Answer:
20m/s^2
Explanation:
Acceleration=Change in velocity/time taken for change
40-20/1
20m/s^2
The initial speed of the shot is 15.02 m/s.
The Shot put is released at a height y<em> </em>from the ground with a speed u. It is released at an angle θ to the horizontal. In a time t, the shot put travels a distance <em>R</em> horizontally.
Pl refer to the attached diagram.
Resolve the velocity u into horizontal and vertical components, u ₓ=ucosθ and uy=u sinθ. The horizontal component remains constant in the absence of air resistance, while the vertical component varies due to the action of the gravitational force.
Write an expression for R.

Therefore,

In the time t, the net displacement of the shotput is y in the downward direction.
Use the equation of motion,

Substitute the value of t from equation (1).

Substitute -2.10 m for y, 24.77 m for R and 38.0° for θ and solve for u.

The shot put was thrown with a speed 15.02 m/s.
Answer:
I Would go with Kye.
Explanation:
Why i would go with her is because she has more life experience with you. She also knows you better. I would usually go with the closest one and the one who knows you the best.
Hope this helps.
<3 Have a good day!
No force is required to lift that balloon. In fact, force is required to hold it down, and if you let go, it's up, up, and away.
Since the balloon's density is less than the density of the air around it, it's lighter than the air it displaces, there is a net upward buoyant force acting on it, and it floats up !
Answer:
q = 2.65 10⁻⁶ C
Explanation:
For this exercise we use Coulomb's law
F =
In this case they indicate that the load is of equal magnitude
q₁ = q₂ = q
the force is attractive because the signs of the charges are opposite
F =
q =
we calculate
q =
q =
Ra 7 10-12
q = 2.65 10⁻⁶ C