You are given a vector in the xy plane that has a magnitude of 84.0 units and a y-component of -67.0 units.
1 answer:
Answer:
Explanation:
a)Magnitude =
84=
x= +50.67 or -50.67 units
b) We are given that the resultant is entirely in the -ve x direction which means that the y-component of the resultant is 0; It means that the y-component of the next vector = -ve of the y component of the initial vector i.e 67.
To make the magnitude 80 units in the negative x direction where the y component is 0, the x component must be -130.67(-50.67 - 80) as the x component is + 50.67units.
Magnitude = = 146.85 units
c) The direction vector = 67/146.85 i - 130.67/146.85 j where i corresponds to the vector in y direction and j corresponds to the vector in x direction. Or this vector is at an angle of 180 - i.e 152.85 degrees from the +ve x-axis.
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Answer:
the unloaded braking efficiency is 84.6 %
Explanation:
Given the data in the question;
by Ignoring aerodynamic resistance; we can find the theoretical stopping distance using the following formula
S = (Y( V₁² - V₂²)) / ( 2g( ηbμ +
± sin∅
))
now given that the tracked is levelled, ∅ = 0, also Y
= 1.04 for level or flat road
Speed V₁ = 60mil/hr = (60×5280)/(1×60×60) = 316800ft/3600s = 88ft/s
now, we substitute in our values to get the braking efficiency;
180ft = (1.04( (88ft/s)² - 0²)) / ( 2(32.2( (ηb/100)(0.80) + (0.018) ± sin(0°)))
180ft = 8053.76 / ( 64.4)(0.008ηb + 0.018)
180ft = 8053.76 / ( 0.5152ηb + 1.1592)
180( 0.5152ηb + 1.1592) = 8053.76
( 0.5152ηb + 1.1592) = 8053.76 /180
0.5152ηb + 1.1592 = 44.7431
0.5152ηb = 44.7431 - 1.1592
0.5152ηb = 43.5839
ηb = 43.5839 / 0.5152
ηb = 84.596 ≈ 84.6 %
Therefore, the unloaded braking efficiency is 84.6 %
Answer:
The circus performer falls back down to the ground
Explanation:
The question parameters are;
The initial velocity of the circus performer = 21 m/s
The angle in which the performer launches himself = 75° towards the platform
The height of the platform above the ground = 20 m
The horizontal distance of the platform from the springboard = 15 m
The vertical motion of the circus performer is given by the following projectile motion relation;
y = y₀ + v₀·sinθ₀t-1/2·g·t²
Where;
y = Height reached by the circus performer
y₀ = Initial height of the the circus performer (the springboard) = 0 m
v₀ = Initial velocity of the the circus performer = 21 m/s
θ₀ = The angle with which the circus performer launches himself = 75°
t = The time of ,light of the circus performer
g = The acceleration due to gravity
Therefore, when the height is 20 m, we have;
20 = 21*sin(75)*t - 1/2*9.81*t²
Which gives;
21*sin(75)*t - 1/2*9.81*t² - 20 = 0
Factorizing using a graphing calculator, gives;
t = 1.623 or t = 2.513
Therefore, the circus performer passes the 20 m mark twice, in his motion, where the first time is when he is on his way up while the second time is when he is on his way down
The horizontal motion of the circus performer is given by the following projectile motion relation;
x = x₀ + v₀*cos(θ₀)* t
Where;
x₀ = The initial position of the circus performer in relation to the final position = 0
Plugging in the value of t when y = 20, we get;
x = 21×cos(75)×1.623 = 8.82 m, which is less than the 15 m platform distance from the spring board
Checking the other time value, we have;
x = 21×cos(75)×2.513 = 13.66 m which is also less than the 15 m platform distance from the spring board
Therefore, the circus performer misses the platform and falls back down to the ground.
