<span>By algebra, d = [(v_f^2) - (v_i^2)]/2a.
Thus, d = [(0^2)-(15^2)]/(2*-7)
d = [0-(225)]/(-14)
d = 225/14
d = 16.0714 m
With 2 significant figures in the problem, the car travels 16 meters during deceleration.</span>
(a) 3.56 m/s
(b) 11 - 3.72a
(c) t = 5.9 s
(d) -11 m/s
For most of these problems, you're being asked the velocity of the rock as a function of t, while you've been given the position as a function of t. So first calculate the first derivative of the position function using the power rule.
y = 11t - 1.86t^2
y' = 11 - 3.72t
Now that you have the first derivative, it will give you the velocity as a function of t.
(a) Velocity after 2 seconds.
y' = 11 - 3.72t
y' = 11 - 3.72*2 = 11 - 7.44 = 3.56
So the velocity is 3.56 m/s
(b) Velocity after a seconds.
y' = 11 - 3.72t
y' = 11 - 3.72a
So the answer is 11 - 3.72a
(c) Use the quadratic formula to find the zeros for the position function y = 11t-1.86t^2. Roots are t = 0 and t = 5.913978495. The t = 0 is for the moment the rock was thrown, so the answer is t = 5.9 seconds.
(d) Plug in the value of t calculated for (c) into the velocity function, so:
y' = 11 - 3.72a
y' = 11 - 3.72*5.913978495
y' = 11 - 22
y' = -11
So the velocity is -11 m/s which makes sense since the total energy of the rock will remain constant, so it's coming down at the same speed as it was going up.
Answer:
B.) to determine that electric beams in cathode ray tubes were actually made of particles
Explanation:
This is the right answer i just took the quiz on edge.
Answer:
Explanation:
Newton's law of universal gravitation states that the force experimented by a satellite of mass m orbiting Mars, which has mass 
at a distance r will be:
where 
is the gravitational constant.
This force is the centripetal force the satellite experiments, so we can write:
Putting all together:
which means:
![r=\sqrt[3]{\frac{GM}{4\pi^2}T^2}](https://tex.z-dn.net/?f=r%3D%5Csqrt%5B3%5D%7B%5Cfrac%7BGM%7D%7B4%5Cpi%5E2%7DT%5E2%7D)
Which for our values is:
![r=\sqrt[3]{\frac{(6.67\times10^{-11}Nm^2/kg^2)(6.39\times10^{23} kg)}{4\pi^2}(1.026\times24\times60\times60s)^2}=20395282m=20395.3km](https://tex.z-dn.net/?f=r%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B%286.67%5Ctimes10%5E%7B-11%7DNm%5E2%2Fkg%5E2%29%286.39%5Ctimes10%5E%7B23%7D%20kg%29%7D%7B4%5Cpi%5E2%7D%281.026%5Ctimes24%5Ctimes60%5Ctimes60s%29%5E2%7D%3D20395282m%3D20395.3km)
Since this distance is measured from the center of Mars, to have the height above the Martian surface we need to substract the radius of Mars R=3389.5 km
, which leaves us with:
Answer:
To both observers, the land opposite them is moving to the right.
Explanation:
I have this class too. and there is also a quizlet with all the answers to the rest of the other questions. Trust me its right .
https://quizlet.com/261219090/oce-1001-chapter-2-flash-cards/
