A skydiver is using wind to land on a target that is 50 m away horizontally. The skydiver starts from a height of 70 m and is fa
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Answer:
(a). The initial velocity is 28.58m/s
(b). The speed when touching the ground is 33.3m/s.
Explanation:
The equations governing the position of the projectile are
where is the initial velocity.
(a).
When the projectile hits the 50m mark, ; therefore,
solving for we get:
Thus, the projectile must hit the 50m mark in 1.75s, and this condition demands from equation (1) that
which gives
(b).
The horizontal velocity remains unchanged just before the projectile touches the ground because gravity acts only along the vertical direction; therefore,
the vertical component of the velocity is
which gives a speed of
Answer:
Part A : E = ε₀ Q₁/R₁² Volt/meter
Part B : V = ε₀ Q₁/R₁ Volt
Explanation:
Given that,
Charge distributed on the sphere is Q₁
The radius of sphere is R
₁
The electric potential at infinity is 0
<em>Part A</em>
The space around a charge in which its influence is felt is known in the electric field. The strength at any point inside the electric field is defined by the force experienced by a unit positive charge placed at that point.
If a unit positive charge is placed at the surface it experiences a force according to the Coulomb law is given by
F = ε₀ Q₁/R₁²
Then the electric field at that point is
E = F/1
E = ε₀ Q₁/R₁² Volt/meter
Part B
The electric potential at a point is defined as the amount of work done in moving a unit positive charge from infinity to that point against electric forces.
Thus, the electric potential at the surface of the sphere of radius R₁ and charge distribution Q₁ is given by the relation
V = ε₀ Q₁/R₁ Volt