Answer:
(C) Decreases by factor of 3
Explanation:
Centripetal acceleration is given by

where <em>v</em> is the linear velocity and <em>r</em> is the radius of the curve.
Let the centripetal acceleration on the curve of radius <em>R</em> be
.
Then

Let the centripetal acceleration on the curve of radius 3<em>R</em> be
.
Then

Here, we see that the acceleration decreases by a factor of 3.
Answer:

Explanation:
The differential equation for given is given as

integrating above equation we have
ln(T-T_s) = -kt + C
At t = 0 , T(0) = 60

2.99 = C
At t =1 , T(1) = 40.49887

- k = -3.687
So we have


Lower mass: 1.20 kg, upper mass: 1.28 kg
Explanation:
In order to solve the problem, we consider the forces acting on the upper mass only first.
The upper mass is acted upon three forces:
- The applied force
, upward - The weight of the mass itself,
, where
is the upper mass and
is the acceleration of gravity, downward - The tension in the string,
, downward
Therefore, the equation of the forces for the upper mass is:

where
is the acceleration (upward)
Solving for
,

Now we can find the lower mass by considering the forces acting on it:
- The tension in the string, T = 16 N, upward
- The weight of the mass itself,
, where
is the lower mass, downward
So the equation of the forces is

And solving for the mass,

Learn more about acceleration and forces:
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U = 80mph = 35.76m/s
S = 0.8m
M = 3500kg
To find the deceleration, we use equation of motion to calculate that
V² = u² + 2as
But v = 0
0 = u² + 2as
-U² = 2as
a = -u² / 2s
a = -(35.76)² / (2 * 0.8)
a = 1278.78 / 1.6 = 799.24 m/s²
The force exerted on the truck during deceleration is equal to the summation of all the forces acting on the truck.
∑F = m. ⃗a
-F = 3500 * 799.24
F = -2797331.25N
F = -2797.33kN
F = -2.797MN.
A vector Quantity has MAGNITUDE and direction, so the ‘what’ is Magnitude.