<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
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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
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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:
Physically, the gas constant is the constant of proportionality that relates the energy scale in physics to the temperature scale, when a mole of particles at the stated temperature is being considered. Thus, the value of the gas constant ultimately derives from historical decisions and accidents in the setting of the energy and temperature scales, plus similar historical setting of the value of the molar scale used for the counting of particles.
Explanation:
Pa follow
The answer is 21m because the motion is in one dimension with constant acceleration.
The initial velocity is 0, because it started from rest, the acceleration <span>ax</span> is <span>4.7<span>m<span>s2</span></span></span>, and the time t is <span>3.0s</span>
Plugging in our known values, we have
<span>Δx=<span>(0)</span><span>(3.0s)</span>+<span>12</span><span>(4.7<span>m<span>s2</span></span>)</span><span><span>(3.0s)</span>2</span>=<span>21<span>m</span></span></span>
Answer:
D
Explanation:
appearance is not a imp factor . location could be imp. becoz a proper environment is need to study such courses.
Answer:
Explanation:
1. We use the conservation of momentum for before the raining and after. And also we take into account that in 0.5h the accumulated water is
100kg/h*0.5h = 50kg
2. the momentum does not conserve because the drag force of water makes that the boat loses velocity
3. If we assume that the force of the boat before the raining is
where we have assumed that the acceleration of the boat is 1m/s{2} just before the rain starts
And if we take the net force as
where we take v=1m/s because we are taking into account tha velocity just after the rain stars.
I hope this is useful for you
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