Answer:
everyone else does this to me so lol
Explanation:
Answer:
position 4
Explanation:
if the north is tilted away it gets no sunlight, so its winter
<span>(a) 12.02 m/s
(b) 52.2 meters
This problem is an example of integral calculus. You've been given an acceleration vector which is usually known as the 2nd derivative. From that you need to calculate the velocity function (1st derivative) and position (actual function) by successively calculating the anti-derivative. So:
A(t) = 6.30 - 2.20t
V(t) = 6.30t - 1.10t^2 + C
We now have a velocity function, but need to determine C. Since we've been given the velocity at t = 0, that's fairly trivial.
V(t) = 6.30t - 1.10t^2 + C
3 = 6.30*0 - 1.10*0^2 + C
3 = 0 + 0 + C
3 = C
So the entire velocity function is:
V(t) = 6.30t - 1.10t^2 + 3
V(t) = -1.10t^2 + 6.30t + 3
Now for the location function which is the anti-derivative of the velocity function.
V(t) = -1.10t^2 + 6.30t + 3
L(t) = -0.366666667t^3 + 3.15t^2 + 3t + C
Now we need to calculate C. And once again, we've been given the location for t = 0, so
L(t) = -0.366666667t^3 + 3.15t^2 + 3t + C
7.3 = -0.366666667*0^3 + 3.15*0^2 + 3*0 + C
7.3 = 0 + 0 + 0 + C
7.3 = C
L(t) = -0.366666667t^3 + 3.15t^2 + 3t + 7.3
Now that we have the functions, they are:
A(t) = 6.30 - 2.20t
V(t) = -1.10t^2 + 6.30t + 3
L(t) = -0.366666667t^3 + 3.15t^2 + 3t + 7.3
let's answer the questions.
(a) What is the maximum speed achieved by the cyclist?
This can only happen at those points that meet either of the following criteria.
1. The derivative is undefined for the point.
2. The value of the derivative is 0 for the point.
As it turns out, the 1st derivative of the velocity function is the acceleration function which we have. So
A(t) = 6.30 - 2.20t
0 = 6.30 - 2.20t
2.20t = 6.30
t = 2.863636364
So one of V(0), V(2.863636364), or V(6) will be the maximum value. Therefore:
V(0) = 3
V(2.863636364) = 12.0204545454545
V(6) = 1.2
So the maximum speed achieved is 12.02 m/s
(b) Total distance traveled?
L(0) = 7.3
L(6) = 59.5
Distance traveled = 59.5 m - 7.3 m = 52.2 meters</span>
Answer:
Explanation:
Given
initial velocity u = 15m/s
Height below the water = 25m
a) Using the equation of motion S = (v+u)/2 * t
25 = 0+15/2 * t
25 = 7.5t
t = 25/7.5
t = 3.33s
Hence it will take 3.33secs to reach the water
b) The horizontal distance is expressed as;
S = Uxt
S = 15(3.33)
S = 50m
Hence the horizontal distance from the cliff where the car entered the water is 50m
c) The velocity of the car v.
Using the equation of motion;
v² = u²+2gS
v² = 15²+2(9.8)(25)
v² = 225+490
v = √715
v = 26.74m/s
Hence the car hit the water at the velocity of 26.74m.s
Answer:
Electrons
Explanation:
Only Protons and Neutrons are found in the nucleus
