Answer: -31.36 m/s
Explanation:
This is a problem of motion in one direction (specifically vertical motion), and the equation that best fulfills this approach is:
(1)
Where:
is the final velocity of the supply bag
is the initial velocity of the supply bag (we know it is zero because we are told it was "dropped", this means it goes to ground in free fall)
is the acceleration due gravity (the negtive sign indicates the gravity is downwards, in the direction of the center of the Earth)
is the time
Knowing this, let's solve (1):
(2)
Finally:
Note the negative sign is because the direction of the bag is downwards as well.
Answer:
0.265
Explanation:
Draw a free body diagram. There are four forces:
Normal force Fn pushing up.
Weight force mg pulling down.
Tension force T at an angle θ.
Friction force Fn μ pushing left.
Sum the forces in the y direction:
∑F = ma
Fn + T sin θ − mg = 0
Fn = mg − T sin θ
Sum the forces in the x direction:
∑F = ma
T cos θ − Fn μ = 0
Fn μ = T cos θ
μ = T cos θ / Fn
μ = T cos θ / (mg − T sin θ)
Given T = 164 N, θ = 10.0°, m = 65.0 kg, and g = 9.8 m/s²:
μ = (164 N cos 10.0°) / (65.0 kg × 9.8 m/s² − 164 N sin 10.0°)
μ = 0.265
Answer:
(a) 0.71 mm
(b) 0.158 cubic cm
Explanation:
The width of one wire is the diameter of the wire.
(a) Let the diameter of each wire is d.
So, 10 d = 14.2 mm
d = 1.42 mm
radius of each wire, r = d/2 = 1.42/2 = 0.71 mm
(b) Length, L = 10 cm
The volume of the single wire is given by
Answer:
Motion with constant velocity of magnitude 1 m/s (uniform motion) for 4 seconds in a positive direction and then for 2 seconds uniform motion with constant velocity of magnitude 3 m/s in reverse direction .
Explanation:
The graph shows a constant velocity of 1 m/s for 4 seconds in the positive direction. After that, between 4 seconds and 6 seconds, the object reverses its motion with constant velocity of magnitude 3m/s.
