Knowing the initial velocity and angle, the horizontal range formula is given by R= V^2sin(2teta) / g, so we can get
sin(2teta)=Rg/V^2
sin(2teta)= (180 x 9.8)/ 80^2= 0.27, sin(2teta)=0.27, 2teta=arcsin(0.27)=15.66, so teta=15.66/2
teta=7.83°
Answer:
μ = 0.436
Explanation:
Given:
Change in diameter, ΔD = 7 × 10⁻³ mm
Original diameter, D = 11.2 mm = 11.2 × 10⁻³ m
Applied force, P = 14100 N
Cross-section area of the specimen, A = 
= 
Now,
elongation due to tensile force is given as:
or
on substituting the values, we get
or
where,
is the strain in the direction of force
Now,
now, the poisson ratio, μ is given as:
on substituting the values we get,
By doing this in your calculator: 4/3 times <span>π and 454 to the third power. good luck :)</span>
Answer: When the worker is on the top rung
Explanation: When the ladder was initially resting on the wall, the torque from the normal reaction on ladder from the horizontal surface is equal to the torque from the vertical surface on ladder.
The weight of the worker produces a torque which is in the direction of the torque from the normal reaction on ladder, produced by the vertical surface. Therefore for the ladder to stay in rotational equilibrium, the torque on ladder from the normal reaction produced by the horizontal surface must increase.
This increase is possible when the worker is on the lower rung, but as the worker goes high, the magnitude of normal reaction from the vertical surface would increase, thereby increasing the risk of slipping of ladder.
