Answer:
Explanation:
a ) Let the distance required in former case be d₁ .
initial velocity u = 30 m /s , final velocity v =0 , deceleration a = 7.00 m /s²
v² = u² - 2 a s
0 = 30² - 2 x 7 x d₁
d₁ = 64.28 m
b) initial velocity u = 30 m /s , final velocity v =0 , deceleration a = 5.00 m /s²
v² = u² - 2 a s
0 = 30² - 2 x 5 x d₂
d₂ = 90 m
c)
t = .5 s
s₁ = ut - .5 at²
= 30 x .5 - .5 x 7 x .5²
= 15 - .875
= 14.125 m
t = .5 s
s₂ = ut - .5 at²
= 30 x .5 - .5 x 5 x .5²
= 15 - .625
= 14.375 m
To solve this problem we will apply the concepts related to electric potential and electric potential energy. By definition we know that the electric potential is determined under the function:

= Coulomb's constant
q = Charge
r = Radius
At the same time

The values of variables are the same, then if we replace in a single equation we have this expression,

If we replace the values, we have finally that the charge is,




Therefore the potential energy of the system is 
Answer:
Kindly check the explanation section.
Explanation:
For the design we are asked for in this question/problem there is the need for us to calculate or determine the strength in fracture and that of the yield. Also, we need to calculate for the block shear strength.
From the question, we have that the factored load = 500kips. Also, note that the tension splice must not slip.
Also, the shear force are resisted by friction, that is to say shear resistance = 1.13 × Tb × Ns.
Assuming our db = 3/4 inches, then the slip critical resistance to shear service load = 18ksi(refer to AISC manual for the table).
If db = 7/8 inches, then the shear force resistance for n bolt = 10.2kips, n > 49.6.
The yielding strength = 0.9 × Aj × Fhb= 736 kips > 500
The fracture strength = .75 × Ah × Fhb = 309 kips.
The bearing strength of 7/8 inches bolt at the edge hole and other holes = 46 kips and 102 kips.
(C). Remember gravity provides an acceleration of 9.81m/s^2, so the y component of velocity initial is zero because it isn’t already falling, and we have the height, so basically we use the kinematic equation vf^2=vi^2+2ad, substitute given values and you get vf^2=2(9.81)(65) which is 1275, when you take the square root you get 35.7m/s for final velocity
(B). Then you use vf=vi+at to get the equation 35.7=(9.81)t, when you divide out you get 3.64s for time t
(A). Finally, since we assume that there is no acceleration or deceleration horizonatally, we just multiply the time taken for it to hit the ground and the initial speed ((3.64)(35.7)) to get 129.96, with significant figures I would round that to 130 metres.
**this is in the order that I felt was easiest to answer**
<span>sound waves are a type of wave sometimes called compression waves, vibrations with enough of an amplitude can compress and decompress the air adjacent to the object causing the waves to form.</span>