Pluto............................
Answer:
A. 14.14 m/s²
B. 127.26 m/s
C. 572.67 m
Explanation:
From the question given above, the following data were obtained:
Mass (m) of jet = 4250 kg
Force (F) = 60100 N
A. Determination of the acceleration of the jet.
Mass (m) of jet = 4250 kg
Force (F) = 60100 N
Acceleration (a) =?
F = ma
60100 = 4250 × a
Divide both side by 4250
a = 60100 / 4250
a = 14.14 m/s²
B. Determination of the speed of the jet after 9 s.
Initial velocity (u) = 0 m/s
Acceleration (a) = 14.14 m/s²
Time (t) = 9 s
Final velocity (v) =?
v = u + at
v = 0 + (14.14 × 9)
v = 0 + 127.26
v = 127.26 m/s
C. Determination of the distance travelled during the time.
Initial velocity (u) = 0 m/s
Acceleration (a) = 14.14 m/s²
Time (t) = 9 s
Distance travelled (s) =?
s = ut + ½at²
s = (0 × 9) + (½ × 14.14 × 9²)
s = 0 + (7.07 × 81)
s = 0 + 572.67
s = 572.67 m.
The answer is D, human.
Happy Friday!!!
Answer:
the required solution is; x(t) = 0.675<em>sin</em>( 2.222t )
Explanation:
Given the data in the question;
Using both Newton's and Hooke's law;
m + k = 0, (0) = 0, (0) = 1.5
given that mass m = 9 kg
= 1.8 m
k is F / x
hence
k = F / x
given that, F = 80 N
we substitute
k = 80 / 1.8
k = 44.44
so
m + k = 0,
we input
9 + 44.44 = 0,
+ 4.9377 = 0
so auxiliary equation is,
r² + 4.9377 = 0
r² = -4.9377
r = √-4.9377
r = ±2.222i
hence, the solution will be;
x(t) = A×cos( 2.222t ) + B×sin( 2.222t )
⇒ (t) = -2.222Asin( 2.222t ) + 2.222Bcos( 2.222t )
using initial conditions
x(0) = 0
⇒ 0 = A
(t) = 1.5
1.5 = 2.222B
so
B = 1.5 / 2.222 = 0.675
Hence, the required solution is; x(t) = 0.675<em>sin</em>( 2.222t )
Answer:
B) Force of gravity
Explanation:
As it is the force which acts downward on the body towards the ground, it is clearly the force gravity represented by the arrow at A.