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miss Akunina [59]

2 years ago

8

The unit of length most suitable for measuring the thickness of a cell phone is a . The unit of length most suitable for measuri

ng the height of a backyard tree is a?

1 answer:

ArbitrLikvidat [17]2 years ago

4 0

The answers include the following:

The unit of length most suitable for measuring the thickness of a cell phone is a meter.
The unit of length most suitable for measuring the height of a backyard tree is a meter.

<h3>What is Meter?</h3>

This is defined as the standard unit for measuring the length of a body and is denoted as m.

Height is a vertical type of length which is why meter was chosen as the most appropriate choice.

Read more about Meter here brainly.com/question/1578784

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You might be interested in

2. How long must a 400 W electrical engine work in order to produce 300 kJ of work?

yanalaym [24]

Answer:

Explanation:

400 W = 400 J/s

300000 J / 400 J/s = 750 s or 12.5 minutes

7 0

3 years ago

Help me yall it due in a few minutes :((()

vaieri [72.5K]

Answer:

B. blocks 2 & 3.

Explanation:

Block 1 has equal & opposite forces acting on it.

Block 2 has 5N on one side, 3N on the other. It will move in the direction the 5N of force is pushing.

Block 3 has no opposing force.

6 0

3 years ago

MATHPHYSSSSSSSS PLEASEEEEEE IM SORRY YOU PROBABLY HATE ME

inysia [295]

Answer:

3.1 m/s

Explanation:

First, find the time it takes for the cat to land. Take down to be positive.

Given:

Δy = 0.61 m

v₀ = 0 m/s

a = 9.81 m/s²

Find: t

Δy = v₀ t + ½ at²

(0.61 m) = (0 m/s) t + ½ (9.81 m/s²) t²

t = 0.353 s

Now find the horizontal velocity needed to travel 1.1 m in that time.

Given:

Δx = 1.1 m

a = 0 m/s²

t = 0.353 s

Find: v₀

Δx = v₀ t + ½ at²

(1.1 m) = v₀ (0.353 s) + ½ (0 m/s²) (0.353 s)²

v₀ = 3.1 m/s

3 0

3 years ago

Use the information from the graph to answer the question.

galina1969 [7]

The displacement of the object as determined from the velocity-time graph is 562.5 m.

<h3>What is a velocity-time graph?</h3>

A velocity-time graph is a graph of the velocity of an object plotted in the vertical or y-axis of the graph against the time taken on the horizontal or x-axis.

The displacement of an object can be obtained from its velocity-time graph by calculating the total area under the graph.

The total area under the graph = area of triangle + area of rectangle

Area of triangle = b*h/2 =

Area of triangle = 25 * (35 - 10)/2 = 312.5 m

Area of rectangle = l * b

Area of rectangle = 10 * 25 = 250 m

Total area = (312.5 + 250) m

Total area = 562.5 m

Therefore, the displacement of the object is 562.5 m

In conclusion, the total area of a velocity-time graph gives the displacement.

Learn more about velocity-time graph at: brainly.com/question/28064297

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5 0

2 years ago

Compare the circular velocity of a particle orbiting in the Encke Division, whose distance from Saturn 133,370 km, to a particle

Ket [755]

Answer:

The particle in the D ring is 1399 times faster than the particle in the Encke Division.

Explanation:

The circular velocity is define as:

$v = \frac{2 \pi r}{T}$

Where r is the radius of the trajectory and T is the orbital period

To determine the circular velocity of both particles it is necessary to know the orbital period of each one. That can be done by means of the Kepler’s third law:

$T^{2} = r^{3}$

Where T is orbital period and r is the radius of the trajectory.

Case for the particle in the Encke Division:

$T^{2} = r^{3}$

$T = \sqrt{(133370 Km)^{3}}$

$T = \sqrt{(2.372x10^{15} Km)}$

$T = 4.870x10^{7} Km$

It is necessary to pass from kilometers to astronomical unit (AU), where 1 AU is equivalent to 150.000.000 Km ( $1.50x10^{8} Km$ )

1 AU is defined as the distance between the earth and the sun.

$\frac{4.870x10^{7} Km}{1.50x10^{8}Km} . 1AU$

$T = 0.324 AU$

But 1 year is equivalent to 1 AU according with Kepler’s third law, since 1 year is the orbital period of the earth.

$T = \frac{0.324 AU}{1 AU} . 1 year$

$T = 0.324 year$

That can be expressed in units of days

$T = \frac{0.324 year}{1 year} . 365.25 days$

$T = 118.60 days$

<em>Circular velocity for the particle in the </em><em>Encke Division</em><em>:</em>

$v = \frac{2 \pi r}{T}$

$v = \frac{2 \pi (133370 Km)}{(118.60 days)}$

For a better representation of the velocity, kilometers and days are changed to meters and seconds respectively.

$118.60 days .\frac{86400 s}{1 day}$ ⇒ 10247040 s

$133370 Km .\frac{1000 m}{1 Km}$ ⇒ 133370000 m

$v = \frac{2 \pi (133370000 m)}{(10247040 s)}$

$v = 81.778 m/s$

Case for the particle in the D Ring:

For the case of the particle in the D Ring, the same approach used above can be followed

$T^{2} = r^{3}$

$T = \sqrt{(69000 Km)^{3}}$

$T = \sqrt{(3.285x10^{14} Km)}$

$T = 1.812x10^{7} Km$

$\frac{1.812x10^{7} Km}{1.50x10^{8}Km} . 1 AU$

$T = 0.120 AU$

$T = \frac{0.120 AU}{1 AU} . 1 year$

$T = 0.120 year$

$T = \frac{0.120 year}{1 year} . 365.25 days$

$T = 43.83 days$

<em>Circular velocity for the particle in </em><em>D Ring</em><em>:</em>

$v = \frac{2 \pi r}{T}$

$v = \frac{2 \pi (69000 Km)}{(43.83 days)}$

For a better representation of the velocity, kilometers and days are changed to meters and seconds respectively.

$43.83 days . \frac{86400 s}{1 day}$ ⇒ 3786912 s

$69000 Km . \frac{1000 m}{ 1 Km}$ ⇒ 69000000 m

$v = \frac{2 \pi (69000000 m)}{(3786912 s)}$

$v = 114.483 m/s$

$\frac{114.483 m/s}{81.778 m/s} = 1.399$

The particle in the D ring is 1399 times faster than the particle in the Encke Division.

7 0

3 years ago