<span>let the fsh jump with initial velocity (u) in direction (angle p) with horizontal
it can cross and reach top of trajectory if its top height h = 1.5m
and horizontal distance d = (1/2) Range
--------------------------------------...
let t be top height time
at top height, vertical component of its velocity =0
vy = 0 = u sin p - gt
t = u sin p/g
h = [u sin p]*t - 0.5 g[t[^2
1.5 = u^2 sin^2 p/g - u^2 sin^2 p/2g
u^2 sin^2 p/2g = 1.5
u^2 sin^2 p = 1.5*2*9.8 = 29.4
u sin p = 5.42 m/s >>>>>>>>>>>>>>> V-component
=====================
t = HALF the time of flight
d = (1/2) Range (R) = (1/2) [2 u^2 sin p cos p/g]
1 = u^2 sin p cos p/g
u sin p * u cos p = 9.8
5.42 * u cos p = 9.8
u cos p = 1.81 m/s >>>>>>>>>>>>> H-component
check>>
u = sqrt[u^2 cos^2 p + u^2 sin^2 p] = 5.71 m/s
u < less than fish's potential jump speed 6.26 m/s
so it will able to cross</span>
Answer:
30N*s
Explanation:
Given the following data;
Force = 10N
Time = 3 seconds
To find the impulse;
Impulse = force * time
Substituting into the equation, we have;
Impulse = 10 * 3
Impulse = 30Ns
Answer:
A box sits stationary on a ramp
Explanation:
Static friction is a force which keeps an object at rest as it is in the case of the box. It has to be overcome for the object to be set into motion.
Static force of friction is calculated as follows:
F= μη
F is static force of friction.
μ is the coefficient of static friction.
η is the normal force.
- Angle (θ) = 60°
- Force (F) = 20 N
- Distance (s) = 200 m
- Therefore, work done
- = Fs Cos θ
- = (20 × 200 × Cos 60°) J
- = (20 × 200 × 1/2) J
- = (20 × 100) J
- = 2000 J
<u>Answer</u><u>:</u>
<u>2</u><u>0</u><u>0</u><u>0</u><u> </u><u>J</u>
Hope you could get an idea from here.
Doubt clarification - use comment section.
What is your choses .becuse it mit be a place
