A cricket starts at the 5 meter mark on a ruler. It jumps to the 10 meter mark. Then, it jumps back to the 6 meter mark. Calcula
1 answer:
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Answer: 1872 N
Explanation:
This problem can be solved by using one of the Kinematics equations and Newton's second law of motion:
(1)
(2)
Where:
is the bullet's final speed (when it leaves the muzzle)
is the bullet's initial speed (at rest)
is the bullet's acceleration
is the distance traveled by the bullet before leaving the muzzle
is the force
is the mass of the bullet
Knowing this, let's begin by isolating from (1):
(3)
(4)
(5)
Substituting (5) in (2):
(6)
Finally:
Answer:
a) See attachment.
b) a_avg = -8.66 i - 2.33 j
c) a = 195 degrees from + x-axis. |a_avg| = 8.97 m/s^2
Explanation:
Given:
- Initial Velocity V_i = (90 i + 110 j)
- Final velocity V_f = (-170 i + 40 j)
- t_i = 0
- t_f = 30.0 s
Find:
(a) Sketch the velocity vectors at t1 and t2. How do these two vectors differ?
(b) the components of the average acceleration
(c) the magnitude and direction of the average acceleration.
Solution:
- The change in velocity is given by:
Δ V = (V_f - V_i)
Δ V = ( -170 i - 90 i + 40 j - 110 j)
Δ V = (-260 i - 70 j)
- The change in time is given as:
Δ t = t_f - t_i = 30 - 0 = 30 s
-The average acceleration is:
a_avg = Δ V / Δ t
a_avg = (-260 i - 70 j) / 30
a_avg = -8.66 i - 2.33 j
- The magnitude of acceleration is:
|a_avg| = sqrt ( 8.66^2 + 2.33^2)
|a_avg| = sqrt (80.4245)
|a_avg| = 8.97 m/s^2
- The direction of acceleration is given by:
tan (Q) = a_y / a_x
Q = arctan(a_y / a_x)
Q = arctan(2.33 / 8.66)
Q = 15 degrees
Hence, 180 + 15 = a. a = 195 degrees from + x-axis.
