All categories

3 years ago

A team of scientists devolpe a robot that moves like a snake when controled by a remote control what kind of robot is this

Engineering

1 answer:

Rufina [12.5K]3 years ago

7 0

I had this question a lot. But The Answer is Electromechanical:)

You might be interested in

He is going ___ in the hot air ballon

Vladimir [108]

no artical shoul be used here

5 0

3 years ago

A sewer pipe 8 inches in diameter can carry 0.662 ft3/s when flowing at a depth of 4 inches.

Rom4ik [11]

Answer:

a) the flow under full capacity is q₂= 1.334 ft³/s

b) the velocity would be v= 3.793 ft/s

Explanation:

a) Since the pipe has 8 inches in diameter but 4 are covered with water flow ( half of a circle in area=A₁) , q₁=0.662 ft³/s then

q₁=A₁*v

then for the same velocity v but area A₂=2*A₁

flow under full capacity= q₂ = A₂*v= 2*A₁*v= 2*q₁=2*0.662 ft³/s= 1.334 ft³/s

b) when flowing at a depth of 4 inches

A₁= (1/2)*(π*D²/4) = π* (1/8)*(8 in)² = 8π in² * (1 ft²/ 144 in²) = π/18 ft² = 0.1745 ft²

then

v=q₁/A₁ = 0.662 ft³/s/0.1745 ft²= 3.793 ft/s

v= 3.793 ft/s

6 0

4 years ago

Show that y = '(t - s)f(s)ds is a solution to my" + ky = f(t). Use g' (0) = 1/m and mg" + kg = 0. 6.1) Derive y' 6.2) Using g(0)

Gre4nikov [31]

Answer:

Explanation:

Given that:

$y = \int^t_og'(t-s) f(s) ds \ \text{is solution to } \ my"ky= f(t)$

where;

$g'(0) = \dfrac{1}{m}$ and $mg"+kg = 0$

$\text{Using Leibniz Formula to prove the above equation:}$

$\dfrac{d}{dt} \int ^{b(t)}_{a(t)} \ f (t,s) \ ds = f(t,b(t) ) * \dfrac{d}{dt}b(t) - f(t,a(t)) *\dfrac{d}{dt}a(t) + \int ^{b(t)}_{a(t)}\dfrac{\partial}{\partial t} f(t,s) \ dt$

So, $y = \int ^t_0 g' (t-s) f(s) \ ds$

$\text{By differentiation with respect to t;}$

$y' = g'(o) f(t) \dfrac{d}{dt}t- 0 + \int^{t}_{0}g'' (t-s) f(s) ds \\ \\ y' = \dfrac{1}{m}f(t) + \int ^t_0 g'' (ts) f(s) \ ds$

$y'' = \dfrac{1}{m} f'(t) + g"(0) f(t) + \int^t_o g"'(t-s) f(s)ds --- (1)$

$Since \ \ mg" (t) +kg (t) = 0 \\ \\ \implies g" (t) = -\dfrac{k}{m} g(t) --- (111) \\ \\ put \ t \ =0 \ we \ get;\\g" (0) = - \dfrac{k}{m } g(0) \\ \\ g"(0) = 0 \ \ \ \ ( because \ g(0) =0) \\ \\$

$Now \ differentiating \ equation (111) \ with \ respect \ to \ t \\ \\ g"'(t) = -\dfrac{k}{m}g(t) \\ \\ replacing \ it \ into \ equation \ (1) \\ \\ y" = \dfrac{1}{m}f' (t) + 0 + \int ^t_o \dfrac{-k}{m}g' (t-s) f(s) \ ds \\ \\ y" = \dfrac{1}{m}f' (t) - \dfrac{k}{m} \int ^t_o g' (t-s) \ f(s) \ ds \\ \\ y" = \dfrac{1}{m}f'(t) - \dfrac{k}{m}y \\ \\ my" = f'(t)-ky \\ \\ \implies \mathbf{ my" +ky = f'(t)}$

7 0

3 years ago

What is the most likely cause a brake light staying on?

Monica [59]

The answer is either a or b

8 0

3 years ago

This is construction plz do the key terms I’ll do anything 5. define key terms make sure you include a) all the definitions b.)

Zanzabum

Answer:

1234567891011121314151617181920

4 0

3 years ago