Answer:
1.05 N
Explanation:
K = 0.7 N/m
e = 1.5 m
F = ?
from Hooke's law of elasticity
F = Ke
= 0.7×1.5
= 1.05 N
A) mass m with F1 acting in the positive x direction and F2 acting perpendicular in the positive y direction<span>
m = 5.00 kg
F1=20.0N ... x direction
F2=15.00N</span><span> ... y direction
Net force ^2 = F1^2 + F2^2 = (20N)^2 + (15n)^2 = 625N^2 =>
Net force = √625 = 25N
F = m*a => a = F/m = 25.0 N /5.00 kg = 5 m/s^2
Answer: 5.00 m/s^2
b) mass m with F1 acting in the positive x direction and F2 acting on the object at 60 degrees above the horizontal.
</span>
<span>m = 5.00 kg
F1=20.0N ... x direction
F2=15.00N</span><span> ... 60 degress above x direction
Components of F2
F2,x = F2*cos(60) = 15N / 2 = 7.5N
F2, y = F2*sin(60) = 15N* 0.866 = 12.99 N ≈ 13 N
Total force in x = F1 + F2,x = 20.0 N + 7.5 N = 27.5 N
Total force in y = F2,y = 13.0 N
Net force^2 = (27.5N)^2 + (13.0N)^2 = 925.25 N^2 = Net force = √(925.25N^2) =
= 30.42N
a = F /m = 30.42 N / 5.00 kg = 6.08 m/s^2
Answer: 6.08 m/s^2
</span>
Answer:
5.843 m
Explanation:
suppose that the arrow leave the bow with a horizontal speed , towards he bull's eye.
lets consider that horizontal motion
distance = speed * time
time = 40/ 37 = 1.081 s
arrow doesnot have a initial vertical velocity component. but it has a vertical motion due to gravity , which may cause a miss of the target.
applying motion equation
(assume g = 10 m/s²)

Arrow misses the target by 5.843m ig the arrow us split horizontally
Answer:
Part a)

Part b)

Part c)

Explanation:
Part a)
As we know that it is having constant torque so here the time taken by it to accelerate is given as



Part b)
angular displacement is given as



Part c)
As we know that the angular deceleration produced by the brakes is given as

now we have



As we know that

so we have
