<span>A person expended 500 newtons to move a full wheelbarrow 30 meters how much work was done?
</span>In physics, a force is said to do work<span> if, when acting, there is a displacement of the point of application in the direction of the force. The work done would be calculated as follows:
</span>
W = F x d = 500 N (30 m) = 15000 J or 15 kJ
From conservation of momentum, the ram force can be calculated similarly to rocket thrust:
F = d(mv)/dt = vdm/dt.
<span>In other words, the force needed to decelerate the wind equals the force that would be needed to produce it.
</span><span> v = 120/3.6 = 33.33 m/s
</span><span> dm/dt = v*area*density
</span> dm/dt = (33.33)*((45)*(75))*(1.3)
dm/dt = <span>
146235.375 </span><span>kg/s
</span><span> F = v^2*area*density
</span> F = (33.33)^2*((45)*(75))*(1.3) = <span>
<span>4874025 </span></span><span>N
</span> This differs by a factor of 2 from Bernoulli's equation, which relates velocity and pressure difference in reference not to a head-on collision of the fluid with a surface but to a fluid moving tangentially to the surface. Also, a typical mass-based drag equation, like Bernoulli's equation, has a coefficient of 1/2; however, it refers to a body moving through a fluid, where the fluid encountered by the body is not stopped relative to the body (i.e., brought up to its speed) like is the case in this problem.
Answer:
The magnitude of Electric Field is 
Explanation:
Given:
- Radius of the solid sphere=R
- Total charge of the sphere=Q
Let consider a Gaussian surface at a distance of r such that 0<r>R in the shape of sphere such that the electric Field due to this E and it is radially outwards.
The charge inside this Gaussian surface volume we have , 
Now using Gauss Law we have
