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3 years ago

A bird flies at a speed of 4 m/s. It flies for 2160 seconds from its nest to the field. How much distance did the bird cover?

Physics

2 answers:

solniwko [45]3 years ago

8 0

Hello :D

For this applicate the formula:

$\boxed{d=vt}$

Data:

d = Distance = ?

v = Velocity = 4 m/s

t = Time = 2160 s

Replace according formula:

d = 4 m/s * 2160 s

Resolving:

d = 8640 m

Result:

The distance is of <u>8640 meters.</u>

Morgarella [4.7K]3 years ago

6 0

Answer:

The bird covers 8,640 m.

Explanation:

Distance= Speed×Time

Distance= 4×2160

Distance= 8,640

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A swimming pool has the shape of a right circular cylinder with radius 21 feet and height 10 feet. Suppose that the pool is full

AysviL [449]

Answer:

The water required to pump all the water to a platform 2 feet above the top of the pool is is 6061310.32 foot-pound.

Explanation:

Given that,

Radius = 21 feet

Height = 10 feet

Weighing = 62.5 pounds/cubic

Work = 4329507.37572

Height = 2 feet

Let's look at a horizontal slice of water at a height of h from bottom of pool

We need to calculate the area of slice

Using formula of area

$A=\pi r^2$

Put the value into the formula

$A=\pi\times21^2$

$A=441\pi\ feet^2$

Thickness of slice $t=\Delta h\ ft$

The volume is,

$V=(441\pi\times\Delta h)\ ft^3$

We need to calculate the force

Using formula of force

$F=W\times V$

Where, W = water weight

V = volume

Put the value into the formula

$F=62.5\times(441\pi\times\Delta h)$

$F=27562.5\pi\times\Delta h\ lbs$

We need to calculate the work done

Using formula of work done

$W=F\times d$

Put the value into the formula

$W=27562.5\pi\times\Delta h\times(10-h)\ ft\ lbs$

We do this by integrating from h = 0 to h = 10

We need to find the total work,

Using formula of work done

$W=\int_{0}^{h}{W}$

Put the value into the formula

$W=\int_{0}^{10}{27562.5\pi\\times(10-h)}dh$

$W=27562.5\pi(10h-\dfrac{h^2}{2})_{0}^{10}$

$W=27562.5\pi(10\times10-\dfrac{100}{2}-0)$

$W=4329507.37572$

To pump 2 feet above platform, then each slice has to be lifted an extra 2 feet,

So, the total distance to lift slice is (12-h) instead of of 10-h

We need to calculate the water required to pump all the water to a platform 2 feet above the top of the pool

Using formula of work done

$W=\int_{0}^{h}{W}$

Put the value into the formula

$W=\int_{0}^{10}{27562.5\pi\\times(12-h)}dh$

$W=27562.5\pi(12h-\dfrac{h^2}{2})_{0}^{10}$

$W=27562.5\pi(12\times10-\dfrac{100}{2}-0)$

$W=1929375\pi$

$W=6061310.32\ foot- pound$

Hence, The water required to pump all the water to a platform 2 feet above the top of the pool is is 6061310.32 foot-pound.

8 0

3 years ago

A machine, modeled as a simple spring-mass system, oscillates in simple harmonic motion. Its acceleration is measured to have an

ANTONII [103]

$a=5000\dfrac{mm}{s}=5\dfrac{m}{s}$

$f=10{Hz}\Longrightarrow t=\dfrac{1}{10}s$

$a_{max}=\dfrac{50\frac{m}{s}}{\frac{s}{10}}\cdot\dfrac{1}{\sqrt{2}}=\dfrac{50\dfrac{m}{s^2}}{\sqrt{2}}\approx\boxed{35.4\dfrac{m}{s^2}}$

Hope this helps.

r3t40

5 0

2 years ago

Rutherford tracked the motion of tiny, positively charged particles shot through a thin sheet of gold foil. Some particles

Anna007 [38]

Question

Rutherford tracked the motion of tiny, positively charged particles shot through a thin sheet of gold foil. Some particles travelled in a straight line and some were deflected at different angles.

Which statement best describes what Rutherford concluded from the motion of the particles?

A) Some particles travelled through empty spaces between atoms and some particles were deflected by electrons.

B) Some particles travelled through empty parts of the atom and some particles were deflected by electrons.

C) Some particles travelled through empty spaces between atoms and some particles were deflected by small areas of high-density positive charge in atoms.

D) Some particles travelled through empty parts of the atom and some particles were deflected by small areas of high-density positive charge in atoms.

Answer:

The right answer is C)

Explanation:

In the experiment described above, a piece of gold foil was hit with alpha particles, which have a positive charge. Alpha particles <em>α</em> were used because, if the nucleus was positive, then it would deflect the positive particles. The principles of physics posit that electric charges of the same orientation repel.

So most as expected some of the alpha particles went right through meaning that the gold atoms comprised mostly empty space except the areas that were with a dense population of positive charges. This area became known as the "nucleus".

Due to the presence of the positive charges in the nucleus, some particles had their paths bent at large angles others were deflected backwards.

Cheers!

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3 years ago

Read 2 more answers

A flux density of 1.2Wb/m^2 is required in the 1 mm air gap of an electromagnet having an iron path of length 1.5 m. Calculate t

Mandarinka [93]

Answer:

The mmf required is $1.125$ × $10^{-3}$ A