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3 years ago

12

an object moving North with an initial velocity of 14 m s accelerates 5 Ms ^2 for 20 seconds what is the final velocity of the o

bject

2 answers:

kogti [31]3 years ago

8 0

Answer:

The final velocity of the object is 114 m/s.

Explanation:

It is given that,

Initial velocity of the object, u = 14 m/s

Acceleration of object, $a=5\ m/s^2$

Time taken, t = 20 s

Let v is the final velocity of the object. It can be calculated using first equation of motion as :

$v=u+at$

$v=14\ m/s+5\ m/s^2\times 20\ s$

v = 114 m/s

So, the final velocity of the object is 114 m/s. Hence, this is the required solution.

bonufazy [111]3 years ago

7 0

The final velocity is 114m/s
$5 = \frac{x - 14}{20} \\ x = \frac{20 \times 5}{1} + 14 = 100 + 14 \\ 114$

good luck

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navik [9.2K]

Answer: A

si, di, are expressed as negative and f is expressed as positive values.

Explanation:

When the object is located at a location in front of the focal point of a convex len, the image will always be located somewhere on the same side of the lens as the object. The image is located behind the object. In this case, the image will be an upright image and the image is enlarged

In this question, the relationship between the focal length, image distance di and object distance is

I/f = 1/di + 1/do that is

I/f - 1/di - 1/do = 0

The image size si is also negative since the image is a virtual image.

Therefore, si, di, are expressed as negative, f is expressed as positive values.

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3 years ago

Neon is compressed from 100 kPa and 24°C to 500 kPa in an isothermal compressor. Determine the change in the specific volume and

trasher [3.6K]

Answer:

ΔV = -0.97 m³/ kg

ΔH = 0 kJ/ kg

Explanation:

<u>To determine the change in the </u><u>specific volume</u><u> we need to </u><u>use the ideal gas law</u><u>:</u>

$PV = RT$

<em>where</em><em> P</em><em>: </em><em>pressure </em><em>of the gas </em><em>V</em><em>: </em><em>volume </em><em>of the gas, </em><em>R</em><em>: i</em><em>deal gas constant</em><em>= 0.4119 kJ/kg.K = 0.4119 kPa.m³/kg.K and </em><em>T</em><em>: </em><em>temperature </em><em>of the gas.</em>

<u />

<u>The </u><u>V₁,</u><u> at a compressed pressure is:</u>

$V_{1}= \frac {RT}{P_{1}}$

$V_{1}= \frac {0.4119 \frac{kPa\cdot m^{3}}{kg\cdot K} \cdot (24 + 273 K)}{100 kPa}$

$V_{1}= 1.22 \frac{m^{3}}{kg}$

<u>Similarly, the </u><u>V₂</u><u> is:</u>

$V_{2}= \frac {RT}{P_{2}}$

$V_{2}= \frac {0.4119 \frac{kPa\cdot m^{3}}{kg\cdot K} \cdot (24 + 273 K)}{500 kPa}$

$V_{2}= 0.25 \frac{m^{3}}{kg}$

Now, the change in the specific volume because the compressor is:

$V_{2} - V_{1} = 0.25 - 1.22 \frac{m^{3}}{kg}$

$V_{2} - V_{1} = -0.97 \frac{m^{3}}{kg}$

Finally, to calculate the change in the specific enthalpy, we need to remember that neon is an ideal gas and that is an isothermal process:

$\Delta H = C_{p} \cdot \Delta T$

$\Delta H = 1.0299 \frac{kJ}{kg \cdot K} \cdot 0$

$\Delta H = 0 \frac{kJ}{kg}$

Have a nice day!

7 0

4 years ago

Why do the planets orbit the Sun in an elliptical shape?

ki77a [65]

Answer:

That's essentially how objects in orbits work as they move closer to the body they orbit, they accelerate faster and faster. Our penny will get so fast that, once it comes around the planet, it will be flung very far away, which will then slow it down. This is what creates an elliptical orbit.

Explanation:

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3 years ago

A rocket is fired vertically upwards with initial velocity 92 m/s at the ground level. Its engines then fire and it is accelerat

Readme [11.4K]

Answer:

The rocket above the ground is in 44 sec.

Explanation:

Given that,

Initial velocity = 92 m/s

Acceleration = 4 m/s²

Altitude = 1200 m

Suppose, How long was the rocket above the ground?

We need to calculate the time

Using equation of motion

$s=ut-\dfrac{1}{2}at^2$

Put the value into the formula

$1200=92t+\dfrac{1}{2}\times4t^2$

$2t^2+92-1200=0$

$t=10\ sec$

We need to calculate the velocity

Using equation of motion

$v=u+at$

Put the value into the formula

$v=92+4\times10$

$v=132\ m/s$

When the rocket hits the ground,

Then, h'=0

We need to calculate the time

Using equation of motion

$h'=h+ut-\dfrac{1}{2}at^2$

Put the value into the formula

$0=1200+132t-\dfrac{1}{2}\times9.8t^2$

$4.9t^2-132t-1200=0$

$t=34\ sec$

When the rocket is in the air it is the sum of the time when it reaches 1000 m and the time when it hits the ground

So, the total time will be

$t'=34+10$

$t'=44\ sec$

Hence, The rocket above the ground is in 44 sec.

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4 years ago

A race car starting from rest accelerates uniformly at a rate of 3.90 m/s^2 what is the cars speed after it has traveled 200 m.

snow_tiger [21]

The formula for accelerational displacement is at^2/2, so we know that 3.9t^2/2 = 200, or 3.9t^2 = 400. t = $\sqrt{400/3.9} \approx 10.12739367$ , at = v, so $3.9 * 10.12739367 \approx 39.5 m/s$

6 0

3 years ago

Read 2 more answers