Answer:
<em>The range is 35.35 m</em>
Explanation:
<u>Projectile Motion</u>
It's the type of motion that experiences an object projected near the Earth's surface and moves along a curved path exclusively under the action of gravity.
Being vo the initial speed of the object, θ the initial launch angle, and 
the acceleration of gravity, then the maximum horizontal distance traveled by the object (also called Range) is:
The projectile was launched at an angle of θ=30° with an initial speed vo=20 m/s. Calculating the range:
The range is 35.35 m
The correct answer is:
<span>B.) At terminal velocity there is no net force
In fact, when the parachutist reaches the terminal velocity, his velocity does not change any more. It means that the acceleration acting on the parachutist is zero, and for Newton's second law, this means the net force acting on him is zero:
</span>
<span>because the acceleration is zero: a=0.
This also means that the two relevant forces acting on the parachutist (gravity, downward, and air resistance, upward) are balanced to produce a net force equal to zero.</span>
The proof that the earth is rotating is the happens of night and day also the seasons, eg. winter, summer, autumn.
i think it’s B. sorry if i’m wrong
Answer:
E=
Explanation:
We are given that
Charge on ring= Q
Radius of ring=a
We have to find the magnitude of electric filed on the axis at distance a from the ring's center.
We know that the electric field at distance x from the center of ring of radius R is given by
Substitute x=a and R=a
Then, we get
Where K=
Hence, the magnitude of the electric filed due to charged ring on the axis of ring at distance a from the ring's center=
