Answer:
h' = 603.08 m
Explanation:
First, we will calculate the initial velocity of the pellet on the surface of Earth by using third equation of motion:
2gh = Vf² - Vi²
where,
g = acceleration due to gravity on the surface of earth = - 9.8 m/s² (negative sign due to upward motion)
h = height of pellet = 100 m
Vf = final velocity of pellet = 0 m/s (since, pellet will momentarily stop at highest point)
Vi = Initial Velocity of Pellet = ?
Therefore,
(2)(-9.8 m/s²)(100 m) = (0 m/s)² - Vi²
Vi = √(1960 m²/s²)
Vi = 44.27 m/s
Now, we use this equation at the surface of moon with same initial velocity:
2g'h' = Vf² - Vi²
where,
g' = acceleration due to gravity on the surface of moon = 1.625 m/s²
h' = maximum height gained by pellet on moon = ?
Therefore,
2(1.625 m/s²)h' = (44.27 m/s)² - (0 m/s)²
h' = (1960 m²/s²)/(3.25 m/s²)
<u>h' = 603.08 m</u>
Answer:
a) 
b) x = 4.47 cm
c) 
d) x = 1.48 cm
Explanation:
a) The center of mass is equal to:

Where m is the mass of beads and x is the distances, if x₁ = d₁, x₂ = d₂ and x₃ = d₃

b) If
m₁ = 23g
m₂ = 15 g
m₃ = 58 g
d₁ = 1.1 cm
d₂ = 1.9 cm
d₃ = 3.2 cm

c) The center of the mass of the beads realtive to the center of bead is:

d) 
Picosecond = 10 ^ -12 seconds.
Zeptosecond = 10^ -18 seconsds
Petaseonds = 10^15 seconds
To express Picoseconds into any of other two, you have to divide 10^-12 by the power index of the one in question
1Picosecond : 10^-12 / 10^-18 = 10^ (-12- 18) = 10^ (-12+18)= 10^6 zeptoseconds
1Picosecond : 10^-12 / 10^15 = 10^ (-12-15) = 10^-27 Petaseconds.
1Picosecond = 10^6 zeptoseconds
1Picosecond = 10^-27 Petaseconds
Answer:
a) 
b) 
c) 
d)
would be the same.
would decrease.
would be the same.
Explanation:
a) On an inclined plane the force of gravity is the sine component of the weight of the block.

b) The friction force is equal to the normal force times coefficient of friction.

c) The work done by the normal force is zero, since there is no motion in the direction of the normal force.
d) The relation between the vertical height and the distance on the ramp is

According to this relation, the work done by the gravity wouldn't change, since the force of gravity includes a term of
.
The work done by the friction force would decrease, because both the cosine term and the distance on the ramp would decline.
The work done by the normal force would still be zero.