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loris [4]
3 years ago
7

if I had 5 toes and 6 toes how many toes will I have NO HINT try your best to get close to the answer give u 105 points

Engineering
2 answers:
SVETLANKA909090 [29]3 years ago
4 0

Answer:

ONE I THANK YEAH CAUSE I'M YOUREEEEEEE SISTERSSSS

Explanation:

Rasek [7]3 years ago
3 0

Answer:

11 toes on one foot? and 5 one the other or just 11 toes?

your NO HINT threw me off

Explanation:

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List 6 different mechanisms in the Rube Goldberg cartoon and predict the purpose of each. Does the mechanism change speed, force
vodka [1.7K]

Answer: Rotary - because it has to

Around in a circle

Explanation:

i hope this helped u

7 0
3 years ago
Three tugboats are used to turn a barge in a narrow channel. To avoid producing any net translation of the barge, the forces app
stiks02 [169]

Answer:

a) Fb= 275.77 lb   Fc= 142.75 lb

b) M = -779.97 lb.ft (i.e. 779.97 lb.ft in clockwise direction)

c) Fax = 195 lb

   Fay = 337.75 lb

   Fbx = 195 lb

   Fby = 195 lb

Explanation:

Question: Three tugboats are used to turn a barge in a narrow channel. To avoid producing any net translation of the barge, the forces applied should be couples. The tugboat at point A applies a 390 lb force.

(a) Determine FB and FC so that only couples are applied.

(b) Using your answers to Part (a), determine the resultant couple moment that is produced.

(c) Resolve the forces at A and B into x and y components, and identify the pairs of forces that constitute couples.

Solution:

<u>For this problem Right hand side is positive X direction and Upwards is positive Y direction. Couples and moments will be considered positive in counterclockwise direction.</u>

<u />

a) For no translation condition

∑ F_{x} = 0      &     ∑F_{y} = 0

Hence,

F_{A}cos(30) - F_{B}cos(45) - F_{C} = 0

F_{A}sin(30) - F_{B} sin(45) = 0

and

F_{A} = 390 lb

Inserting the value of F_{A} and solving the remaining equations simultaneously yields (magnitudes),

F_{B} = 275.77 lb\\F_{C} = 142.75 lb

b) Summing up moments

M=45 ( -F_{Ay}-F_{Cy}) +5 (-F_{By})+22(-F_{Ax}-F_{Bx})\\ =45(-390cos(30)-142.75)+5(-275.77cos(45))+22 (-390sin(30)-275.77sin(45))

M = -779.97 lb.ft (i.e. 779.97 lb.ft clockwise)

c)

F_{Ax} = 390 sin(30)  = 195 lb

F_{Ay} = 390 cos(30) = 337.75lb\\

F_{Bx} = 275.77 sin(45) = 195lb\\F_{By} = 275.77 cos(45) = 195 lb

8 0
4 years ago
When the 2.8-kg bob is given a horizontal speed of 1.5 m/s, it begins to rotate around the horizontal circular path A. The force
Rama09 [41]

Answer:

The speed is the same at 1.5 m/s while

The work done by the force F is 0.4335 J

Explanation:

Here we have angular acceleration α = v²/r

Force = ma = 2.8 × 1.5²/r₁

and ω₁ = v₁/r₁ = ω₂ = v₁/r₂

The distance moved by the force = 600 - 300 = 300 mm = 0.3 m

If the velocity is constant

The speed is 1.5 m/s while the work done is

2.8 × 1.5²1/(effective radius) ×0.3

r₁ = effective radius

2.8*9.81 = 2.8 × 1.5²/r₁

r₁ = 0.229

The work done by the force = 2.8 × 1.5²*1/r₁ *0.3 = 0.4335 J

4 0
3 years ago
Find the power and the rms value of the following signal square: x(t) = 10 sin(10t) sin(15t)
ArbitrLikvidat [17]

Answer:

\mathbf{P_x =25 \ watts}

\mathbf{x_{rmx} = 5 \ unit}

Explanation:

Given that:

x(t) = 10 sin(10t) . sin (15t)

the objective is to find the power and the rms value of the following signal square.

Recall that:

sin (A + B) + sin(A - B) = 2 sin A.cos B

x(t) = 10 sin(15t) . cos (10t)

x(t) = 5(2 sin (15t). cos (10t))

x(t) = 5 × ( sin (15t + 10t) +  sin (15t-10t)

x(t) = 5sin(25 t) + 5 sin (5t)

From the knowledge of sinusoidial signal  Asin (ωt), Power can be expressed as:

P= \dfrac{A^2}{2}

For the number of sinosoidial signals;

Power can be expressed as:

P = \dfrac{A_1^2}{2}+ \dfrac{A_2^2}{2}+ \dfrac{A_3^2}{2}+ ...

As such,

For x(t), Power  P_x = \dfrac{5^2}{2}+ \dfrac{5^2}{2}

P_x = \dfrac{25}{2}+ \dfrac{25}{2}

P_x = \dfrac{50}{2}

\mathbf{P_x =25 \ watts}

For the number of sinosoidial signals;

RMS = \sqrt{(\dfrac{A_1}{\sqrt{2}})^2+(\dfrac{A_2}{\sqrt{2}})^2+(\dfrac{A_3}{\sqrt{2}})^2+...

For x(t), the RMS value is as follows:

x_{rmx} =\sqrt{(\dfrac{5}{\sqrt{2}} )^2 +(\dfrac{5}{\sqrt{2}} )^2 }

x_{rmx }=\sqrt{(\dfrac{25}{2} ) +(\dfrac{25}{2} ) }

x_{rmx }=\sqrt{(\dfrac{50}{2} )}

x_{rmx} =\sqrt{25}

\mathbf{x_{rmx} = 5 \ unit}

8 0
3 years ago
Fluid fills the space between two parallel plates. The differential equation that describes instantaneous fluid velocity for uns
USPshnik [31]

Answer:

u/v = S (y²w) / m sinwt + y/h

Explanation:

see attached image

3 0
4 years ago
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