Answer:
radial acceleration is 41.8 m / s²
Explanation:
The acceleration for circular motion is
a = v² / r
They also give us the X and Y position where the body falls when the rope breaks, let's write the projectile launch equations
x = vox t
y = v₀ₓ t - ½ g t2
Since the circle is horizontally the v₀ₓ is zero (v₀ₓ = 0)
x = v₀ₓ t
t = x / v₀ₓ
y = - ½ g t²
Let's replace and calculate the initial velocity on the X axis
y = - ½ g (x / vox)²
v₀ₓ = √ (g x² / 2 y)
v₀ₓ = √ [- (-9.8) 1.6² / (2 1.00)]
v₀ₓ = 3.54 m / s
This is the horizontal velocity, but since it circle is in horizontal position it is also the velocity of the body at the point of rupture.
Now we can calculate the radial acceleration
a = v² / r
a = 3.54² / 0.300
a = 41.8 m / s²
Explanation:
For this problem, use the first law of thermodynamics. The change in energy equals the increase in heat energy minus the work done.
ΔU=Q−W
We are not given a value for work, but we can solve for it using the force and distance. Work is the product of force and displacement.
W=FΔx
W=3N×2m
W=6J
Now that we have the value of work done and the value for heat added, we can solve for the total change in energy.
ΔU=Q−W
ΔU=10J−6J
ΔU=4J
Answer is 4J
i think this may help you very much
The answer is 18000 J
I hope this helps!^^ , if you need the work to be shown please tell me, I hope you have a great day!^^
Answer:
1.5 m
Explanation:
Length. L = 12 m
Width, W = 16 m
Area, A = 12 x 16 = 192 m^2
Let the width of pavement be d.
The new length, L' = 12 + 2d
the new width, W' = 16 + 2d
New Area, A' = L' x W' = (12 + 2d)(16 + 2d) = 192 + 56 d + 4d^2
Difference in area = A' - A
285 = 192 + 56 d + 4d^2 - 192
93 = 56 d + 4d^2
4d^2 + 56 d - 93 = 0

\
d = 1.5 m
Thus, the width of the pavement is 1.5 m.