Answer:
we agree with
Edgar: The net force on the ball at the top position is 9 N. Both the tension and the weight are acting downward so you have to add them.
Explanation:
Weight of the ball is given as
![W = 2N](https://tex.z-dn.net/?f=W%20%3D%202N)
so we have
![m = \frac{W}{g}](https://tex.z-dn.net/?f=m%20%3D%20%5Cfrac%7BW%7D%7Bg%7D)
![m = 0.204 kg](https://tex.z-dn.net/?f=m%20%3D%200.204%20kg)
now tension force at the top is given as
![T_{top} = 7 N](https://tex.z-dn.net/?f=T_%7Btop%7D%20%3D%207%20N)
![T_{bottom} = 15 N](https://tex.z-dn.net/?f=T_%7Bbottom%7D%20%3D%2015%20N)
Now at the top position by force equation we can say that ball will have two downwards forces
1) Tension force
2) Weight of the ball
so net force on the ball is given as
![F_{net} = T + W](https://tex.z-dn.net/?f=F_%7Bnet%7D%20%3D%20T%20%2B%20W)
![F_{net} = 7 + 2 = 9 N](https://tex.z-dn.net/?f=F_%7Bnet%7D%20%3D%207%20%2B%202%20%3D%209%20N)
So we agree with
Edgar: The net force on the ball at the top position is 9 N. Both the tension and the weight are acting downward so you have to add them.
This question can be solved by using the equations of motion.
a) The initial speed of the arrow is was "9.81 m/s".
b) It took the arrow "1.13 s" to reach a height of 17.5 m.
a)
We will use the second equation of motion to find out the initial speed of the arrow.
![h= v_it + \frac{1}{2}gt^2\\](https://tex.z-dn.net/?f=h%3D%20v_it%20%2B%20%5Cfrac%7B1%7D%7B2%7Dgt%5E2%5C%5C)
where,
vi = initial speed = ?
h = height = 35 m
t = time interval = 2 s
g = acceleration due to gravity = 9.81 m/s²
Therefore,
![35\ m = (v_i)(2\ s)+\frac{1}{2}(9.81\ m/s^2)(2\ s)^2\\\\v_i(2\ s)=19.62\ m\\\\v_i = \frac{19.62\ m}{2\ s}](https://tex.z-dn.net/?f=35%5C%20m%20%3D%20%28v_i%29%282%5C%20s%29%2B%5Cfrac%7B1%7D%7B2%7D%289.81%5C%20m%2Fs%5E2%29%282%5C%20s%29%5E2%5C%5C%5C%5Cv_i%282%5C%20s%29%3D19.62%5C%20m%5C%5C%5C%5Cv_i%20%3D%20%5Cfrac%7B19.62%5C%20m%7D%7B2%5C%20s%7D)
<u>vi = 9.81 m/s</u>
b)
To find the time taken by the arrow to reach 17.5 m, we will use the second equation of motion again.
![h= v_it + \frac{1}{2}gt^2\\](https://tex.z-dn.net/?f=h%3D%20v_it%20%2B%20%5Cfrac%7B1%7D%7B2%7Dgt%5E2%5C%5C)
where,
g = acceleration due to gravity = 9.81 m/s²
h = height = 17.5 m
vi = initial speed = 9.81 m/s
t = time = ?
Therefore,
![17.5 = (9.81)t+\frac{1}{2}(9.81)t^2\\4.905t^2+9.81t-17.5=0](https://tex.z-dn.net/?f=17.5%20%3D%20%289.81%29t%2B%5Cfrac%7B1%7D%7B2%7D%289.81%29t%5E2%5C%5C4.905t%5E2%2B9.81t-17.5%3D0)
solving this quadratic equation using the quadratic formula, we get:
t = -3.13 s (OR) t = 1.13 s
Since time can not have a negative value.
Therefore,
<u>t = 1.13 s</u>
Learn more about equations of motion here:
brainly.com/question/20594939?referrer=searchResults
The attached picture shows the equations of motion in the horizontal and vertical directions.
10km/10min is a legitimate speed. So is meters/sec, km/hour (kph), etc.
Kph is very common for vehicles:
10 km/10 min (60 min/hr) = 60 kph
Explanation:
Speed or velocity (V) = 35 m/s
Kinetic energy (K. E) = 1500 Joule
mass (m) = ?
We know
K.E = 1/2 * m * v²
1500 = 1/2 * m * 35²
1500 * 2 = 1225m
m = 3000 / 1225
m = 2.45 kg
The mass of the object is 2.45 kg
Hope it will help :)