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amm1812
3 years ago
11

Which of the following is one part of a chemical formula?

Physics
2 answers:
Korvikt [17]3 years ago
8 0

Answer:

C

Explanation:

The symbol of each element in the substance

a_sh-v [17]3 years ago
5 0

Answer:A

Explanation: number that shows the total atomic mass of the substance

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A cello string 0.75 m long has a 220 hz fundamental frequency. find the wave speed along the vibrating string. answer in units o
maxonik [38]
For fundamental frequency of a string to occur, the length of the string has to be half the wavelength. That is,

1/2y = L, where L = length of the string, y = wavelength.

Therefore,
y = 2L = 2*0.75 =1.5 m

Additionally,
y = v/f Where v = wave speed, and f = ferquncy

Then,
v = y*f = 1.5*220 = 330 m/s
4 0
3 years ago
A bag of sand has a density of 45 g/cm3 and a mass of 15 kg. How much space does the sand take up?
Aleks [24]

Volume = mass/density

 

Volume = 15000 g/45 g/cm3 ≈ 333.3 cm<span>3</span>

3 0
3 years ago
Read 2 more answers
Consider the nearly circular orbit of Earth around the Sun as seen by a distant observer standing in the plane of the orbit. Wha
ikadub [295]

We have that the spring constant is mathematically given as

k=2.37*10^{11}N/m

Generally, the equation for angular velocity is mathematically given by

\omega=\sqrt{k}{m}

Where

k=spring constant

And

\omega =\frac{2\pi}{T}

Therefore

\frac{2\pi}{T}=\sqrt{k}{n}

Hence giving spring constant k

k=m((\frac{2 \pi}{T})^2

Generally

Mass of earth m=5.97*10^{24}

Period for on complete resolution of Earth around the Sun

T=365 days

T=365*24*3600

Therefore

k=(5.97*10^{24})((\frac{2 \pi}{365*24*3600})^2

k=2.37*10^{11}N/m

In conclusion

The effective spring constant of this simple harmonic motion is

k=2.37*10^{11}N/m

For more information on this visit

brainly.com/question/14159361

8 0
3 years ago
Consider a projectile of mass 20 kg launched with a speed 9 m/s at an elevation angle of 45 degrees. Taking the launch point as
viva [34]

Answer:

a) L=0. b) L = 262 k ^   Kg m²/s and c)  L = 1020.7 k^   kg m²/s

Explanation:

It is angular momentum given by

      L = r x p

Bold are vectors; where L is the angular momentum, r the position of the particle and p its linear momentum

One of the easiest ways to make this vector product is with the use of determinants

{array}\right] \left[\begin{array}{ccc}i&j&k\\x&y&z\\px&py&pz\end{array}\right]

Let's apply this relationship to our case

Let's start by breaking down the speed

      v₀ₓ = v₀ cosn 45

      voy =v₀ sin 45

      v₀ₓ = 9 cos 45

      voy = 9 without 45

      v₀ₓ = 6.36 m / s

      voy = 6.36 m / s

a) at launch point r = 0 whereby L = 0

. b) let's find the position for maximum height, we can use kinematics, at this point the vertical speed is zero

   vfy² = voy²- 2 g y

   y = voy² / 2g

   y = (6.36)²/2 9.8

   y = 2.06 m

Let's calculate the angular momentum

L= \left[\begin{array}{ccc}i&j&k\\x&y&0\\px&0&0\end{array}\right]

L = -px y k ^

L = - (m vox) (2.06) k ^

L = - 20 6.36 2.06 k ^

L = 262 k ^   Kg m² / s

The angular momentum is on the z axis

c) At the point of impact, at this point the height is zero and the position on the x-axis is the range

     R = vo² sin 2θ / g

     R = 9² sin (2 45) /9.8

     R = 8.26 m

L = \left[\begin{array}{ccc}i&j&k\\x&0&0\\px&py&0\end{array}\right]

L = - x py k ^

L = - x m voy

L = - 8.26 20 6.36 k ^

L = 1020.7 k^   kg m² /s

5 0
3 years ago
Please Help me in this question..​
Vlad [161]

Answer:

1 different

Explanation:

2 light

3 sunny

4 a shadow

5dark

<h2 />
8 0
3 years ago
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