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tensa zangetsu [6.8K]

3 years ago

10

The distance and direction between starting and ending positions

1 answer:

rodikova [14]3 years ago

8 0

Answer:

The length traveled by an object moving in any direction or even changing direction is called distance. The location of an object in a frame of reference is called position. For straight line motion, positions can be shown using a number line. The separation between original and final position is called displacement

Explanation:

good luck

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Please help if you can

denis23 [38]

Answer:

The answer is B

Explanation:

Because when the both sides aren't balanced one side has to cause motion. (fall down)

6 0

2 years ago

Pweese help one more timeeee<br><br>pweese look at the image below

4vir4ik [10]

Answer:

increasing; speeding up is my answer

7 0

3 years ago

Read 2 more answers

An amusement park ride consists of a rotating circular platform 8.26 m in diameter from which 10 kg seats are suspended at the e

VashaNatasha [74]

To solve this problem we will begin by finding the necessary and effective distances that act as components of the centripetal and gravity Forces. Later using the same relationships we will find the speed of the body. The second part of the problem will use the equations previously found to find the tension.

PART A) We will begin by finding the two net distances.

$r = \frac{8.26}{2} = 4.13m$

And the distance 'd' is

$d = lsin\theta$

$d = 1.14 sin 16.2\°$

$d = 0.318m$

Through the free-body diagram the tension components are given by

$Tcos\theta = mg$

$Tsin\theta = \frac{mv^2}{R}$

Here we can watch that,

$R = r+d$

Dividing both expression we have that,

$tan\theta = \frac{v^2}{Rg}$

Replacing the values,

$tan(16.2) = \frac{v^2}{(4.13+0.318)(9.8)}$

$v = 4.83371m/s$

PART B) Using the vertical component we can find the tension,

$Tcos\theta = mg$

$T = \frac{mg}{cos\theta}$

$T = \frac{(10+26.2)(9.8)}{cos(16.2)}$

$T = 369.42N$

6 0

3 years ago

A jet aircraft is traveling at 262 m/s in hor-

NeTakaya

Solution :

Speed of the air craft, $S_a$ = 262 m/s

Fuel burns at the rate of, $S_b$ = 3.92 kg/s

Rate at which the engine takes in air, $S_{air}$ = 85.9 kg/s

Speed of the exhaust gas that are ejected relative to the aircraft, $S_{exh}$ =921 m/s

Therefore, the upward thrust of the jet engine is given by

$F=S_{air}(S_{exh}-S_a)+(S_b \times S_{exh})$

F = 85.9(921 - 262) + (3.92 x 921)

= 4862635.79 + 3610.32

= $4.8 \times 10^6 \ N$

Therefore thrust of the jet engine is $4.8 \times 10^6 \ N$ .

3 0

3 years ago

Mathematically speaking, the energy of a wave is directionally proportional to the square of the amplitude of the wave Calculate

kenny6666 [7]

The energy of the wave is D) 16.0 units squared

Explanation:

The energy of a wave is directly proportional to the square of the amplitude of the wave.

Mathematically, this can be written as

$E\propto A^2$

where

E is the energy of the wave

A is its amplitude

For the wave in this problem, the amplitude is

A = 4 units

Therefore, assuming that the constant of proportionality is 1, the energy of this wave is

$E=A^2=4^2=16$ units squared

Learn more about waves:

brainly.com/question/5354733

brainly.com/question/9077368

#LearnwithBrainly

6 0

3 years ago

Read 2 more answers