Answer:
A) 580m
B) 0 m/s
C) 9.8m/s^2
D) downward
E) 10.87s
F) 106.62 m/s
Explanation:
A) The distance traveled by the rocket is calculated by using the following expression:
a: acceleration of the rocket = 2.90 m/s^2
t: time of the flight = 20.0 s
B) In the highest point the rocket has a velocity with magnitude zero v = 0m/s because there the rocket stops.
C) The engines of the rocket suddenly fails in the highest point. There, the acceleration of the rocket is due to the gravitational force, that is 9.8 m/s^2
D) The acceleration points downward
E) The time the rocket takes to return to the ground is given by:
10.87 seconds
F) The velocity just before the rocket arrives to the ground is:
Answer:
(a) a= 0.139 m/s²
(b) d= 4.45 m
(c) vf= 1.1 m/s
Explanation:
a) We apply Newton's second law:
∑F = m*a Formula (1)
∑F : algebraic sum of the forces in Newton (N)
m : mass (kg)
a : acceleration (m/s²)
Data
F₁= +2.05 * 10³ N : forward push by a motor
F₂= -1.87* 10³ N : resistive force due to the water.
m= 1300 kg
Calculation of the acceleration of the boat
We replace data in the formula (1):
∑F = m*a
F₁+F₂= m*a
a= 0.139 m/s²
b) Kinematics of the boat
Because the boat moves with uniformly accelerated movement we apply the following formulas:
d= v₀t+ (1/2)*a*t² Formula (2)
vf= v₀+at Formula (3)
Where:
d:displacement in meters (m)
t : time interval (s)
v₀: initial speed (m/s)
vf: final speed (m/s)
a: acceleration (m/s²
)
Data
v₀ = 0
a= 0.139 m/s²
t = 8 s
Calculation of the distance traveled by the boat in 8 s
We replace data in the formula (2)
d= v₀t+ (1/2)*a*t²
d= 0+ (1/2)*(0.139)*(8)²
d= 4.45 m
c) Calculation of the speed of the boat in 8 s
We replace data in the formula (3):
vf= v₀+at
vf= 0+( 0.139)*(8)
vf= 1.1 m/s
B, good luck, hope this helps :)
Is this science??? What subject is that??
