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3 years ago

12

Create a class Suzuki with following characteristics Initialize member variable of Car class using the super constructor which c

ontains two parameters model and speed. Implement getModel

1 answer:

Ganezh [65]3 years ago

5 0

Answer:

See attached file.

Explanation:

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15- Vipsana's Gyros House sells gyros. The cost of ingredients (pita, meat, spices, etc.) to make a gyro is $2.00. Vipsana pays

sineoko [7]

D is a great answer

4 0

3 years ago

Elijah wants to make some changes to a game that he is creating. Elijah wants to move the bleachers to the right, the coaches to

Ahat [919]

Answer:

the editor

Explanation:

the variable is usually a number, and the conditional loop has nothign to do with movement/ where something is located

7 0

3 years ago

Read 2 more answers

A coil of wire 8.6 cm in diameter has 15 turns and carries a current of 2.7 A. The coil is placed in a magnetic field of 0.56 T.

SVEN [57.7K]

Answer:

Explanation:

it is given that diameter = 8.6 cm

$radius =\frac{8.6}{2}=4.3\ cm=4.3\times 10^{-2}\ m$

current =2.7 ampere

number of turns = 15

$area =\pi r^2=3.14\times \left ( 4.3\times 10^{-2} \right )^{2}=0.005806 m^{2}$

magnetic field =0.56 T

maximum torque= BINASINΘ for maximum torque sinΘ=1

so maximum torque==0.56×2.7×0.005806×15=0.13174 Nm

4 0

3 years ago

What is the magnitude of the maximum stress that exist at the tip of an internal crack having a radius of curvature of 1.9 x 10-

Hitman42 [59]

Answer:

2800 [MPa]

Explanation:

In fracture mechanics, whenever a crack has the shape of a hole, and the stress is perpendicular to the orientation of such, we can use a simple formula to calculate the maximum stress at the crack tip

$\sigma_{m} = 2 \sigma_{p} (\frac{l_{c}}{r_{c}})^{0.5}$

Where $\sigma_{m}$ is the magnitude of he maximum stress at the tip of the crack, $\sigma_{p}$ is the magnitude of the tensile stress, $l_{c}$ is $1/2$ the length of the internal crack, and $r_{c}$ is the radius of curvature of the crack.

We have:

$r_{c}=1.9*10^{-4} [mm]$

$l_{c}=3.8*10^{-2} [mm]$

$\sigma_{c}=140 [MPa]$

We replace:

$\sigma_{m} = 2*(140 [MPa])*(\frac{\frac{3.8*10^{-2} [mm]}{2}}{1.9*10^{-4} [mm]})^{0.5}$

We get:

$\sigma_{m} = 2*(140 [MPa])*(\frac{\frac{3.8*10^{-2} [mm]}{2}}{1.9*10^{-4} [mm]})^{0.5}=2800 [MPa]$

5 0

3 years ago

when removing the pistons and rods assemblies from a cylinder block technician a positions the throw of the crankshaft at the to

defon

Answer: Technician A

Reason: Piston and rod are connected with crank shaft and connecting rod. Smaller end of connecting rod is connected with piston and bigger end is connected with crankshaft.

<u>Explanation: </u>

When removing the piston and rod assemblies from a cylinder block: The technician with correct approach. So to remove piston from cylinder technician must throw crankshaft first and the remove connecting rod by losing nuts and caps...So technician A is in right way. Technician B using hammer to remove piston from the rod not possible because connecting rod and piston connected by nuts and caps can’t be separate by hammer.

Technician A positions the throw of the crankshaft at the top of its stroke and removes the connecting rod nuts and cap.

Technician B covers the rod bolts with hammers and pushes the piston and rod assembly out with the wooden hammer handle or wooden drift and supports the piston as it comes out of the cylinder.

4 0

3 years ago