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3 years ago

Please help! Can give brainliest too.

Physics

1 answer:

Step2247 [10]3 years ago

7 0

Explanation:

$v = \sqrt{2ax}$

Take the square of both sides:

$v^2 = \left(\sqrt{2ax}\right)^2 = 2ax$

Divide both sides by 2a and you will get

$x = \dfrac{v^2}{2a}$

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(20 points) An engineer is tasked with developing a model to study a cylindrical heat exchanger in a steam system. The prototype

wel

This question is incomplete, the complete question is;

An engineer is tasked with developing a model to study a cylindrical heat exchanger in a steam system. The prototype cylinder is 2.54 cm in diameter and the steam properties are: velocity = 30 m/s; density = 0.6 kg/m³; and absolute viscosity = 1.197 X 10⁻⁵ N-s/m², respectively. The model is going to be tested in a water tunnel where the water temperature is 20°C and the velocity is 3 m/s

a) what is the prototype Reynolds number, based on using the cylinder diameter as the characteristic length?

b) what should be the diameter of the model be to ensure dynamic similitude?

Answer:

a) the prototype Reynolds number is 38195.488

b) 0.01278 m or 1.278 cm should be the diameter of the model be to ensure dynamic similitude

Explanation:

Given that;

1 prototype ; d = 2.54 cm = 0.0254 m, Vp = 30 m/s, Sp = 0.6 kg/m³, Up = 1.197 × 10⁻⁵ N-s/m²

Model{ water at 20°C}; dm = ?, Vm = 3 m/s, Pm = 998.23 kg/m³, Um = 1.002 × 10⁻³ N-s/m²

a) what is the prototype Reynolds number,

to calculate prototype Reynolds number we use the expression;

(Re)p = SpVpdP / Up

we substitute our value

(Re)p = (0.6 × 30 × 0.0254) / 1.197 × 10⁻⁵

(Re)p = 38195.488

Therefore the prototype Reynolds number is 38195.488

what should be the diameter of the model be to ensure dynamic similitude?

i.e dm = ?

so dynamic similarity [ viscous flow]

(Re)m = (Re)p

[PmVmdm / Um] = [SpVpdP / Up]

we substitute

[998.23 × 3 × dm / 1.002 × 10⁻³] = (0.6 × 30 × 0.0254) / 1.197 × 10⁻⁵

2994.69dm / 1.002 × 10⁻³ = 38195.488

2994.69dm = 38.2718

dm = 38.2718 / 2994.69

dm = 0.01278 m or 1.278 cm

Therefore 0.01278 m or 1.278 cm should be the diameter of the model be to ensure dynamic similitude

4 0

3 years ago

A student walks 144 m west, and then turns around and walks 89 m east If this takes place in a 7.5-minute interval, what is the

Darina [25.2K]

The average speed is 0.517 m/s

The average velocity is 0.12 m/s west

Explanation:

When dividing the total distance covered by object by time, we get the value for average speed.

$\text { speed }=\frac{\text {distance}}{\text {time}}$

Calculate the distance without consideration of motion’s direction. So, the distance walks 144 m first, and then another 89 m, so the total distance covered is

d = 144 + 89 = 233 meters

Given t = 7.5 minutes. Convert minute into seconds,

$7.5 \times 60=450 \text { seconds }$

So, the average speed can be calculates as below,

$\text { speed }=\frac{233}{450}=0.517 \mathrm{m} / \mathrm{s}$

When dividing the object’s displacement by time taken, we can calculate average velocity.

$\text {velocity}=\frac{\text {displacement}}{\text {time}}$

In given case, the students walks 144 m west first, and then 89 m east. The displacement is the distance in a straight line between the initial and final position: therefore, in this case, the displacement is

d = 144 (west) - 89 (east) = 55 m (west)

The time taken is t = 450 s

So, the average velocity is

$\text {velocity}=\frac{55}{450}=0.12 \mathrm{m} / \mathrm{s}$

And the direction is west (the same as the displacement).

8 0

4 years ago

A 5.00 kg object oscillates back and forth at the end ofa spring whose spring constant is 49.3 N/m. An obersever istraveling at

defon

Answer:

5.571 sec

Explanation:

angular frequency = √ (k/m) = √ (49.3 / 5) = 3.14 rad/s

Period To = 2π / angular frequency

Period To = 2π/3.14 = 2 × 3.14 / 3.142 = 2.00 sec which you got

T measured by the observer = To / (√ (1 - (v²/c²))) = 2 / √( 1 - 0.871111) = 2 / 0.35901 = 5.571 sec

t=2.00/(1-√((2.80*10^8)^2/(3.00*10^8)^2))= should have been ( To / (√ (1 - (v²/c²))). where To = 2.00 sec

8 0

3 years ago

Which of the following is an example of a luminous object? Dog Tree Candle Moon

Keith_Richards [23]

Candle is the correct answer to the answer

7 0

3 years ago

A 2kg ball is thrown with an acceleration of 15m/s2. A 2kg ball is thrown with an acceleration of 10m/s2. Which ball

DerKrebs [107]

A :-) for this question , we should apply
F = ma
( i ) Given - m = 2 kg
a = 15 m/s^2
Solution :
F = ma
F = 2 x 15
F = 30 N

( ii ) Given - m = 2 kg
a = 10 m/s^2
Solution :
F = ma
F = 2 x 10
F = 20 N

.:. The net force of object ( i ) has greater force compared to object ( ii ) by
( 30 - 20 ) 10 N

5 0

3 years ago