the radius of the axle.
Mathematically, it can be written as :
.................(1)
On wheel and axle A, the wheel and axle are very close in size while on wheel and axle B, the radius of the wheel is much greater than the radius of the axle.
From equation (1) it is clear that the mechanical advantage is directly proportional to the radius of wheel.
So, the mechanical advantage of wheel and axle B is greater.
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Answer:
175.3 N
Explanation:
The motion of the ball is a uniform circular motion, therefore the net force on it must be equal to the centripetal force.
There are two forces acting on the ball at the lowest point of motion:
- The tension in the string, T , upward
- The weight of the ball, , downward
The net force (centripetal force) has the same direction as the tension (upward, towards the centre of the circular path), so we can write:
where the term on the right is the expression for the centripetal force, and where:
T is the tension in the string
is the weight of the ball
is the mass of the ball
v = 5.7 m/s is the speed of the ball at the lowest point
r = 4.4 m is the length of the rope, so the radius of the circle
Solving for T, we find the tension in the string:
Answer:
The correct statement is "The electric field is directed toward the electron and has a magnitude of ".
Explanation:
According to Coulomb's law, the magnitude of the electric field due to a static point charge q at a point r distance away from it is given by
- k is the Coulmob's constant.
The direction of the electric field along the line joining the charge and the point where electric field is to be found and it is directed from positive charge to negative charge.
Conventionally, we assume a positive test charge placed at the point where electric field is to be found, the test charge has very small charge such that its charge does not affect the electric field due to the given charge.
The charge on the electron = -e.
The electric field due to an electron is given by
The direction of this electric field is from positive test charge, placed at the point where electric field is to be found, towards the electron along the line joining the two.
Thus, the correct statement is "The electric field is directed toward the electron and has a magnitude of ".