1A)
x=v0x*t=v0cosθ*t
x=52co31*3.2=142.6 m
1B)
y0=1/2gt^2-v0y*t=1/2gt^2-v0sinθt
y=0.5*9.8*3.2^2-52*sin31*3,2=23.4 m
2A)
x=2v0^2sin(2θ)/g
v0=[xg/2sin(2θ)]^1/2=14.4 m/s
the initial speed relative to the ground is
v=v0-4.4=10 m/s
2B)
fly time is
t=2voy/g
t=2*14.4/9.8=2.94
2C)
mgy=1/2mv0y^2
y=v0y^2/(2g)=10.58 m
To develop the problem it is necessary to take into account the concepts related to Torque and sum of moments.
By torque it is understood that
Where,
F= Force
d = Distance
The value of the given Torque acts from the center of mass causing it to rotate clockwise.
The Force must then be located at the other end down to make a movement opposite the Torque in the center of mass.
I enclose a graph that allows us to understand the problem in a more didactic way.
The correct answer is D.
-- The boiling points of the first group are all at temperatures
that are way lower than a comfortable room.
. . . . . Nitrogen . . . -320° F
. . . . . Helium . . . . -452° F
. . . . . Neon . . . . . -411° F
The freezing points of the second group are all at temperatures
that are way higher than a comfortable room.
. . . . . Lithium. . . . . . 357° F
. . . . . Sodium . . . . . 208° F
. . . . . Potassium . . . 146 °F
Answer:
Given
mᴀ = mʙ = mᴄ = 62 kg
Fᴀ = 480N
Fʙ = 240 N
Fᴄ = 600 N
Asked
aᴀ, aʙ, aᴄ = ?
Solution
Based on the question, we know that the mass of road A, road B and road C are same. And, the formula of acceleration is
so, we can calculate the acceleration of road A, road B and road C.
• for road A
aᴀ = 7,7419 m/s²
aᴀ = 7,75 m/s²
• for road B
aʙ = 3,8709 m/s²
aʙ = 3,87 m/s²
• for road C
aᴄ = 9,6774 m/s²
aᴄ = 9,68 m/s²
Explanation:
The answer is a, series circuit.
