Answer:
The horizontal component of the velocity is 21.9 m/s.
Explanation:
Please see the attached figure for a better understanding of the problem.
Notice that the vector v and its x and y-components (vx and vy) form a right triangle. Then, we can use trigonometry to find the magnitude of vx, the horizontal component of the velocity.
To find vx, let´s use the following trigonometric rule of right triangles:
cos α = adjacent / hypotenuse
cos 5.7° = vx / 22 m/s
22 m/s · cos 5.7° = vx
vx = 21.9 m/s
The horizontal component of the velocity is 21.9 m/s.
We know that either team is winning which means that the cumulative forces (sum of each girl force) on both teams are equal.
This means,
being x the force of the fourth girl on the east team.
Solving to x, you get
The fourth girl on the east is pulling with a force of 220N.
Answer:
a) v = 54.7m/s
b) v = (58 - 1.66a) m/s
c) t = 69.9 s
d) v = -58.0 m/s
Explanation:
Given;
The height equation of the arrow;
H = 58t - 0.83t^2
(a) Find the velocity of the arrow after two seconds. m/s;
The velocity of the arrow v can be given as dH/dt, the change in height per unit time.
v = dH/dt = 58 - 2(0.83t) ......1
At t = 2 seconds
v = dH/dt = 58 - 2(0.83×2)
v = 54.7m/s
(b) Find the velocity of the arrow when t = a. m/s
Substituting t = a into equation 1
v = 58 - 2(0.83×a)
v = (58 - 1.66a) m/s
(c) When will the arrow hit the surface? (Round your answer to one decimal place.) t = s
the time when H = 0
Substituting H = 0, we have;
H = 58t - 0.83t^2 = 0
0.83t^2 = 58t
0.83t = 58
t = 58/0.83
t = 69.9 s
(d) With what velocity will the arrow hit the surface? m/s
from equation 1;
v = dH/dt = 58 - 2(0.83t)
Substituting t = 69.9s
v = 58 - 2(0.83×69.9)
v = -58.0 m/s
As kinetic energy increases, substance temperature increases
Answer:
Explanation:
Initial velocity, u = 85 m/s
Final velocity, v = 45 m/s
Time, t = 4.5 s
We need to find the acceleration of the dragster. The rate at which the velocity of an object changes is called the acceleration of an object. It can be given by :
So, the acceleration of the dragster is 
and it is deaccelerating.
