Data:
h=25 m
v₀=0 m/s
g=9,8 m/s²
we use this formula
h=h₀+v₀.t+(1/2).g.t²
25 m=0 m+0 m/s.t+(1/2).(9,8 m/s²).t²
25 m=4,9 m/s².t²
t²=25 m /4,9 m/s²
t²=5,1 s²
t=√(5,1 s²)=2,26 s
Now, we calculate the última velocity
v=v₀+gt
v=0 m/s+9,8 m/s².(2,26 s)=22,14 m/s≈22 m/s
solution: c.) 22 m/s
The answer to the question is choice 2
Answer:
C
Explanation:
Because everything on Earth falls at the same speed, the masses of the balls do not matter. Since the acceleration due to gravity is constant, their speeds will both be increasing at the same rate, and therefore the difference in speeds would remain constant until they hit the ground. Hope this helps!
Answer:
The angle of launch is 52.49 Degree.
Explanation:
The Range R and Height H of a thrown object is calculated using the formula,
R=V₀² sin(2φ)/g
H=V₀²sin²(φ)/g
From these equations it can be written,
V₀²=R g/ sin(2φ)
V₀²=H g/ sin²(φ)
These values are equal so it can be written by equating these equations,
R g/sin(2φ)=H g/sin²(φ)
tan(φ)= 2H/R
Given H=72.3 m and R=111 m, the angle of launch is,
tan(φ)= 2*72.3/111
φ= 52.49 Degree.
Check out other solutions,
brainly.com/question/1495042
#SPJ10
Answer:
a) τmax = 586.78 P.S.I.
b) σmax = 15942.23 P.S.I
Explanation:
D = 3.81 in
d = 3.24 in
P = 930 lb
L = 3.7 ft = 44.4 in
a) The maximum horizontal shear stress can be obtained as follows
τ = V*Q / (t*I)
where
V = P = 930 lb
Q = (2/3)*(R³- r³) = (1/12)*(D³- d³) = (1/12)*((3.81 in)³- (3.24 in)³)
⇒ Q = 1.7745 in³
t = D - d = 3.81 in - 3.24 in = 0.57 in
I = (π/64)*(D⁴-d⁴) = (π/64)*((3.81 in)⁴- (3.24 in)⁴) = 4.9341 in⁴
then
τ = (930 lb)*(1.7745 in³) / (0.57 in*4.9341 in⁴)
⇒ τmax = 586.78 P.S.I.
b) We can apply the following equation in order to get the maximum tension bending stress in the pipe
σmax = Mmax *y / I
where
Mmax = P*L = 930 lb*44.4 in = 41292 lb-in
y = D/2 = 3.81 in /2 = 1.905 in
I = 4.9341 in⁴
then
σmax = (41292 lb-in)*(1.905 in) / (4.9341 in⁴) = 15942.23 P.S.I
