Answer: vf1/vf2= 1/ sqrt(2)
Explanation :on the moon no drag force so we have only the force of gravity. aceleration is g(moon)= 1.62m/s2.the rest is basic kinematics
if the rock travels H to the bottom we can calculate velocity:
vo=0m/s (drops the rock) , yo=0
vf*vf= vo*vo+2g(y-yo)
when the rock is halfway y = H/2 so:
vf1*vf1=2*g*H/2 so vf1 = sqrt(gH)
when the rock reach the bottom y=H so:
vf2*vf2=2*g*H so vf2 = sqrt(2gH)
so vf1/vf2= 1/ sqrt(2)
good luck from colombia
The volume of the rod is 1.26×10⁻⁵ m³, and the linear charge density of the rod is 3.64 C/m
<h3>What is volume?:</h3>
This is the product of the height of a solid object and its crossectional area.
The Volume of the rod is can be calculated using the formula below.
Note: A rod has the shape of a cylinder.
Formula:
- V = πr²h............... Equation 1
Where:
- V = Volume of the rod
- r = radius of the rod
- h = height of the rod.
From the question,
Given:
- r = 4mm = 0.004 m
- h = 25 cm = 0.25 m
- π = 3.14
Substitute these values into equation 1
- V = 3.14(0.004²)(0.25)
- V = 1.26×10⁻⁵ m³
<h3>What is linear charge density:</h3>
This is the ratio of the charge on an object to the length of the object.
The linear charge density of the rod can be calculated using the formula below.
- D = Q/h.................... Equation 2
Where:
- D = Linear charge density of the rod
- Q = Charge on the rod.
- h = height or length of the rod
From the question
Given:
- Q = 0.91 C
- h = 25 cm = 0.25 m
Substitute these values into equation 2
- D = 0.91/0.25
- D = 3.64 C/m
Hence, The volume of the rod is 1.26×10⁻⁵ m³, and the linear charge density of the rod is 3.64 C/m
Learn more about charge density here: brainly.com/question/14568868
distance d = 5 km = 5 x 1000 m = 5000 m
time taken = 25 minute = 25 x 60 sec = 1500 sec
average velocity V = d/t
V = 5000/1500
V = 3.33 m/s towards east
