Answer:
The speed of the automobile after 1.43s is 10
Explanation:
≅ 10
Answer:
insect B by 12m
Explanation:
30min = 1800s (times by 60)
5m/min x 30min = 150m
9cm/s x 1800s = 16,200cm = 162m
162m - 150m = 12m
Answer:
F = 39.2 N
Explanation:
Since, the object is in uniform motion. Therefore, the frictional force on object will be:
Frictional Force = μk N = μk mg
where,
μk = coefficient of kinetic friction = 0.2
m = mass of crate = 10 kg
g = 9.8 m/s²
Therefore,
Frictional Force = (0.2)(10 kg)(9.8 m/s²)
Frictional Force = 19.6 N
The horizontal component of force must be equal to this frictional force to continue the uniform motion:
F Sin 30° = 19.6 N
F = 19.6 N/Sin 30°
<u>F = 39.2 N</u>
Answer:
a) W = 6.75 J and b) v = 3.87 m / s
Explanation:
a) In the problem the force is nonlinear and they ask us for work, so we must use it's definition
W = ∫ F. dx
Bold indicates vectors. In a spring the force is applied in the direction of movement, whereby the scalar product is reduced to the ordinary product
W = ∫ F dx
We replace and integrate
W = ∫ (-60 x - 18 x²) dx
W = -60 x²/2 -18 x³/3
Let's evaluate between the integration limits, lower W = 0 for x = -0.50 m, to the upper limit W = W for x = 0 m
W = -30 [0- (-0.50) 2] -6 [0 - (- 0.50) 3]
W = 7.5 - 0.75
W = 6.75 J
b) Work is equal to the variation of kinetic energy
W = ΔK
W = ΔK = ½ m v² -0
v =√ 2W/m
v = √(2 6.75/ 0.90)
v = 3.87 m / s
Answer:
if you stretch a spring with k = 2, with a force of 4N, the extension will be 2m. the work done by us here is 4x2=8J. in other words, the energy transferred to the spring is 8J. but, the stored energy in the spring equals 1/2x2x2^2=4J (which is half of the work done by us in stretching it).