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2 years ago

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A spool of thread has an average radius of 1.00 cm. If the spool contains 62.8 m of thread, how many turns of thread are on the

spool? "Average radius" allows us to not need to treat the layering of threads on lower layers.

1 answer:

Simora [160]2 years ago

7 0

To solve this problem we will start from the given concept in which the number of turns is equivalent to the length of the thread per circumference of spool. That is:

$N = \frac{l}{\phi}$

Where,

l = length of the thread

$\phi$ = circumference of spool

For \phi we have that,

$\phi = 2\pi r \rightarrow 2\pi (0.01)$

For l we have that

l = 62.8m

Finally the number of Turns would be,

$N = \frac{l}{\phi}$

$N = \frac{62.8}{2\pi (0.01)}$

$N = 1000turns$

Therefore the number of turns of thread on the spool are 1000turns.

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Betelgeuse is how many times larger than the sun

Aleksandr-060686 [28]

Betelgeuse is one of the largest known stars and is probably at least the size of the orbits of Mars or Jupiter around the sun. That's a diameter about 700 times the size of the Sun or 600 million miles. For a star it has a rather low surface temperature (6000 F compared to the Sun's 10,000 F).

3 0

3 years ago

A vector → A has a magnitude of 56.0 m and points in a direction 30.0° below the negative x axis. A second vector, → B , has a m

MissTica

Answer:

The magnitude of the vector $\vec{C}$ is 107.76 m

Explanation:

To find the components of the vectors we can use:

$\vec{A} = | \vec{A} | \ ( \ cos(\theta) \ , \ sin (\theta) \ )$

where $| \vec{A} |$ is the magnitude of the vector, and θ is the angle over the positive x axis.

The negative x axis is displaced 180 ° over the positive x axis, so, we can take:

$\vec{A} = 56.0 \ m \ ( \ cos( 180 \° + 30 \°) \ , \ sin (180 \° + 30 \°) \ )$

$\vec{A} = 56.0 \ m \ ( \ cos( 210 \°) \ , \ sin (210 \°) \ )$

$\vec{A} = ( \ -48.497 \ m \ , \ - 28 \ m \ )$

$\vec{B} = 82.0 \ m \ ( \ cos( 180 \° - 49 \°) \ , \ sin (180 \° - 49 \°) \ )$

$\vec{B} = 82.0 \ m \ ( \ cos( 131 \°) \ , \ sin (131 \°) \ )$

$\vec{B} = ( \ -53.797 \ m \ , \ 61.886\ m \ )$

Now, we can perform vector addition. Taking two vectors, the vector addition is performed:

$(a_x,a_y) + (b_x,b_y) = (a_x+b_x,a_y+b_y)$

So, for our vectors:

$\vec{C} = ( \ -48.497 \ m \ , \ - 28 \ m \ ) + ( \ -53.797 \ m \ , ) = ( \ -48.497 \ m \ -53.797 \ m , \ - 28 \ m \ + \ 61.886\ m \ )$

$\vec{C} = ( \ - 102.294 \ m , \ 33.886 m \ )$

To find the magnitude of this vector, we can use the Pythagorean Theorem

$|\vec{C}| = \sqrt{C_x^2 + C_y^2}$

$|\vec{C}| = \sqrt{(- 102.294 \ m)^2 + (\ 33.886 m \)^2}$

$|\vec{C}| =107.76 m$

And this is the magnitude we are looking for.

5 0

3 years ago

Your brother is insisting that you’ll never hear a sound produced behind a barrier wall at the end of your yard you notice that

Tresset [83]

Answer

D.Diffraction

Explanation

Diffraction is a property that is experienced by waves when they come across a barrier when they are in motion.

The ways tends to curve behind the barrier. This is called diffraction of waves.

Now, sound is a wave and it also experience diffraction. . So the brother will be able to hear the sound due to diffraction

8 0

2 years ago

Read 2 more answers

A 1 200-kg car traveling initially at vCi 5 25.0 m/s in an easterly direction crashes into the back of a 9 000-kg truck moving i

sukhopar [10]

Answer:

The velocity of the truck after the collision is 20.93 m/s

Explanation:

It is given that,

Mass of car, m₁ = 1200 kg

Initial velocity of the car, $v_{Ci}=25\ m/s$

Mass of truck, m₂ = 9000 kg

Initial velocity of the truck, $v_{Ti}=20\ m/s$

After the collision, velocity of the car, $v_{Cf}=18\ m/s$

Let $v$ is the velocity of the truck immediately after the collision. The momentum of the system remains conversed.

$initial\ momentum=final\ momentum$

$1200\ kg\times 25\ m/s+9000\ kg\times 20\ m/s=1200\ kg\times 18+9000\ kg\times v$

$210000-21600=9000\ kg\times v$

$v=20.93\ m/s$

So, the velocity of the truck after the collision is 20.93 m/s. Hence, this is the required solution.

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2 years ago

Is the temperature and density of Earth's crust high or low? Why?

Setler [38]

Answer:

The crust is the first layer of the earth. It is split up into two parts the continental crust, and the oceanic crust. The oceanic crust takes up 71% of the earths crust, and the other 29% of the crust is continental. The continental is made up of igneous rocks, and the oceanic crust is made up of sedimentary and basalt rocks. The continental crust is older than the oceanic crust, some of the rocks are 3.9 billion years old. The density average of the oceanic crust is 3g/cm. The average density of the continental earth is 2.7g/cm. The temperature of the crust is around 200-400 degrees Celsius. The crust is about 60 km thick under a continent and 5 km thick under the ocean. The crust is constantly moving. The crust doesn't even make up 1% of the earth! The crust is the layer were tectonic plates can be found.

Explanation:

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3 years ago