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azamat
3 years ago
12

Everybody can you post the lyrics to your favorite song? I'll give brainliest if I know the song or if I like the lyrics, I'll b

e doing a few of these so.... yeah have fun!>
Physics
2 answers:
Setler [38]3 years ago
5 0

Answer:

I cant post da lyrics lol but this is the song

Explanation:

chill bill by rob stone feat. J. Davis and spooks

andrezito [222]3 years ago
4 0

Answer:

Show and tell

I'm on display for all you (bad word) to see

Show and tell

Harsh words if you don't get a pic with me

Buy and sell (buy and sell me, baby)

Like I'm a product to society

Art don't sell

Unless you (bad word) every authority

Explanation:

You might be interested in
What is the voltage of a motor that draws a current of 2 a and produces 240 w of power?
MA_775_DIABLO [31]
Here Power = Voltage * Current

So, Voltage = Power/Current

Put the values, 

V = 240/2

V = 120 V

In short, Your Final Answer would be: 120 Volts

Hope this helps!
8 0
3 years ago
A magnetic dipole <img src="https://tex.z-dn.net/?f=%5Cvec%7Bm%7D" id="TexFormula1" title="\vec{m}" alt="\vec{m}" align="absmidd
ser-zykov [4K]
We are asked to know the reason why magnetic dipole can be written as U = -m *B. The reason for this was that the negative sign is just showing or indicating the direction of the force. If it if negative, it means that the work done is opposite to the force field.
7 0
3 years ago
Give two mathematical examples of Newton's third law and how you get the solution​
bagirrra123 [75]

Answer:

1) Any particle moving in a horizontal plane slowed by friction, deceleration = 32 μ

2) The particle moving by acceleration = P/m - 32μ OR The external force = ma + 32μm

Explanation:

* Lets revise Newton’s Third Law:

- For every action there is a reaction, equal in magnitude and opposite

 in direction.

- Examples:

# 1) A particle moving freely against friction in a horizontal plane

- When no external forces acts on the particle, then its equation of

  motion is;

∵ ∑ forces in direction of motion = mass × acceleration

∵ No external force

∵ The friction force (F) = μR, where μ is coefficient of the frictional force

   and R is the normal reaction of the weight of the particle on the

   surface

∵ The frictional force is in opposite direction of the motion

∴ ∑ forces in the direction of motion = 0 - F

∴ 0 - F = mass × acceleration

- Substitute F by μR

∴ - μR = mass × acceleration

∵ R = mg where m is the mass of the particle and g is the acceleration

  of gravity

∴ - μ(mg) = ma ⇒ a is the acceleration of motion

- By divide both sides by m

∴ - μ(g) = a

∵ The acceleration of gravity ≅ 32 feet/sec²

∴ a = - 32 μ

* Any particle moving in a horizontal plane slowed by friction,

 deceleration = 32 μ

# 2) A particle moving under the action of an external force P in a

  horizontal plane.

- When an external force P acts on the particle, then its equation

 of motion is;

∵ ∑ forces in direction of motion = mass × acceleration

∵ The external force = P

∵ The friction force (F) = μR, where μ is coefficient of the frictional force

   and R is the normal reaction of the weight of the particle on the

   surface

∵ The frictional force is in opposite direction of the motion

∴ ∑ forces in the direction of motion = P - F

∴ P - F = mass × acceleration

- Substitute F by μR

∴ P - μR = mass × acceleration

∵ R = mg where m is the mass of the particle and g is the acceleration

  of gravity

∴ P - μ(mg) = ma ⇒ a is the acceleration of motion

∵ The acceleration of gravity ≅ 32 feet/sec²

∴ P - 32μm = ma ⇒ (1)

- divide both side by m

∴ a = (P - 32μm)/m ⇒ divide the 2 terms in the bracket by m

∴ a = P/m - 32μ

* The particle moving by acceleration = P/m - 32μ

- If you want to fin the external force P use equation (1)

∵ P - 32μm = ma ⇒ add 32μm to both sides

∴ P = ma + 32μm

* The external force = ma + 32μm

7 0
3 years ago
Approximately how many degrees does high latitude ocean water change from the ocean surface to a depth of 1 km
saveliy_v [14]

There is no change in ocean water temperature when it is from the ocean surface to a depth of 1 km.


3 0
3 years ago
Two taut strings of identical mass and length are stretched with their ends fixed, but the tension in one string is 1.10 times g
ollegr [7]

Answer:

The  beat frequency when each string is vibrating at its fundamental frequency is 12.6 Hz

Explanation:

Given;

velocity of wave on the string with lower tension, v₁ = 35.2 m/s

the fundamental frequency of the string, F₁ = 258 Hz

<u>velocity of wave on the string with greater tension;</u>

v_1 = \sqrt{\frac{T_1}{\mu }

where;

v₁ is the velocity of wave on the string with lower tension

T₁ is tension on the string

μ is mass per unit length

v_1 = \sqrt{\frac{T_1}{\mu} } \\\\v_1^2 = \frac{T_1}{\mu} \\\\\mu = \frac{T_1}{v_1^2} \\\\ \frac{T_1}{v_1^2} =  \frac{T_2}{v_2^2}\\\\v_2^2 = \frac{T_2v_1^2}{T_1}

Where;

T₁ lower tension

T₂ greater tension

v₁ velocity of wave in string with lower tension

v₂ velocity of wave in string with greater tension

From the given question;

T₂ = 1.1 T₁

v_2^2 = \frac{T_2v_1^2}{T_1}  \\\\v_2 = \sqrt{\frac{T_2v_1^2}{T_1}} \\\\v_2 = \sqrt{\frac{1.1T_1*(35.2)^2}{T_1}}\\\\v_2 = \sqrt{1.1(35.2)^2} = 36.92 \ m/s

<u>Fundamental frequency of wave on the string with greater tension;</u>

<u />f = \frac{v}{2l} \\\\2l = \frac{v}{f} \\\\thus, \frac{v_1}{f_1}  =\frac{v_2}{f_2} \\\\f_2 = \frac{f_1v_2}{v_1} \\\\f_2 =\frac{258*36.92}{35.2} \\\\f_2 = 270.6 \ Hz<u />

Beat frequency = F₂ - F₁

                          = 270.6 - 258

                          = 12.6 Hz

Therefore, the  beat frequency when each string is vibrating at its fundamental frequency is 12.6 Hz

6 0
3 years ago
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