Three-dimensional measuring references all of these EXCEPT:

Provide one example of a bad collision, and suggest an engineering solution to avoid the collision.

JulsSmile [24]

Answer:

1). Keep your distance. Drive far enough behind the car in front of you so you can stop safely. ...

Drive strategically. Avoid situations that could force you to suddenly use your brakes. ...

Don't get distracted. ...

Don't drive when drowsy or under the influence.

2). By far the deadliest accident type is the head-on collision. Head-on collisions consider both vehicle's speed at the time of the crash, which means even an accident at lower speeds can be catastrophic

Explanation:

first is how to avoid the collision and second is bad collision

7 0

2 years ago

Given the strings s1 and s2, not necessarily of the same length, create a new string consisting of alternating characters of s1

andrew11 [14]

Answer:

THE ANSA IZ A

Explanation:

4 0

4 years ago

Consider a thin suspended hotplate that measures 0.25 m × 0.25 m. The isothermal plate has a mass of 3.75 kg, a specific heat of

Orlov [11]

Answer:

Heat losses by convection, Qconv = 90W

Heat losses by radiation, Qrad = 5.814W

Explanation:

Heat transfer is defined as the transfer of heat from the heat surface to the object that needs to be heated. There are three types which are:

1. Radiation

2. Conduction

3. Convection

Convection is defined as the transfer of heat through the actual movement of the molecules.

Qconv = hA(Temp.final - Temp.surr)

Where h = 6.4KW/m2K

A, area of a square = L2

= (0.25)2

= 0.0625m2

Temp.final = 250°C

Temp.surr = 25°C

Q = 64 * 0.0625 * (250 - 25)

= 90W

Radiation is a heat transfer method that does not rely upon the contact between the initial heat source and the object to be heated, it can be called thermal radiation.

Qrad = E*S*(Temp.final4 - Temp.surr4)

Where E = emissivity of the surface

S = boltzmann constant

= 5.6703 x 10-8 W/m2K4

Qrad = 5.6703 x 10-8 * 0.42 * 0.0625 * ((250)4 - (25)4)

= 5.814 W

7 0

4 years ago

.) If the charges attracting each other in the problem above have equal magnitude, what is the magnitude of each charge?

Sedaia [141]

Answer:

Not seeing any other information, the best answer I can give is 2m.

Explanation:

M = magnitude

You see, if they have an equal charge, and you add them, it'd be 2 * m, or 2m.

8 0

3 years ago

A basketball has a 300-mm outer diameter and a 3-mm wall thickness. Determine the normal stress in the wall when the basketball

faltersainse [42]

Answer:

2.65 MPa

Explanation:

To find the normal stress (σ) in the wall of the basketball we need to use the following equation:

$\sigma = \frac{p*r}{2t}$

<u>Where:</u>

p: is the gage pressure = 108 kPa

r: is the inner radius of the ball

t: is the thickness = 3 mm

Hence, we need to find r, as follows:

$r_{inner} = r_{outer} - t$

$r_{inner} = \frac{d}{2} - t$

<u>Where:</u>

d: is the outer diameter = 300 mm

$r_{inner} = \frac{300 mm}{2} - 3 mm = 147 mm$

Now, we can find the normal stress (σ) in the wall of the basketball:

$\sigma = \frac{p*r}{2t} = \frac{108 kPa*147 mm}{2*3 mm} = 2646 kPa = 2.65 MPa$

Therefore, the normal stress is 2.65 MPa.

I hope it helps you!

3 0

3 years ago