Will give correct answer brainliest

(1) In nondestructive testing, a discontinuity may be defined as an interruption in the normal physical structure or configurati

QveST [7]

Answer: BOTH ARE TRUE

Explanation: Nondestructive testing or Evaluation is a term used in the field of science and technology to describe the evaluations, analysis or testing carried out on components of materials without destroying any part or components of the test materials. It is very useful in scientific research or industrial engineering environments. When any disruption of physical structure or configuration of a component will lead to discontinuing of the test, and it may not affect the usefulness of the affected parts.

4 0

3 years ago

A stunt driver wants to make his car jump over 8 cars parked side by side below a horizontal ramp.

Genrish500 [490]

Answer:

a) Minimum speed must he drive off the horizontal ramp = 39.78 m/s

b) Minimum speed must he drive off the horizontal ramp with 7° above the horizontal = 23.93 m/s

Explanation:

a) The height of ramp = 1.5 meter

Horizontal distance he must clear = 22 meter

The car is having horizontal motion and vertical motion. In case of vertical motion the acceleration on the car is acceleration due to gravity.

We have equation of motion, $s= ut+\frac{1}{2} at^2$ , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

In case of vertical motion initial velocity = 0 m/s, acceleration = 9.8 $m/s^2$ , we need to calculate time when displacement = 1.5 meter.

$1.5=0*t+\frac{1}{2} *9.8*t^2\\ \\ t = 0.553 seconds$

So the car has to cover a distance of 22 meter in 2.119 seconds.

So minimum speed required = 22/0.553 = 39.78 m/s

Minimum speed must he drive off the horizontal ramp = 39.78 m/s

b) When the take of angle is 7⁰ the vertical speed of car is not zero = V sin 7 = 0.122 V

So the in case of vertical motion we have initial velocity = 0.122 V, S = -1.5 meter( below ramp), acceleration = -9.8 $m/s^2$

Substituting

$-1.5=0.122V*t-\frac{1}{2} *9.8*t^2\\ \\ 4.9t^2-0.122Vt-1.5=0$

In case of horizontal motion

Horizontal speed of car = V cos 7 = 0.993V

So it has to travel 22 meter in t seconds

0.993Vt = 22, Vt = 22.155 m

Substituting in the equation $4.9t^2-0.122Vt-1.5=0$

We will get $4.9t^2-0.122*22.155-1.5=0\\ \\ t = 0.926 seconds$

Speed required = 22.155/0.926 = 23.93 m/s

Minimum speed must he drive off the horizontal ramp with 7° above the horizontal = 23.93 m/s

7 0

4 years ago

This position-time graph describes an object's motion. Use it to determine the

kirza4 [7]

We're happy that you're asking for the "displacement", because displacement is simply the straight-line distance between the start-point and end-point, and we don't care about any of the motions or gyrations along the way.

From the graph:

-- The location of the object at time-zero, when time begins, is 10 meters.

-- The location of the object after 6.0 seconds is 4 meters.

-- The distance between the start-point and end-point is

(final location) - (initial location)

-- So Displacement = (4 meters) - (10 meters)

Displacement = -6 meters

3 0

3 years ago

An aircraft flying at a steady velocity of 70 m/s eastwards at a height of 800 m drops a package of supplies.

iren2701 [21]

The motion of the package can be described as the motion of a projectile,

given that it has an horizontal velocity and it is acted on by gravity.

a) $\vec{v}$ = 70·i

b) The package will reach ground in approximately 12.77 seconds.

c) The speed of the package as it lands is approximately 145.51 m/s.

d) The path of the package based on a stationary frame of reference is parabolic

e) The path of the package as seen from the plane is directly vertical downwards

Reasons:

Velocity of the aircraft = 70 m/s

Direction of flight of the aircraft = Eastward

Height from which the aircraft drops the package, h = 800 m

a) The initial velocity of the package, $\vec{v}$ = 70·i

b) The time it will take the package to reach the ground, t, is given by the formula;

$\displaystyle h = \mathbf{\frac{1}{2} \cdot g \cdot t^2}$

Where;

g = The acceleration due to gravity ≈ 9.81 m/s²

Therefore;

$\displaystyle t = \mathbf{\sqrt{ \frac{2 \cdot h}{g} }}$

Which gives;

$\displaystyle t = \sqrt{ \frac{2 \times 800}{9.81} } \approx \mathbf{12.77}$

The time it will take the package to reach the ground, t ≈ 12.77 seconds

c) The vertical velocity just before the package reaches the ground, $v_y$ , is given as follows;

$v_y^2$ = 2·g·h

Therefore;

$v_y$ = √(2·g·h)

Which gives;

$v_y$ = √(2 × 9.81 × 800) ≈ 125.28

$v_y$ ≈ 125.28 m/s

Which gives; $\vec{v}$ = 70·i - 125.28·j

Therefore, |v| = √(70² + (-125.28)²) ≈ 143.51

The speed of the package as it lands, |v| ≈ 143.51 m/s

d) The motion of the package that includes both horizontal and vertical motion is a projectile motion.

Therefore;

The path of the package is the path of a projectile, which is a parabolic shape.

e) As seen by someone on the aeroplane, the horizontal velocity will be

zero, therefore, the package will appear as accelerating directly vertical

downwards.

Learn more about projectile motion here:

brainly.com/question/1130127

6 0

2 years ago

What's the temperature in Enterprise Alabama

Fynjy0 [20]

As of now, the current temperature in Enterprise, Alabama is 58 degrees Fahrenheit. However, it may change based off of the climate and the day. Hope this helped!

6 0

3 years ago

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