Answer:
less than the weight of the block.
Explanation:
From the free body diagram, we get.
The normal force is N = Mg cosθ
The tension in the string is T = Mg sinθ
Wight of the block when the block is static, W = Mg
Now since the magnitude of cosθ is in the range of : 0 < cosθ < 1,
therefore, the normal force is less than the weight of the static block.
Answer:
We are given the trajectory of a projectile:
y=H+xtan(θ)−g2u2x2(1+tan2(θ)),
where H is the initial height, g is the (positive) gravitational constant and u is the initial speed. Since we are looking for the maximum range we set y=0 (i.e. the projectile is on the ground). If we let L=u2/g, then
H+xtan(θ)−12Lx2(1+tan2(θ))=0
Differentiate both sides with respect to θ.
dxdθtan(θ)+xsec2(θ)−[1Lxdxdθ(1+tan2(θ))+12Lx2(2tan(θ)sec2(θ))]=0
Solving for dxdθ yields
dxdθ=xsec2(θ)[xLtan(θ)−1]tan(θ)−xL(1+tan2(θ))
This derivative is 0 when tan(θ)=Lx and hence this corresponds to a critical number θ for the range of the projectile. We should now show that the x value it corresponds to is a maximum, but I'll just assume that's the case. It pretty obvious in the setting of the problem. Finally, we replace tan(θ) with Lx in the second equation from the top and solve for x.
H+L−12Lx2−L2=0.
This leads immediately to x=L2+2LH−−−−−−−−√. The angle θ can now be found easily.
A, D is the correct answers
5 newton forward because the force of the engine is greater than the force of friction so that means the car is going forward.
You can think of this: when friction is greater than the force of an object its slowing down but when the force of the car is greater than the force of friction its speeding up.
Hope this helps
Answer: 3 meters
Explanation:
If you were to use a 3 line answer then it would look like this:
d = W/F
d = 3000/1000
d = 3 meters
