Answer:
The ramp lands at an horizontal distance of 3.989 m
Explanation:
Range: Range is defined as the horizontal distance of a projectile from the point of projection to the point where the projectile hits the projection plane again. It is measure in meters (m)
R = U²sin2∅/g..................................... Equation 1
Where R = The horizontal distance from the end of the ramp, U = rider's speed, ∅ = the ramp's angle to the horizontal.
<em>Given: U = 6.3 m/s, ∅ = 40°, g = 9.8 m/s²</em>
<em>Substituting these values into equation 1</em>
<em>R = [6.3²sin2(40)]/9.8</em>
<em>R = (39.69×sin80)/9.8</em>
<em>R = 39.69×0.985/9.8</em>
R = 3.989 m
Thus the ramp lands at an horizontal distance of 3.989 m
Given:
Time taken for the travel: 40 mins=2400 secs
Velocity: 1.2m/s
Speed and velocity have the same magnitude. Speed is a scalar where as velocity is a vector quantity.
Distance traveled= speed x time.
Distance traveled: 1.2 x 2400=2880m
Answer:
a) the spring will stretch 60.19 mm
with the same box attached as it accelerates upwards
b) spring will be relaxed when the elevator accelerates downwards at 9.81 m/s²
Explanation:
Given that;
Gravitational acceleration g = 9.81 m/s²
Mass m = 5 kg
Extension of the spring X = 50 mm = 0.05 m
Spring constant k = ?
we know that;
mg = kX
5 × 9.81 = k(0.05)
k = 981 N/m
a)
Given that; Acceleration of the elevator a = 2 m/s² upwards
Extension of the spring in this situation = X1
Force exerted by the spring = F
we know that;
ma = F - mg
ma = kX1 - mg
we substitute
5 × 2 = 981 × X1 - (5 ×9.81 )
X1 = 0.06019 m
X1 = 60.19 mm
Therefore the spring will stretch 60.19 mm
with the same box attached as it accelerates upwards
B)
Acceleration of the elevator = a
The spring is relaxed i.e, it is not exerting any force on the box.
Only the weight force of the box is exerted on the box.
ma = mg
a = g
a = 9.81 m/s² downwards.
Therefore spring will be relaxed when the elevator accelerates downwards at 9.81 m/s²
There are no probably-true statements on the list of choices that you provided.