An object is in motion when it undergoes a continuous change of what?
1 answer:
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Answer:
Option B, 93 cm
Explanation:
An diagram of the seed's motion is attached to this solution.
This is very close to a projectile motion question. And the quantity to be calculated, how far along the grant a seed released would travel is called the Range.
And this would be obtained from the equations of motion,
First of, the height of the plant is related to some quantities of the motion with this relation.
H = u(y) t + 0.5g(t^2)
U(y) = initial vertical component of velocity = 0 m/s, H = height at which motion began, = 20cm = 0.2 m
That means t = √(2H/g)
The horizontal distance covered, R,
R = u(x) t + 0.5g(t^2) = u(x) t (the second part of the equation goes to zero as the vertical component of the acceleration of this motion is 0)
(substituting the t = √(2H/g) derived from above
R = u(x) √(2H/g)
Where u(x) = the initial horizontal component of the bomb's velocity = maximum initial speed, that is, 4.6 m/s, H = vertical height at which the seed was released = 20 cm = 0.2 m, g = acceleration due to gravity = 9.8 m/s2
R = 4.6 √(2×0.2/9.8) = 0.929 m = 0.93 m = 93 cm. Option B.
QED!
Answer:
5.7 m
Explanation:
AD = length of the ladder = L = 8 m
AB = distance of the center of mass of the ladder = (0.5) L = (0.5) 8 = 4 m
AC = distance of person on the ladder from the bottom end = x
W = weight of the ladder = 240 N
= weight of the person = 710 N
F = force by the wall on the ladder
N = normal force by ground on the ladder = ?
Using equilibrium of force along the vertical direction
N = + W
N = 710 + 240
N = 950 N
μ = Coefficient of static friction = 0.55
f =static frictional force on the ladder
Static frictional force is given as
f = μ N
f = (0.55) (950)
f = 522.5 N
Force equation along the horizontal direction is given as
F = f
F = 522.5 N
using equilibrium of torque about point A
F Sin50 (AD) = W Cos50 (AB) + ( Cos50 (AC))
(522.5) Sin50 (8) = (240) Cos50 (4) + (710) Cos50 (x)
x = 5.7 m
