Question

Question 2

Answer 1

Answer:

Centripetal acceleration, a = 0.03125 m/s²

Explanation:

Given that,

Distance from center of a rotating platter, r = 0.08 m

Tangential speed of the pickle,                v = 0.05 m/s

The centripetal acceleration is given by the formula

                                         a = v²/r

Substituting the given values in the above equation

                                         a =  (0.05 m/s)² / 0.08 m

                                            =  0.03125 m/s²

Hence, the the pickle's centripetal acceleration a = 0.03125 m/s²

Answer 2

The pickle's centripetal acceleration is 0.03125 m/s²

Further explanation

Centripetal Acceleration can be formulated as follows:

$\large {\boxed {a = \frac{ v^2 } { R } }$

a = Centripetal Acceleration ( m/s² )

v = Tangential Speed of Particle ( m/s )

R = Radius of Circular Motion ( m )

$\texttt{ }$

Centripetal Force can be formulated as follows:

$\large {\boxed {F = m \frac{ v^2 } { R } }$

F = Centripetal Force ( m/s² )

m = mass of Particle ( kg )

v = Tangential Speed of Particle ( m/s )

R = Radius of Circular Motion ( m )

Let us now tackle the problem !

$\texttt{ }$

To find the pickle's centripetal acceleration can be carried out in the following way:

Given:

Radius of Circular Motion = 0.08 m

Tangential Speed of Pickle = 0.05 m/s

Unknown:

Centripetal Acceleration of Pickle = a = ?

Solution:

$a = v^2 \div R$

$a = 0.05^2 \div 0.08$

$a = 0.03125 \texttt{ m/s}^2$

$\texttt{ }$

Learn more

• Impacts of Gravity : brainly.com/question/5330244
• Effect of Earth’s Gravity on Objects : brainly.com/question/8844454
• The Acceleration Due To Gravity : brainly.com/question/4189441

Answer details

Grade: High School

Subject: Physics

Chapter: Circular Motion

Keywords: Gravity , Unit , Magnitude , Attraction , Distance , Mass , Newton , Law , Gravitational , Constant