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3 years ago

What is it's potential energy when it is 1000m above the ground

Physics

1 answer:

ycow [4]3 years ago

5 0

Explanation:

stresses within itself or other factors. since 1000m is above the ground, it would rely on gravity, such as; " what goes up must come down," the gravity is present because it is above ground level, example, a mud slide on a small island, needs h2o to become heavier to collapse...

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An aircraft is at a standstill. It accelerates to 140kts in 28 seconds. The aircraft weighs 28,000 lbs. How many feet of runway

stiv31 [10]

Answer:

runway use is 3307.8 feet

Explanation:

given data

velocity = 140 kts = 140 × 0.5144 m/s = 72.016 m/s

time = 28 seconds

weight = 28000 lbs

to find out

How many feet of runway was used

solution

we will use here first equation of motion for find acceleration

v = u + at ..............1

here v is velocity given and u is initial velocity that is 0 and a is acceleration and t is time

put here value in equation 1

72.016 = 0 + a(28)

a = 2.572 m/s²

and

now apply third equation of motion

s = ut + 0.5×a×t² .......................2

here s is distance and u is initial speed and t is time and a is acceleration

put here all value in equation 2

s = 0 + 0.5×2.572×28²

s = 1008.24 m = 3307.8 ft

so runway use is 3307.8 feet

5 0

4 years ago

All visible light (light that our eyes can detect) has a wavelength between 400-700 nanometers.  Wavelengths just smaller than 4

den301095 [7]

The Balmer light series comes under the visible light.

<u>Explanation:</u>

The transition of electrons from higher to energy level with 2 as principal quantum number results in the spectral emission lines of hydrogen atom and this series of lines are known as Balmer series.

Mostly, these lines has the wavelength of more than 400 nm but lesser than 700 nm. Generally of the four categories namely, 410, 434, 486, 656 nm which comes under the type of visible light. So, it can be concluded that the Balmer series light falls under visible light.

In astronomy, Balmer lines occur in various stellar (celestial or astronomical) objects due to the higher content of hydrogen in the universe. Therefore, they are commonly seen and relatively strong when compared to other element lines.

Note: nm is nanometer (one billionth of a meter in length)

4 0

3 years ago

There is a 50 g sample of Ra-229. It has a half-life of 4 minutes.

Sloan [31]

Via half-life equation we have:

$A_{final}=A_{initial}(\frac{1}{2})^{\frac{t}{h} }$

Where the initial amount is 50 grams, half-life is 4 minutes, and time elapsed is 12 minutes. By plugging those values in we get:

$A_{final}=50(\frac{1}{2})^\frac{12}{4}=50(\frac{1}{2})^{3}=50(\frac{1}{8})=6.25g$

There is 6.25 grams left of Ra-229 after 12 minutes.

4 0

4 years ago

If a beam of red light and a beam of violet light are incident at the same angle on the inclined surface of a prism from air med

Kaylis [27]

Answer: The red light bends the least while the violet the most.

Explanation:

i think this is it

3 0

3 years ago

A person walks at a speed of 6 km/h from point A to point B. If he improves his pace by 1.5 km/h, he will arrive 1 hour earlier.

Fofino [41]

Answer:

a) The distance is 30 km

The time duration is 5 hours

b) s₁ is approximately 28.142 km or s₁ is approximately 1.505 km

Explanation:

The initial speed with of the person, v₁ = 6 km/h

The distance the walked by the person, d = From point A to point B

The rate at which the person increases the speed, Δv = 1.5 km/h

The time it takes for the person to arrive at point B from point A at the new speed, t₂ = 1 hour earlier than when walking at 6 km/h

a) Let t₁ represent the time it takes the person walking from point A to point B at 6 km/h, we have;

t₂ = t₁ - 1...(1)

d/t₁ = 6...(2)

d/t₂ = 6 + 1.5 = 7.5

∴ d/t₂ = 7.5...(3)

From equation (2), we have;

d = 6 × t₁ = 6·t₁

Plugging in d = 6·t₁, and t₂ = t₁ - 1 in equation (3) gives;

d/t₂ = 7.5

∴ 6·t₁/t₁ - 1 = 7.5

6·t₁ = 7.5 × (t₁ - 1) = 7.5·t₁ - 7.5

7.5·t₁ - 6·t₁ = 7.5

1.5·t₁ = 7.5

t₁ = 7.5/1.5 = 5

t₁ = 5

The time it takes the person walking from point A to point B at 6 km/h, t₁ = 5 hours

The distance from point A to point B, d = 6 km/h × 5 hours = 30 km

b) The distance the person travels at the initial speed, v₁ (6 km/h) = s₁

The duration the person pauses for a rest = 15 minutes = 1/4 hours

The speed with which he walks the rest of the journey, v₂ = 7.5 km/h

The time earlier than expected that he arrives, Δt = 30 minutes = 0.5 hours

We note that the total distance, d = 30 km

The expected time, t₁ = 5 hours

Therefore, we have;

s₁ + s₂ = 30 km

s₂ = 30 - s₁

v₁/s₁ + 1/4 + v₂/s₂ = t₁ - 0.5

Therefore;

6/s₁ + 1/4 + 7.5/(30 - s₁) = 5 - 0.5 = 4.5

6/28.142+ 1/4 + 7.5/(30 - 28.142) = 5 - 0.5 = 4.5

6/s₁ + 7.5/(30 - s₁) = 4.5 - 1/4 = 4.25

-(3·s₁ + 360)/(2·s₁²- 60·s₁) = 4.25

2·s₁²- 60·s₁) × 4.25 + 3·s₁ + 360 = 0

17·s₁²- 504·s₁ + 720 = 0

s₁ = (504 ± √((-504)² - 4 × 17 × 720))/(2 × 17)

s₁ ≈ 28.142 or s₁ = 1.505

The distance the individual travels at v₁ = 6 km/h, s₁ ≈ 28.142 km or 1.505 km

4 0

3 years ago