Answer:
C. A rocket traveling to the Moon
Answer:
a) a = 8m/s^2
b) a = 9.8m/s^2
c) a = 200m/s^2
d) a = 9.8m/s^2
Explanation:
Using the laws of newton:
F = ma
where F is the force, m the mass and a the aceleration
a) Solving for the aceleration:
a = F/m
so, replacing the force and the mass of Blanche, we get:
a = 800N/100kg
a = 8m/s^2
b) By the laws of newton, the sumatory of vertical force for Blanche is equal to:
mg = ma
Where g is the gravity
Then, solving for a:
a = g
a = 9.8m/s^2
c) At the same way, solving for the aceleration:
a = F/m
so, replacing the values of the force and the mass of the shot, we get:
a = 800N/4kg
a = 200m/s^2
d) Then, sumatory of vertical force for the shot is equal to:
mg = ma
where g is the gravity
solving for a:
a = g
a = 9.8m/s^2
Calculate the circumference of the circular path of the wheel by the equation,
C = 2πr
where C is the circumference,
r is the radius
Substituting the known values,
C = 2(π)(0.487 m) = 3.06 m
Then, period of the motion will be determined by dividing the number of rotations done,
T = (3.06 m) / (5.75 /s) = 0.53 s
<em>ANSWER: 0.53 s</em>
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Answer:
xf = 5.68 × 10³ m
yf = 8.57 × 10³ m
Explanation:
given data
vi = 290 m/s
θ = 57.0°
t = 36.0 s
solution
firsa we get here origin (0,0) to where the shell is launched
xi = 0 yi = 0
xf = ? yf = ?
vxi = vicosθ vyi = visinθ
ax = 0 ay = −9.8 m/s
now we solve x motion: that is
xf = xi + vxi × t + 0.5 × ax × t² ............1
simplfy it we get
xf = 0 + vicosθ × t + 0
put here value and we get
xf = 0 + (290 m/s) cos(57) (36.0 s)
xf = 5.68 × 10³ m
and
now we solve for y motion: that is
yf = yi + vyi × t + 0.5 × ay × t
² ............2
put here value and we get
yf = 0 + (290 m/s) × sin(57) × (36.0 s) + 0.5 × (−9.8 m/s2) × (36.0 s) ²
yf = 8.57 × 10³ m
