Well, you didn't ask a question, and 4 m/s is not an angular speed.
So all I can offer is a couple of observations:
1). The tension in the rope is
M V² / R = (1.5 kg) x (4 m/s)² / R
= (24 kg-m²/s²) / (distance of the ball from the pole).
2). Tetherball was the only thing I played at camp,
more than 60 years ago, and I loved it !
It was a tough game, because we had to skin
our own T.Rex and use his hide to make the ball
and his guts for the rope.
The period of the orbit would increase as well
Explanation:
We can answer this question by applying Kepler's third law, which states that:
"The square of the orbital period of a planet around the Sun is proportional to the cube of the semi-major axis of its orbit"
Mathematically,

Where
T is the orbital period
a is the semi-major axis of the orbit
In this problem, the question asks what happens if the distance of the Earth from the Sun increases. Increasing this distance means increasing the semi-major axis of the orbit,
: but as we saw from the previous equation, the orbital period of the Earth is proportional to
, therefore as
increases, T increases as well.
Therefore, the period of the orbit would increase.
Learn more about Kepler's third law:
brainly.com/question/11168300
#LearnwithBrainly
Answer:
t = 1.4[s]
Explanation:
To solve this problem we must use the principle of conservation of linear momentum, which tells us that momentum is conserved before and after applying a force to a body. We must remember that the impulse can be calculated by means of the following equation.

where:
P = impulse or lineal momentum [kg*m/s]
m = mass = 50 [kg]
v = velocity [m/s]
F = force = 200[N]
t = time = [s]
Now we must be clear that the final linear momentum must be equal to the original linear momentum plus the applied momentum. In this way we can deduce the following equation.

where:
m₁ = mass of the object = 50 [kg]
v₁ = velocity of the object before the impulse = 18.2 [m/s]
v₂ = velocity of the object after the impulse = 12.6 [m/s]
![(50*18.2)-200*t=50*12.6\\910-200*t=630\\200*t=910-630\\200*t=280\\t=1.4[s]](https://tex.z-dn.net/?f=%2850%2A18.2%29-200%2At%3D50%2A12.6%5C%5C910-200%2At%3D630%5C%5C200%2At%3D910-630%5C%5C200%2At%3D280%5C%5Ct%3D1.4%5Bs%5D)
Their is going to be 4 sig figs
F=ma
F=125 N
m= 50 kg
125=50a
a=2.5 m/s^2