Considering that the grasshopper starts on the ground and lands on the ground, there is no change in elevation. The range of the grasshopper is given by: <span>R=<span><span><span>v2</span>sin2θ</span>g</span></span>
If we now take the information that this is a maximal distance, the maximum range is given when the angle of launch is 45 degrees. This yields the equation: <span><span>R<span>max</span></span>=<span><span>v2</span>g</span></span>
We know R_max=0.4 m and g=9.8 m/s^2, so we can solve for v: <span>v=1.98m/s</span>
We are asked for the horizontal component, which by trigonometry is: <span><span>vx</span>=vcosθ</span>
We again take our angle of launch as 45 degrees and arrive at: <span><span><span>vx</span>=1.40m/s</span></span>
Answer:
The essential characteristics of an object in equilibrium is acceleration of the object must be zero.
Explanation:
Equilibrium: An object acted upon by several forces is said to be in equilibrium if it does no it moves or rotate. Equilibrium can be defined as the state of stability of an object.
Condition for equilibrium
(i) The resultant force acting on the object must me equal to zero.
(ii) The sum of moment acting on the object must be equal to zero.
But.
F = ma..................... Equation 1
Where F = net force , m = mass, a = acceleration.
from the equation above,
For F = 0 N, a must be equal to zero.
<em>From the above, the essential characteristics of an object in equilibrium is acceleration must be zero</em>
Answer:
4334.4 J
Explanation:
Work done equals to kinetic energy change
KE=½mv²
Change in KE is given by
∆KE=½m(v²-u²)
Where m is mass of water-skier, KE is kinetic energy, ∆KE is the change in kinetic energy, v is final velocity and u is initial velocity.
Substituting 72 kg for m, 12.1 m/s for v and 5.10 m/s for u then
∆KE=½*72(12.1²-5.10²)=4334.4J
Therefore, the work done by the net external force acting on the skier is equal to 4334.4 J