Answer: & Explanation:
small scaled building models are very rare to make/build nowadays.
due to advance technology it is more wiser and more convenient to use 3d programs like revit in creating building model to scale.
so It is always a good idea to make models of your ideas, before deciding on the final design.
This will help you determine whether your idea is going to work or needs modifying. A model allows you to test your solution quickly and cheaply.
Reworking a massive building takes years, millions of pounds, and meticulous planning and project management.
Answer:
<h3>The answer is 3 kg</h3>
Explanation:
The mass of the object can be found by using the formula

f is the force
a is the acceleration
From the question we have

We have the final answer as
<h3>3 kg</h3>
Hope this helps you
Answer:
s = 0.9689 m
Explanation:
given,
Height of fall of paratroopers = 362 m
speed of impact = 52 m/s
mass of paratrooper = 86 Kg
From from snow on him = 1.2 ✕ 10⁵ N
now using formula
F = m a
a = F/m


Using equation of motion
v² = u² + 2 a s


s = 0.9689 m
The minimum depth of snow that would have stooped him is s = 0.9689 m
Answer:
0.705 m/s²
Explanation:
a) The sprinter accelerates uniformly from rest and reaches a top speed of 35 km/h at the 67-m mark.
Using newton's law of motion:
v² = u² + 2as
v = final velocity = 35 km/h = 9.72 m/s, u = initial velocity = 0 km/h, s = distance = 67 m
9.72² = 0² + 2a(67)
134a = 94.484
a = 0.705 m/s²
b) The sprinter maintains this speed of 35 km/h for the next 88 meters. Therefore:
v = 35 km/h = 9.72 m/s, u = 35 km/h = 9.72 m/s, s = 88 m
v² = u² + 2as
9.72² = 9.72² + 2a(88)
176a = 9.72² - 9.72²
a = 0
c) During the last distance, the speed slows down from 35 km/h to 32 km/h.
u = 35 km/h = 9.72 m/s, v = 32 km/h = 8.89 m/s, s = 200 - (67 + 88) = 45 m
v² = u² + 2as
8.89² = 9.72² + 2a(45)
90a = 8.89² - 9.72²
90a = -15.4463
a = -0.1716 m/s²
The maximum acceleration is 0.705 m/s² which is from 0 to 67 m mark.
Answer:
F ’= 1/32 F
We see that the value of the force is the initial force over 32
Explanation:
In this problem the sphere that is touching the others is connected to ground, after each touch,
Let's analyze the charge of the gray sphere, when you touch it for the first time, the charge is divided between the two spheres each having Q / 2, when the sphere separates and touches ground, its charge passes zero. When I touch the gray dial again, its charge is reduced by half
½ (Q / 2) = ¼ Q
For the red dial repeat the same scheme
with the first touch the charge is reduced to Q / 2
with the second touch e reduce to ½ (Q / 2) = ¼ Q
with the third toce it is reduced to ½ (¼ Q) = ⅛ Q
Now let's analyze what happens to the electric force
if the force is F for when the charge of each sphere is Q
F = k Q Q / r²
with the remaining charge strength is
F ’= k (¼ Q) (⅛ Q) / r²
F ’= 1/32 k Q Q / r²
F ’= 1/32 F
We see that the value of the force is the initial force over 32