Suppose that the space shuttle Columbia accelerates at 14.0 m/s2 for 8.50 minutes after takeoff.
1 answer:
Answer:
A. speed = 7.14 Km/s
B. distance = 1820.7 Km
Explanation:
Given that: a = 14.0 m/, t = 8.50 minutes.
But,
t = 8.50 = 8.50 x 60
= 510 seconds
A. By applying the first equation of motion, the speed of the shuttle at the end of 8.50 minutes can be determined by;
v = u + at
where: v is the final velocity, u is the initial velocity, a is the acceleration and t is the time.
u = 0
So that,
v = 14 x 510
= 7140 m/s
The speed of the shuttle at the end of 8.50 minute is 7.14 Km/s.
B. the distance traveled can be determined by applying second equation of motion.
s = ut + a
where: s is the distance, u is the initial velocity, a is the acceleration and t is the time.
u = 0
s = a
= x 14 x
= 7 x 260100
= 1820700 m
The distance that the shuttle has traveled during the given time is 1820.7 Km.
You might be interested in
Answer:
to overcome the out of friction we must increase the angle of the plane
Explanation:
To answer this exercise, let's propose the solution of the problem, write Newton's second law. We define a coordinate system where the x axis is parallel to the plane and the other axis is perpendicular to the plane.
X axis
fr - Wₓ = m a (1)
Y axis
N- = 0
N = W_{y}
let's use trigonometry to find the components of the weight
sin θ = Wₓ / W
cos θ = W_{y} / W
Wₓ = W sin θ
W_{y} = W cos θ
the friction force has the formula
fr = μ N
fr = μ Wy
fr = μ mg cos θ
from equation 1
at the point where the force equals the maximum friction force
in this case the block is still still so a = 0
F = fr
F = (μ mg) cos θ
We can see that the quantities in parentheses with constants, so as the angle increases, the applied force must be less.
This is the force that balances the friction force, any force slightly greater than F initiates the movement.
Consequently, to overcome the out of friction we must increase the angle of the plane
the correct answer is to increase the angle of the plane
Answer:
a). 53.75 N and 101.92 N
b). 381.44 N and 723.25 N
Explanation:
a).
ρ,
,
b).
ρ,
,