Answer:
v = 8.72 m/s
Explanation:
To find the speed of the raindrop joint to the mosquito, you take into account the momentum conservation law for an inelastic collision. Before the collision the total momentum of raindrop and mosquito must be equal to the total momentum of both raindrop and mosquito after the collision.
(1)
v1: speed of the mosquito before the collision= 0 m/s (it is at rest)
v2: speed of the raindrop before the collision = 8.9 m/s
m1: mass of the mosquito
m2: mass of the raindrop = 50m1 (50 time more massive that the mosquito)
v: speed of both raindrop and mosquito after the collision
You solve the equation (1) for v and replace the values of the rest of the parameters:
hence, after the inelastic collision the speed of the raindrop andmosquito is 8.72 m/s
Answer:
V₀ₓ = 10.94 m/s
V₀y = 18.87 m/s
Explanation:
To find the launch velocity, we use 1st equation of motion.
Vf = Vi + at
where,
Vf = Final Velocity of Ball = Launch Speed = V₀ = ?
Vi = Initial Velocity = 0 m/s (Since ball was initially at rest)
a = acceleration = 376 m/s²
t = time = 0.058 s
Therefore,
V₀ = 0 m/s + (376 m/s²)(0.058 s)
V₀ = 21.81 m/s
Now, for x-component:
V₀ₓ = V₀ Cos θ
where,
V₀ₓ = x-component of launch velocity = ?
θ = Angle with horizontal = 59.9⁰
V₀ₓ = (21.81 m/s)(Cos 59.9°)
<u>V₀ₓ = 10.94 m/s</u>
<u></u>
for y-component:
V₀ₓ = V₀ Sin θ
where,
V₀y = y-component of launch velocity = ?
θ = Angle with horizontal = 59.9⁰
V₀y = (21.81 m/s)(Sin 59.9°)
<u>V₀y = 18.87 m/s</u>
<u></u>
The correct answer would be compound
Answer:
Explanation:
The acceleration of an object is defined as:
where
v is the final velocity
u is the initial velocity
t is the time
For the plane in this problem, we have:
u = 0
v = 60 m/s
t = 20 s
Substituting into the equation,
1) The buoyant force acting on an object immersed in a fluid is:
where
is the density of the fluid,
is the volume of displaced fluid, and
is the gravitational acceleration.
2) We must calculate the volume of displaced fluid. Since the titanium object is completely immersed in the fluid (air), this volume corresponds to the volume of 1 Kg of titanium, whose density is
. Using the relationship between density, volume and mass, we find
3) Now we can recall the formula written at step 1) and calculate the buoyant force. The air density is
, so we have
4) The weight of 1 Kg of titanium is instead:
So, the buoyant force is negligible compared to the weight.