The period of a simple pendulum is given by:

where L is the pendulum length, and g is the gravitational acceleration of the planet. Re-arranging the formula, we get:

(1)
We already know the length of the pendulum, L=1.38 m, however we need to find its period of oscillation.
We know it makes N=441 oscillations in t=1090 s, therefore its frequency is

And its period is the reciprocal of its frequency:

So now we can use eq.(1) to find the gravitational acceleration of the planet:
Answer:
no they can't talk to each other bcoz of the lack of atmosphere.
Explanation:
l hope it helps you
Answer:

Explanation:
Given:
- mass of the object on a horizontal surface,

- coefficient of static friction,

- coefficient of kinetic friction,

- horizontal force on the object,

<u>Now the value of limiting frictional force offered by the contact surface tending to have a relative motion under the effect of force:</u>

where:
normal force of reaction acting on the body= weight of the body


As we know that the frictional force acting on the body is always in the opposite direction:
So, the frictional force will not be at its maximum and will be equal in magnitude to the applied external force and hence the body will not move.
so, the frictional force will be:

Answer:
a) 2.063*10^-4
b) 1.75*10^-4
Explanation:
Given that: d= 1.628 mm = 1.628 x 10-3 I= 12 mA = 12.0 x 10-8 A The Cross-sectional area of the wire is:

a) <em>The Potential difference across a 2.00 in length of a 14-gauge copper </em>
<em> wire: </em>
L= 2.00 m
From Table Copper Resistivity
= 1.72 x 10-8 S1 • m The Resistance of the Copper wire is:

=0.0165Ω
The Potential difference across the copper wire is:
V=IR
=2.063*10^-4
b) The Potential difference if the wire were made of Silver: From Table: Silver Resistivity p= 1.47 x 10-8 S1 • m
The Resistance of the Silver wire is:

=0.014Ω
The Potential difference across the Silver wire is:
V=IR
=1.75*10^-4