Suppose you have a pendulum clock which keeps correct time on Earth (acceleration due to gravity = 1.6 m/s2). For ever hour interval (on Earth) the Moon clock:

A) (9.8/1.6) h

B) 1 h

C) the square root of 9.8/1.6 h

D) (1.6/9.8) h

E) the square root of 1.6/9.8 h

Question

Angel Stephenson

Physics

7 September, 15:16

Suppose you have a pendulum clock which keeps correct time on Earth (acceleration due to gravity = 1.6 m/s2). For ever hour interval (on Earth) the Moon clock:

A) (9.8/1.6) h

B) 1 h

C) the square root of 9.8/1.6 h

D) (1.6/9.8) h

E) the square root of 1.6/9.8 h

+2

Answers (1)

Know the Answer?

Freckles · Answer 1

The time-period of a simple pendulum is

Time = 2 π √ (length/grav-accel)

After unraveling the question, then completing it, and working out what I believe it's trying to ask, the choice that correctly answers the question that I have invented is choice-E.

Suppose you have a pendulum clock which keeps correct time on Earth (acceleration due to gravity = 1.6 m/s2). For ever hour interval (on Earth) the Moon clock:A) (9.8/1.6) hB) 1 hC) the square root of 9.8/1.6 hD) (1.6/9.8) hE) the square root of 1.6/9.8 h

Suppose you have a pendulum clock which keeps correct time on Earth (acceleration due to gravity = 1.6 m/s2). For ever hour interval (on Earth) the Moon clock:

A) (9.8/1.6) h

B) 1 h

C) the square root of 9.8/1.6 h

D) (1.6/9.8) h

E) the square root of 1.6/9.8 h