Answer:
The velocity of motion at which the occupants of the car appear to weigh 20% less than their normal weight is approximately 19.81 m/s
Explanation:
The given parameters are;
The curvature of the hill, r = 200 m
Due to the velocity, v, the occupants weight = 20% less than the normal weight
The outward force of an object due to centripetal (motion) force is given by the following equation;
Where;
r = The radius of curvature of the hill = 200 m
Given that the weight of the occupants, W = m × g, we have;
v = √(0.2 × g × r)
By substitution, we have;
v = √(0.2 × 9.81 × 200) ≈ 19.81
The velocity of motion at which the occupants of the car appear to weigh 20% less than their normal weight ≈ 19.81 m/s.
<span>The process shown in this diagram contributed great amounts of heat to the young planet Earth and is best known as radioactive
decay. Decay is known to release large amounts of heat. </span>
Consider horizontal component:
Using the formula:v2=u2+2a<span>s</span>
<span>0=(16sin<span>470</span><span>)2</span>+2(−9.81)s</span><span>s=6.98m</span><span>=7.0m(2s.f.)</span>The maximu height from the ground is 8.5m
To start with solving this
problem, let us assume a launch angle of 45 degrees since that gives out the
maximum range for given initial speed. Also assuming that it was launched at
ground level since no initial height was given. Using g = 9.8 m/s^2, the
initial velocity is calculated using the formula:
(v sinθ)^2 = (v0 sinθ)^2
– 2 g d
where v is final
velocity = 0 at the peak, v0 is the initial velocity, d is distance = 11 m
Rearranging to find for
v0: <span>
v0 = sqrt (d * g/ sin(2 θ)) </span>
<span>v0 = 10.383 m/s</span>
Well, first of all, let's not make something complicated ... for which we're missing an important component anyway ... out of something simple. There's more to "velocity" than 'meters per second'. So let's just call that what it is: " Speed ".
Here's what you've told us:
-- Something is moving along a ruler or meter stick.
-- At every instant from time=0 to time=5, its speed was constant at 4 m/s.
-- At time=0, it was located at 6m .
We can calculate:
-- Moving steadily at 4 m/s for 5 seconds, it moved (4 x 5) = 20 meters.
-- If it was located at 6m when time=0, then at time=6, it had advanced
to (6 + 20) = 26 meters.
