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:
Examples of Newton's third law of motion are ubiquitous in everyday life. For example, when you jump, your legs apply a force to the ground, and the ground applies and equal and opposite reaction force that propels you into the air. Engineers apply Newton's third law when designing rockets and other projectile devices.
Answer:
13 m/s
Explanation:
I assume we are ignoring friction.
The boy's PE will all be converted to KE at the bottom of the hill.
to find PE = mgh we need to know h
h = 50 sin 10 = 8.68 meters
then: PE = 20 * 9.81 * 8.68 =<u> 1703.49</u> j
KE = 1/2 m v^2 = <u>1703 .49</u>
v = 13 m/s
Yes it is a chemical reaction
Answer:
0.0034 sec
Explanation:
L = initial length
T = initial time period = 2.51 s
Time period is given as
L = 1.56392 m
L' = new length
ΔT = Rise in temperature = 142 °C
α = coefficient of linear expansion = 19 x 10⁻⁶ °C
New length due to rise of temperature is given as
L' = L + LαΔT
L' = 1.56392 + (1.56392) (19 x 10⁻⁶) (142)
L' = 1.56814 m
T' = New time period
New time period is given as
T' = 2.5134 sec
Change in time period is given as
ΔT = T' - T
ΔT = 2.5134 - 2.51
ΔT = 0.0034 sec
