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Natali [406]
3 years ago
15

The number of significant figures in 0.060900 is

Physics
1 answer:
Lady bird [3.3K]3 years ago
8 0

Answer:

There are <u>5</u> significant figures.

Explanation:

You must start conting your sig figs until you continue to hit zeros at the end. Those zeroes at the end are disregarded. So 0.0609 is where you get your <em>sig figs</em> from.

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Each of the four expansion models (recollapsing, critical, coasting, and accelerating) predict different ages for the universe,
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Answer:

This is because the age of the universe is determined by the pace of expansion in the past, and each model forecasts a different pace.

Explanation:

The age of the universe is determined by the pace of expansion in the past, and each model forecasts a different pace.

This is due to the fact that the expansion rate in the coasting model is constant and never changes. Because the cosmos is growing faster now than during the old days, recollapsing and critical models give shorter ages. According to the accelerating model, the universe is growing at a slower rate currently than in the past, implying an older age.

4 0
3 years ago
What results from splitting an atomic nucleus
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A uranium-235 atom<span> absorbs a neutron and fissions into two new </span>atoms<span> (fission fragments), releasing three new neutrons and some binding energy. ... Several heavy elements, such as uranium, thorium, and plutonium, undergo both spontaneous fission, a form of radioactive decay and induced fission, a form of </span>nuclear<span> reaction.</span>
7 0
3 years ago
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Scheduling comes before sequencing. In scheduling, we develop plans for our work centers where production activities occur. We l
ser-zykov [4K]

Answer: 89

Explanation:

4 0
3 years ago
The base of a right prism is a rhombus with diagonals of 6 and 8. If the altitude of the prism is 12, what is the total surface
Agata [3.3K]

Answer:

total surface area is 432

Explanation:

Given data

base  = 6

diagonals = 8

altitude = 12

to find out

total surface area

solution

we know total surface area of prism is

total surface area = lateral surface area + 2base area  ..............1

so

first we calculate base perimeter i.e = 2 length + 2 width

so perimeter = 2(8) + 2(6) = 25

and area  = length * width = 8*6 = 48

so lateral surface area is perimeter * height i.e

lateral surface area = 28* 12

lateral surface area = 336

put this value in equation 1 we get

total surface area = lateral surface area + 2base area

total surface area = 336 + 2(48)

total surface area is 432

3 0
3 years ago
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Initially, a 2.00-kg mass is whirling at the end of a string (in a circular path of radius 0.750 m) on a horizontal frictionless
drek231 [11]

Answer:

v_f = 15 \frac{m}{s}

Explanation:

We can solve this problem using conservation of angular momentum.

The angular momentum \vec{L} is

\vec{L}  = \vec{r} \times \vec{p}

where \vec{r} is the position and \vec{p} the linear momentum.

We also know that the torque is

\vec{\tau} = \frac{d\vec{L}}{dt}  = \frac{d}{dt} ( \vec{r} \times \vec{p} )

\vec{\tau} =  \frac{d}{dt}  \vec{r} \times \vec{p} +   \vec{r} \times \frac{d}{dt} \vec{p}

\vec{\tau} =  \vec{v} \times \vec{p} +   \vec{r} \times \vec{F}

but, as the linear momentum is \vec{p} = m \vec{v} this means that is parallel to the velocity, and the first term must equal zero

\vec{v} \times \vec{p}=0

so

\vec{\tau} =   \vec{r} \times \vec{F}

But, as the only horizontal force is the tension of the string, the force must be parallel to the vector position measured from the vertical rod, so

\vec{\tau}_{rod} =   0

this means, for the angular momentum measure from the rod:

\frac{d\vec{L}_{rod}}{dt} =   0

that means :

\vec{L}_{rod} = constant

So, the magnitude of initial angular momentum is :

| \vec{L}_{rod_i} | = |\vec{r}_i||\vec{p}_i| cos(\theta)

but the angle is 90°, so:

| \vec{L}_{rod_i} | = |\vec{r}_i||\vec{p}_i|

| \vec{L}_{rod_i} | = r_i * m * v_i

We know that the distance to the rod is 0.750 m, the mass 2.00 kg and the speed 5 m/s, so:

| \vec{L}_{rod_i} | = 0.750 \ m \ 2.00 \ kg \ 5 \ \frac{m}{s}

| \vec{L}_{rod_i} | = 7.5 \frac{kg m^2}{s}

For our final angular momentum we have:

| \vec{L}_{rod_f} | = r_f * m * v_f

and the radius is 0.250 m and the mass is 2.00 kg

| \vec{L}_{rod_f} | = 0.250 m * 2.00 kg * v_f

but, as the angular momentum is constant, this must be equal to the initial angular momentum

7.5 \frac{kg m^2}{s} = 0.250 m * 2.00 kg * v_f

v_f = \frac{7.5 \frac{kg m^2}{s}}{ 0.250 m * 2.00 kg}

v_f = 15 \frac{m}{s}

8 0
3 years ago
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