Answer:
1- The acceleration of the object is larger in magnitude the smaller the radius of the circle.
Explanation:
The acceleration of an object in a circular path is
As can be seen from the equation, if the radius of the circle is decreases, the magnitude of the acceleration increases.
As for the direction of the acceleration, it is always towards the center, and it is always perpendicular to the direction of the velocity.
Answer:
Diameter of Newton’s 5th ring = 0.30 cm
Diameter of Newton’s 15th ring = 0.62 cm
Diameter of Newton’s 25th ring = ?
From Newton’s rings experiment we infer that
D2n+m − D2n = 4λmR
For the 5th and 15th rings we have
D215 − D25 = 4λ * 10 * R _______ (1) (m = 10)
For 15th and 25th rings
D225 − D215 = 4λ * 10 * R _______ (2) (m = 10)
We equate the two derivatives
Equation (2) = Equation (1)
D225 − D215 = D215 − D25
D225 = 2D215 – D25
Substituting the values into the equation
D225 = 2 * 0.62 * 0.62 – 0.3 * 0.3 =0.6788 cm2
D25 = 0.8239 cm
We can solve the problem by using Ohm's law, which states that an Ohmic conductor the following relationship holds:
where
is the potential difference applied to the resistor
I is the current flowing through it
R is the resistance
In our problem, I=4.00 A and
, so the potential difference is
Answer:
The minimum speed =
Explanation:
The minimum speed that the rocket must have for it to escape into space is called its escape velocity. If the speed is not attained, the gravitational pull of the planet would pull down the rocket back to its surface. Thus the launch would not be successful.
The minimum speed can be determined by;
Escape velocity =
where: G is the universal gravitational constant, M is the mass of the planet X, and R is its radius.
If the appropriate values of the variables are substituted into the expression, the value of the minimum speed required can be determined.