That would be 0 degrees Celsius aka the melting point of water.... If you look at the diagram I attached you notice that at 0 degrees Celsius it is flat, this is because much heat is needed at this point for water to rise to 1 degree... It is the same for the boiling point (100)
Hope this helps! (If correct, please rank as brainliest answer) :)
There's so much going on here, in a short period of time.
<u>Before the kick</u>, as the foot swings toward the ball . . .
-- The net force on the ball is zero. That's why it just lays there and
does not accelerate in any direction.
-- The net force on the foot is 500N, originating in the leg, causing it to
accelerate toward the ball.
<u>During the kick</u> ... the 0.1 second or so that the foot is in contact with the ball ...
-- The net force on the ball is 500N. That's what makes it accelerate from
just laying there to taking off on a high arc.
-- The net force on the foot is zero ... 500N from the leg, pointing forward,
and 500N as the reaction force from the ball, pointing backward.
That's how the leg's speed remains constant ... creating a dent in the ball
until the ball accelerates to match the speed of the foot, and then drawing
out of the dent, as the ball accelerates to exceed the speed of the foot and
draw away from it.
Answer:
The velocity will be v1 = 0.58[m/s]
Explanation:
This problem can be solved by the law of conservation of the moment, which explains that the moment of a system remains constant because there are no external forces acting on it.
We have the following initial data:
m1 = mass of the skater = 55 [kg]
m2 = mass of the ball = 3 [kg]
v2 = velocity of the ball = 8 [m/s]
Therefore:
![m_{1}*v_{1}+m_{2}*v_{2}=m_{1}*v_{1}+m_{2}*v_{2}\\(50*0)+(3*0)=(50*v_{1})+(3*8)\\50+3-24=50*v_{1}\\v_{1}= 0.58[m/s]](https://tex.z-dn.net/?f=m_%7B1%7D%2Av_%7B1%7D%2Bm_%7B2%7D%2Av_%7B2%7D%3Dm_%7B1%7D%2Av_%7B1%7D%2Bm_%7B2%7D%2Av_%7B2%7D%5C%5C%2850%2A0%29%2B%283%2A0%29%3D%2850%2Av_%7B1%7D%29%2B%283%2A8%29%5C%5C50%2B3-24%3D50%2Av_%7B1%7D%5C%5Cv_%7B1%7D%3D%200.58%5Bm%2Fs%5D)
Answer:
See the answers below.
Explanation:
To solve this problem we must use the following equation of kinematics.
where:
Vf = final velocity = 10 [m/s]
Vo = initial velocity = 40 [m/s]
t = time = 5 [s]
a = acceleration [m/s²]
Now replacing:
![10=40-a*5\\40-10=a*5\\30=5*a\\a=6[m/s^{2}]](https://tex.z-dn.net/?f=10%3D40-a%2A5%5C%5C40-10%3Da%2A5%5C%5C30%3D5%2Aa%5C%5Ca%3D6%5Bm%2Fs%5E%7B2%7D%5D)
Note: The negative sign in the above equation means that the velecity is decreasing.
2)
To solve this second part we must use the following equation of kinematics.
where:
x = distance [m]
![(10)^{2} =(40)^{2} -2*6*x\\100=1600-12*x\\12*x=1600-100\\12*x=1500\\x=125[m]](https://tex.z-dn.net/?f=%2810%29%5E%7B2%7D%20%3D%2840%29%5E%7B2%7D%20-2%2A6%2Ax%5C%5C100%3D1600-12%2Ax%5C%5C12%2Ax%3D1600-100%5C%5C12%2Ax%3D1500%5C%5Cx%3D125%5Bm%5D)
