Hi there!
In this instance, the object spinning in a horizontal circle will experience a net force in the horizontal direction due to tension.
The net force is equivalent to the centripetal force, so:
∑F = T
mv²/r = T
Solve for v:
v = √rT/m
v = 13.96 m/s
Answer:
To maintain enough time to prevent a collision, a system operating in air traffic where aircraft speed does not
fall below 100 km/h (most medium-sized UAVs and GA aircraft) will need to be able to detect obstacles which
subtend an arc-width of as small as 0.125 mra
Answer:
1.a) 1 kJ
1.b) 4 kJ
ratio 1:4
1.c) 4 times as before
2.a) 3.33 m/s2
Explanation:
1.a) bicycle's velocity =Displacement/time
=100/20 m/s
=5 m/s
bicycler's KE =1/2 *mass*(velocity)^2
=1/2*80*5^2
=1000 J = 1 kJ
1.b) bicycle's new velocity =200/20 m/s
=10 m/s
bicycler's new KE =1/2*80*10^2
=4000 J = 4 kJ
Ratio= KE 1 :KE new
= 1 :4
1.c) when bicycler's speed was doubled it increased the KE by 4 times (2^2). because In KE we consider the square of the speed , so the factor we increase the speed , the KE will get increased with the square value of it
ex : speed is triple the prior value , then the KE is as 3^2 times as before. that is 9 times
2.a) car acceleration = (20-0)/6 m/s2
= 3.33 m/s2
Answer:
<h2>a) Time elapsed before the bullet hits the ground is 0.553 seconds.</h2><h2>b)
The bullet travels horizontally 110.6 m</h2>
Explanation:
a) Consider the vertical motion of bullet
We have equation of motion s = ut + 0.5 at²
Initial velocity, u = 0 m/s
Acceleration, a = 9.81 m/s²
Displacement, s = 1.5 m
Substituting
s = ut + 0.5 at²
1.5 = 0 x t + 0.5 x 9.81 xt²
t = 0.553 s
Time elapsed before the bullet hits the ground is 0.553 seconds.
b) Consider the horizontal motion of bullet
We have equation of motion s = ut + 0.5 at²
Initial velocity, u = 200 m/s
Acceleration, a = 0 m/s²
Time, t = 0.553 s
Substituting
s = ut + 0.5 at²
s = 200 x 0.553 + 0.5 x 0 x 0.553²
s = 110.6 m
The bullet travels horizontally 110.6 m
Answer:
21.59 m/s
Explanation:
recall that one of the equations of motions can be expressed as
v² = u² + 2as
where,
v = final velocity (we are asked to find this)
u = initial velocity = 0m/s (because it says that it starts from rest)
a = acceleration = 3.7m/s²
s = distance travelled = 63 m
simply substitute the known values above into the equation:
v² = u² + 2as
v² = 0² + 2(3.7)(63)
v² = 466.2
v = √466.2
v = 21.59 m/s
