Idk<span>Around what year can Carbon-14 dating be used up to and why?</span>
Answer:
W = 8.92 10² kJ
Explanation:
For this exercise they give us the strength, we must calculate the distance traveled, for this we need the rocket acceleration let's use Newton's second law
F = m a
a = F / m
a = 20 103/1400
a = 14.29 m/s²
With kinematics we can find the distance traveled
² = v₀² + 2 a x
x = ( ²-v₀²) / 2 a
x = (50² -35²) / 2 14.29
x = 1275 / 28.58
x = 44.61 m
Let's calculate the work
W = F.d
The bold is vector; as indicated by the force is in the direction of movement the scalar product is reduced to the ordinary product
W = F d
W = 20 10³ 44.61
W = 8.92 10⁵ J
W = 8.92 10² kJ
Answer:
i dont see an atom. or are you talking about any atom?
Given Information:
Force = f = 1 pound
Stretched length = x = 0.1 ft
Required Information:
Work done = W = ?
Answer:
Work done = 6.05 ft.lb
Explanation:
From the Hook's law we know that
f(x) = kx
Where f is the applied force, k is spring constant and x is length of spring being stretched.
k = F/x
k = 1/0.1
k = 10 lb/ft
f(x) = 10x
The work done is given by
W = ∫ f(x) dx
Where f(x) = 10x and limits of integration are (1.1, 0)
W = ∫ 10x dx
W = 10*x²/2
W = 5x²
Evaluating the limits,
W = 5(1.1)² - 5(0)²
W = 6.05 - 0
W = 6.05 ft.lb
Therefore, 6.05 ft.lb work has been done in stretching the spring from its natural length to 1.1 feet beyond it's natural length.
Answer:
The final speed of the bicyclist is 15.44 m/s.
Explanation:
Given;
initial velocity, u = 0
time of motion, t = 4.8 s
distance covered, d = 37.0 m
The acceleration of the bicyclist is calculated as;
d = ut + ¹/₂at²
37 = 0 + ¹/₂(4.8)²a
37 = 11.52a
a = 37 / 11.5
a = 3.22 m/s²
The final speed of the bicyclist is given as;
v² = u² + 2ad
v² = 0 + 2(3.22)(37)
v² = 238.28
v = √238.28
v = 15.44 m/s
Therefore, the final speed of the bicyclist is 15.44 m/s.