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2 years ago

List some ways houses would be built differently if gravity were much stronger or much weaker ?two

Physics

2 answers:

tia_tia [17]2 years ago

8 0

Answer:

If gravity were stronger, houses and buildings would have to be more sturdy so they do not get crushed to the ground. If gravity were weaker, buildings and houses would have to be built stronger so they do not float away.

Hope this helps you!

Serggg [28]2 years ago

3 0

If gravity where much weaker, then the houses wouldn’t need as much structural support, and then could be built as much more taller as they wouldn’t go through as much stress. and if gravity was stronger, buildings would need to be a lot more structurally sound and wouldn’tbe able to maintain a tall structure as the force of gravity would be too much for them

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a baseball is hit 3 feet above ground level at 100 feet per second and at an angle of 45 with respect to the ground. (g=32 feet/

LiRa [457]

Answer:

hmax=81ft

Explanation:

Maximum height of the object is the highest vertical position along its trajectory.

The vertical velocity is equal to 0 (Vy = 0)

$0=V_{y}-g*t=v_{0}*sin(\alpha)-g*th\\$

we isolate th (needed to reach the maximum height hmax)

$th = \frac{v_{0}*sin(\alpha)}{g}$

The formula describing vertical distance is:

$y = Vy * t-g* t^{2} / 2$

So, given y = hmax and t = th, we can join those two equations together:

$hmax = Vy* th-g*th^{2}/2$

$hmax =Vo^{2}*sin(\alpha )^{2}/(2*g)$

if we launch a projectile from some initial height h all you need to do is add this initial elevation

$hmax =h+Vo^{2}*sin(\alpha)^{2}/(2*g)$

$hmax =3+100^{2}*sin(45)^{2}/(2 * 32)=81 ft$

6 0

3 years ago

If I drop a watermelon from the top of one of the tower dorms at CSU, and it takes 3.34 seconds to hit the ground, calculate how

WINSTONCH [101]

T= 3.34

Vi= 0

A= 9.81

D= ?

d=Vit+1/2at^2

d= 1/2(9.81)(3.34)2

d= 54.7 or 55 meters tall

4 0

3 years ago

A small metal object is tied to the end of a string and whirled round a circular path of radius 30cm. The object makes 20 oscill

LUCKY_DIMON [66]

Answer:

(i) The angular speed of the small metal object is 25.133 rad/s

(ii) The linear speed of the small metal object is 7.54 m/s.

Explanation:

Given;

radius of the circular path, r = 30 cm = 0.3 m

number of revolutions, n = 20

time of motion, t = 5 s

(i) The angular speed of the small metal object is calculated as;

$\omega = \frac{20 \ rev}{5 \ s} \times \frac{2 \pi \ rad}{1 \ rev} = \frac{40\pi \ rad}{5 \ s} = 8\pi \ rad/s = 25.133 \ rad/s$

(ii) The linear speed of the small metal object is calculated as;

$v = \omega r\\\\v = 25.133 \ rad/s \ \times \ 0.3 \ m\\\\v = 7.54 \ m/s$

6 0

2 years ago

As the proton approaches the uranium nucleus, the repulsive force slows down the proton until it comes momentarily to rest, afte

Klio2033 [76]

Answer:

$r^2 = \frac{ 2 k \ Ze^2}{ 2m}$

Explanation:

For this exercise we must use the principle of conservation of energy

starting point. The proton very far from the nucleus

Em₀ = K = ½ m v²

final point. The point where the proton is stopped (v = 0)

Em_f = U = q V

where the potential is

V = k Ze / r²

Let us consider that all the charge of the nucleus is in the center, therefore r is the distance from this point to the proton that is approaching

Energy is conserved

Em₀ = Em_f

½ m v² = e ( $k \frac{Ze}{r^2}$ )

$r^2 = \frac{ 2 k \ Ze^2}{ 2m}$

with this expression we can find the closest approach distance (r)

3 0

3 years ago

PLEASE HELP ME I NEED THIS ASAP!!!!

Zina [86]

Answer:

loud bangs

Explanation:

the pots for cooking fell

3 0

2 years ago

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