<span>a) write a polynomial expression for the position of the particle at any time t greater or equal to zero.
</span>Position is found by integrating velocity:
<span>s(t) = (t^3)/3 - 4t^2 + 7t + c
</span>where c is a constant corresponding to the position at t=0. <span>
b) at what time(s) is the particle changing direction
</span>the particle changes direction whenever the velocity is zero; the velocity function equals
<span>(t-1)(t-7) a difference of squares so the zeros are 1 and 7, it changes direction at 1 second and 7 seconds. </span><span>
c) find the total distance traveled by the particle from t=0 and t=4
</span><span>s(0) = c
s(1) = 8/3 + c
s(4) = 64/3 - 64 + 28 + c.
</span>
from 0 to 1 the particle travels 8/3 units. From 1 to 4 it travels -(64/3 - 36 - 8/3) = (-(56/3 - 108/3))
<span>=-(-52/3) = 52/3 units
</span>
<span>so in total it travels 52/3 + 8/3 =20 units</span>
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3.92N
Explanation:
Given mass:
Mass of bag = 400g
The mass of a substance is the amount of matter it contains.
Weight of a substance is the vertical force acting on a body in the presence of gravity.
Weight = mass x gravity
Since the mass of the bag is 400g;
Now we convert to kg
1000g = 1kg
400g = 
= 0.4kg
Weight of bag = 0.4kg x 9.8m/s²
Weight of bag = 3.92N
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Answer:
D. 2^(3/2)
Explanation:
Given that
T² = A³
Let the mean distance between the sun and planet Y be x
Therefore,
T(Y)² = x³
T(Y) = x^(3/2)
Let the mean distance between the sun and planet X be x/2
Therefore,
T(Y)² = (x/2)³
T(Y) = (x/2)^(3/2)
The factor of increase from planet X to planet Y is:
T(Y) / T(X) = x^(3/2) / (x/2)^(3/2)
T(Y) / T(X) = (2)^(3/2)
Answer:
I think the answer is half a second cause Time period = Time taken by the pendulum/Number of oscillations made by the pendulum so 5 divided by 10 equals 0.5 which is half a second
