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3 years ago

6

Like the alto in a choir the octavina plays the lower melody . what musical element does this tell?

1 answer:

Katena32 [7]3 years ago

7 0

<em>L</em><em>ike the alto in a choir the octavina plays the lower melody . what musical element does this tell?</em>

answer : melody....

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What are 5 different ways to use a fidget spinner?

butalik [34]

Answer:

Balancing on one finger.

Tossing between hands.

Using a different body part.

Building a tower.

Throwing it high in the air.

Flipping it over.

Switching between fingers.

Going behind the back.

Explanation:

Just stuff I used to do ^^

5 0

3 years ago

Car A with a mass of 725 kilograms is traveling east at an initial velocity of 15 meters/second. It collides head–on with car B,

ikadub [295]

Answer:

$p_t_o_t_a_l=250kg\frac{m}{s}$

Explanation:

<u>The total momentum of a system is defined by:</u>

$(mv)_t_o_t=m_1v_1+m_2v_2+...$

Where,

$(mv)_t_o_t$ is the total momentum or it could be expressed also as $p_t_o_t_a_l$ .

$m_1$ and $m_2$ represents the masses of the objects interacting in the system.

$v_1$ and $v_2$ are the velocities of the objects of the system.

<em>Remember: </em><em>The momentum is a fundamental physical magnitude of vector type.</em>

We have:

$m_1=725 kg$

$v_1=15\frac{m}{s}\\m_2=625 kg$

We are going to take the east side as positive, and the west side as negative. Then the velocity of the car B, has to be <u>negative</u>. It goes in a different direction from car A.

$v_2=-17\frac{m}{s}$

Then the total momentum of the system is:

$p_t_o_t_a_l=m_1v_1+m_2v_2\\p_t_o_t_a_l=(725kg)(15\frac{m}{s})+(625kg)(-17\frac{m}{s})\\p_t_o_t_a_l=10875kg\frac{m}{s}-10625kg\frac{m}{s}\\p_t_o_t_a_l=250kg\frac{m}{s}$

8 0

4 years ago

10 POINTS!!! Determine the pressure of your book in pascals (Pa). Show your work! (the pressure of the book in psi is 0.03 psi)

Musya8 [376]

You've listed a lot of data here, in both metric and customary units,
and I'm not even sure it's all needed. Let me try and boil it down:

Pressure on a surface =
   (total force on a surface) divided by (area of the surface).

The answer to the question is the pressure expressed in pascals.
There's actually enough information here to answer the question
in 2 different ways. We could ...

-- simply convert (0.03 pound per inch²) to pascals, or
-- go through the whole calculation of force, area, and then their quotient.

To me, converting 0.03 psi to Pa looks easier.

-- 1 pascal = 1 newton / 1 meter²

-- On Earth, 1 kilogram of mass weighs 9.8 Newtons and 2.2 pounds.
From this, we can calculate that

2.2 pounds of force = 9.8 newtons of force.

   1 pound = 4.45 newtons

(0.03 pound/inch²) x (4.45 newton/pound) x (1inch/2.54cm)² x (100cm/1m)² =

(0.03 x 4.45 x 1² x 100²) / (2.54² x 1²)    newton/meter² = 206.9 Pa .

7 0

4 years ago

In an electron microscope, electrons are accelerated to great velocities. Calculate the wavelength of an electron traveling with

Diano4ka-milaya [45]

Answer:

Wavelength, $\lambda=1.04\times 10^{-13}\ m$

Explanation:

It is given that,

Velocity of an electron, $v=7\times 10^6\ m/s$

Mass of an electron, $m=9.1\times 10^{-28}\ kg$

We need to find the wavelength of an electron. It can be calculated using the De- Broglie wavelength as :

$\lambda=\dfrac{h}{mv}$

$\lambda=\dfrac{6.63\times 10^{-34}}{9.1\times 10^{-28}\times 7\times 10^6}$

$\lambda=1.04\times 10^{-13}\ m$

So, the wavelength of an electron is $1.04\times 10^{-13}\ m$ . Hence, this is the required solution.

7 0

3 years ago

Suppose we could shrink the Earth without changing its mass. At what fraction of its current radius would the free-fall accelera

spin [16.1K]

Answer:

$R' = \frac{1}{\sqrt{3}}R$

Explanation:

The acceleration due to gravity on the surface of the Earth is given by:

$g=\frac{GM}{R^2}$

where

G is the gravitational constant

M is the mass of the Earth

R is the radius of the Earth

Here we want to find the new Earth radius R' for which the gravitational acceleration at the surface, g', would be 3 times the current value of g:

$g' = 3g$

So we would have

$\frac{GM}{R'^2}=3(\frac{GM}{R^2})$

Solving the equation for R', we find

$R'^2 = \frac{1}{3}R^2\\R' = \frac{1}{\sqrt{3}}R$

7 0

3 years ago

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