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2 years ago

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]2 years ago

4 0

Answer:

ONE I THANK YEAH CAUSE I'M YOUREEEEEEE SISTERSSSS

Explanation:

Rasek [7]2 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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Traffic at a roundabout moves

sweet-ann [11.9K]

Answer:

b-counter-clockwise

Explanation:

3 0

3 years ago

Tech A says you can find the typical angle of a V-block engine by dividing the number of cylinders by 720

Lady_Fox [76]

Answer:

Tech A is correct

Explanation:

Tech A is right as its V- angle is identified by splitting the No by 720 °. Of the piston at the edge of the piston.

Tech B is incorrect, as the V-Angle will be 720/10 = 72 for the V-10 motor, and he says 60 °.

8 0

2 years ago

A charge of +2.00 μC is at the origin and a charge of –3.00 μC is on the y axis at y = 40.0 cm . (a) What is the potential at po

Nimfa-mama [501]

a) Potential in A: -2700 V

b) Potential difference: -26,800 V

c) Work: $4.3\cdot 10^{-15} J$

Explanation:

The electric potential at a distance r from a single-point charge is given by:

$V(r)=\frac{kq}{r}$

where

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

q is the charge

r is the distance from the charge

In this problem, we have a system of two charges, so the total potential at a certain point will be given by the algebraic sum of the two potentials.

Charge 1 is

$q_1=+2.00\mu C=+2.00\cdot 10^{-6}C$

and is located at the origin (x=0, y=0)

Charge 2 is

$q_2=-3.00 \mu C=-3.00\cdot 10^{-6}C$

and is located at (x=0, y = 0.40 m)

Point A is located at (x = 0.40 m, y = 0)

The distance of point A from charge 1 is

$r_{1A}=0.40 m$

So the potential due to charge 2 is

$V_1=\frac{(8.99\cdot 10^9)(+2.00\cdot 10^{-6})}{0.40}=+4.50\cdot 10^4 V$

The distance of point A from charge 2 is

$r_{2A}=\sqrt{0.40^2+0.40^2}=0.566 m$

So the potential due to charge 1 is

$V_2=\frac{(8.99\cdot 10^9)(-3.00\cdot 10^{-6})}{0.566}=-4.77\cdot 10^4 V$

Therefore, the net potential at point A is

$V_A=V_1+V_2=+4.50\cdot 10^4 - 4.77\cdot 10^4=-2700 V$

Here we have to calculate the net potential at point B, located at

(x = 0.40 m, y = 0.30 m)

The distance of charge 1 from point B is

$r_{1B}=\sqrt{(0.40)^2+(0.30)^2}=0.50 m$

So the potential due to charge 1 at point B is

$V_1=\frac{(8.99\cdot 10^9)(+2.00\cdot 10^{-6})}{0.50}=+3.60\cdot 10^4 V$

The distance of charge 2 from point B is

$r_{2B}=\sqrt{(0.40)^2+(0.40-0.30)^2}=0.412 m$

So the potential due to charge 2 at point B is

$V_2=\frac{(8.99\cdot 10^9)(-3.00\cdot 10^{-6})}{0.412}=-6.55\cdot 10^4 V$

Therefore, the net potential at point B is

$V_B=V_1+V_2=+3.60\cdot 10^4 -6.55\cdot 10^4 = -29,500 V$

So the potential difference is

$V_B-V_A=-29,500 V-(-2700 V)=-26,800 V$

The work required to move a charged particle across a potential difference is equal to its change of electric potential energy, and it is given by

$W=q\Delta V$

where

q is the charge of the particle

$\Delta V$ is the potential difference

In this problem, we have:

$q=-1.6\cdot 10^{-19}C$ is the charge of the electron

$\Delta V=-26,800 V$ is the potential difference

Therefore, the work required on the electron is

$W=(-1.6\cdot 10^{-19})(-26,800)=4.3\cdot 10^{-15} J$

4 0

3 years ago

A material point in equilibrium has 1 independent component of shear stress in the xz plane. a)True b)- False

ozzi

Answer:

True

Explanation:

For point in xz plane the stress tensor is given by $\left[\begin{array}{ccc}Dx_{} &txz\\tzx&Dz\\\end{array}\right]$

where Dx is the direct stress along x ; Dz is direct stress along z ; tzx and txz are the shear stress components

We know that the stress tensor matrix is symmetrical which means that tzx = txz ( obtained by moment equlibrium )

thus we require only 1 independent component of shear stress to define the whole stress tensor at a point in 2D plane

8 0

2 years ago

"At 195 miles long, and with 7,325 miles of coastline, the Chesapeake Bay is the largest and most complex estuary in the United

Paraphin [41]

Answer:

see explaination

Explanation:

Part a) Width of bay at Potomac River:

Given Data:

· Actual Width at Potomac River = 30 miles

· Bay Model Length Ratio Lr = 1/1000

In fluid mechanics models of real structures are prepared in simulation so that they can be analyzed accurately. A model is known to have simulation if model carries same geometric, kinematic and dynamic properties at a small scale.

Length of any part in model = Actual length x Lr

Hence,

Model Width of bay at Potomac River = 30 x 1/1000 = 0.03 miles

Since 1 mile = 5280 ft

Model Width of bay at Potomac River = 0.03 x 5280 = 158.4 ft

Part b) Model Length of bay bridge in model:

Given Data:

· Actual Length of bay bridge = 4.3 miles

· Bay Model Length Ratio Lr = 1/1000

Model Length = Actual Length x Lr = 4.3 x 1/1000 = 0.0043 miles

Since 1 mile = 5280 ft

Model Length in feet = 0.0043 x 5280 = 22.704 ft

Part c) Model Length of bay bridge in model:

Given Data:

· Model Area = 8 acre

· Bay Model Length Ratio Lr = 1/1000

Model Area = Actual Area x Lr x Lr

8 Model Area :: Actual Area =- (Lp)2 2 = 8,000,000 acre 1000

Since 1 square mile = 640 acre,

Actual Area in square miles = 8,000,000/640 = 12,500 square miles

Part d) Average and maximum depth of model:

Given Data:

· Actual Average depth = 28 ft

· Actual Maximum depth = 174 ft

· Bay Model Length Ratio Lr = 1/1000

Model average depth = Actual average depth x Lr = 28 x 1/1000 = 0.028 feet

Since 1 ft = 12 inch

Model average depth in inch = 0.028 x 12 = 0.336 in

Model maximum depth = Actual maximum depth x Lr = 174 x 1/1000 = 0.174 feet

Since 1 ft = 12 inch

Model maximum depth in inch = 0.174 x 12 = 2.088 in

4 0

3 years ago