Q: ken, 0.75 kg, moves toward a wall (his path normal to the wall) at 52 m/s. 13.0 ms after he touches the wall he pushes himself off in the opposite direction at 60 m/s. What is the magnitude of the average force the wall exerts on Ken during this rapid maneuver
Answer:
-6461.54 N
Explanation:
From Newton's Fundamental equation,
F = m(v-u)/t.................... Equation 1
Where F = Force exerted in sonic, m = mass of ken, v = final velocity, u = initial velocity, t = time.
Given: m = 0.75 kg, v = - 60 m/s (opposite direction), u = 52 m/s, t = 13 ms = 0.013 s
Substitute into equation 1
F = 0.75(-60-52)/0.013
F = 0.75(-112)/0.013
F = -84/0.013
F = -6461.54 N
Note: The negative sign tells that the force act in opposite direction to the initial motion of ken.
Hence the magnitude of the average force of the wall = -6461.54 N
Answer:B. two larger, less stable nuclei
Explanation: They collied and don't combine
Answer:
1) A. 0.44 m/s East + 0.33 m/s North
2) A. 0 m/s²
3) A. a scalar calculated as distance divided by time.
4) B. 31 km per hour
Explanation:
1) Velocity is DISPLACEMENT over time.
at 1 m/s, total time of walking is 9000 seconds
displacement is 3000 m north and 5000 - 1000 = 4000 m east
4000 m/ 9000 s = 0.44 m/s E
3000 m/ 9000s = 0.33 m/s N
2) constant speed means no acceleration
3) A. a scalar calculated as distance divided by time.
4) displacement 50 km N and 80 km W
v = √(50² + 80²) / (1 + 2) = 31.446603... km/hr
Answer:
The correct relationships are T-fg=ma and L-fg=0.
(A) and (C) is correct option.
Explanation:
Given that,
Weight Fg = mg
Acceleration = a
Tension = T
Drag force = Fa
Vertical force = L
We need to find the correct relationships
Using balance equation
In horizontally,
The acceleration is a
...(I)
In vertically,
No acceleration
Put the value of mg
....(II)
Hence, The correct relationships are T-fg=ma and L-fg=0.
(A) and (C) is correct option.
Force = mass * acceleration
10 N - 2 N = 20 kg * acceleration
8 N = 20 kg * acceleration
8 / 20 = acceleration
2/5 m/s^2 = acceleration
