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3 years ago

Arrange the values according to magnitude

Physics

1 answer:

Talja [164]3 years ago

7 0

Answer:

$59000$

$1.8\times 10^5$

$4.3\times 10^{-2}$

$9.0\times 10^{-6}$

$8.4\times 10^{-6}$

Explanation:

The given numbers are

$1.8\times 10^5=18000$

$4.3\times 10^{-2}=0.043$

$8.4\times 10^{-6}=0.0000084
$

$9.0\times 10^{-6}=0.000009$

$59000$

The arrangement of the numbers from largest to smallest is

$59000$

$1.8\times 10^5$

$4.3\times 10^{-2}$

$9.0\times 10^{-6}$

$8.4\times 10^{-6}$ .

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3 years ago

A cylindrical bucket, open at the top, is 28.0 cm high and 11.0 cm in diameter. A circular hole with a cross-sectional area 1.55

svlad2 [7]

Answer:

so height is 0.1283 m

Explanation:

given data

height = 28 cm

diameter = 11 cm

cross-sectional area = 1.55 cm2

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to find out

How high will the water in the bucket rise

solution

we know that here

potential energy = kinetic energy

mgh = 1/2 mv²

multiply both sides by the 2 and we get

2mgh=mv²

solve it we get

√(2gh) = v ....................1

h = v²/2g ...............2

and

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4 years ago

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3 years ago

Read 2 more answers

nder some circumstances, a star can collapse into an extremely dense object made mostly of neutrons and called a neutron star. T

Alexeev081 [22]

Answer:

2.92 * 10³ rad/s

Explanation:

Given:

Initial Radius of Original Star (Ri) = 6.0 * 10^5 km

Final Radius of Neutron Star (Rf) = 16km

Angular Speed = 1 revolution in 35 days

We need to convert this to rad/s

To do that, we first convert to rad/day

i.e (1 rev/35 days) * (2π rad/ 1 rev)

We then convert the days to hour

i.e. (1 rev / 35 days) * (2π rad/ 1 rev) * (1 day / 24 hours)

Finally, we convert the hour to seconds (3600 seconds makes an hour)

i.e. (1 rev / 35 days) * (2π rad/ 1 rev) * (1 day / 24 hours) * (1 hour/ 3600 sec)

Angular Speed = 2π rad/ 3024000 secs

Angular Speed (wi) = 2.079 * 10^-6rad/s

From the question, we're asked to calculate the angular speed of the neutron star (wf)

Applying law of conservation of angular momentum to a system whose moment of Inertia changes, we have

Ii*wi = If * wf ----------------- Formula

Where Ii and If are the initial and final Inertia of the star

The relationship between Inertia and Radius of each object is I = 2/5MR²

So, Ii = 2/5(MRi²) and If = 2/5(MRf²)

Substitute the above in the formula quoted

We have 2/5(MRi²)wi = 2/5(MRf²)wf ---------------- Divide through by 2M/5

We are left with, Ri²wi = Rf²wf

Make wf the subject of the formula

wf = wi * (Ri/Rf)²

wf = 2.079 * 10^-6rad/s * (6.0 * 10^5 km/16km)²

wf = 2.079 * 10^-6rad/s * (6.0 * 10^5 km/16km) * (6.0 * 10^5 km/16km)

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3 years ago

What average force is needed to accelerate a 7.00-gram pellet from rest to 155 m/s over a distance of 0.600 m along the barrel o

leonid [27]

Answer: The force needed is 140.22 Newtons.

Explanation:

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F = m a

In order to calculate the acceleration, given the displacement d,

$d = \frac{1}{2}at^2\implies a=\frac{2d}{t^2}$

we will need to determine the time t it took for the pellet to make the distance through the barrel of 0.6m. That time can be determined using the average velocity of the pellet while traveling through the barrel. Since the velocity is a linear function of time, as mentioned above, the average is easy to calculate as:

$\overline{v}=\frac{1}{2}(v_{end}-v_{start})=\frac{1}{2}(155-0)\frac{m}{s}=77.5\frac{m}{s}$

This value can be used to determine the time for the pellet through the barrel:

$t = \frac{d}{\overline{v}}=\frac{0.6m}{77.5\frac{m}{s}}\approx0.00774s$

Finally, we can use the above to calculate the force:

$F = ma = m\frac{2d}{t^2} = 0.007kg\cdot \frac{2\cdot 0.6 m}{0.00774^2 s^2}\approx 140.22N$

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