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3 years ago

Can someone help me please

Physics

2 answers:

iogann1982 [59]3 years ago

6 0

Examine the water molecules in the animation. After the wave has passed which of the following is true?

Eva8 [605]3 years ago

4 0

Explanation:

<h3>Density </h3>

unit name : kilogram per cubic metres. ρ = m/V,

symbol : kg/m³.

<h3>Pressure </h3>

unit name : Pascal.

symbol : Pa.

<h3>Work/Energy</h3>

unit name : Joule

symbol : J

<h3>Power </h3>

unit name : Watt

symbol : W

<h3>Force </h3>

unit name : Newton

symbol : N.

please try to do the derivation on your own, I left it for you.

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Please please help me please please help please please help me please

Kaylis [27]

Answer:

Electromagnetic waves do not require any medium to travel whereas mechanical waves must have a medium to propagate.

So, Basically, it is B I believe.

Hope It Helps!

6 0

3 years ago

water vapor contained in a piston–cylinder assembly undergoes an isothermal expansion at 240°c from a pressure of 7 bar to a pre

mafiozo [28]

The ideal gas constant is a proportionality constant that is added to the ideal gas law to account for pressure (P), volume (V), moles of gas (n), and temperature (T) (R). R, the global gas constant, is 8.314 J/K-1 mol-1.

According to the Ideal Gas Law, a gas's pressure, volume, and temperature may all be compared based on its density or mole value.

The Ideal Gas Law has two fundamental formulas.

PV = nRT, PM = dRT.

P = Atmospheric Pressure

V = Liters of Volume

n = Present Gas Mole Number

R = 0.0821atmLmoL K, the Ideal Gas Law Constant.

T = Kelvin-degree temperature

M stands for Molar Mass of the Gas in grams Mol d for Gas Density in gL.

Learn more about Ideal gas law here-

brainly.com/question/28257995

#SPJ4

7 0

1 year ago

A long ramp made of cast iron is sloped at a constant angle θ = 52.0∘ above the horizontal. Small blocks, each with mass 0.42 kg

dezoksy [38]

Answer:

For cast iron we have

$h = 0.92 m$

For copper

$h = 1.05 m$

For Lead

$h = 1.23 m$

For Zinc

$h = 2.43 m$

Explanation:

As we know that final speed of the block is calculated by work energy theorem

$W_f + W_g = \frac{1}{2}mv^2$

now we have

$-\mu_k mg cos\theta(\frac{h}{sin\theta}) + mgh = \frac{1}{2}mv^2$

now we have

$v^2 = 2gh - 2\mu_k g h cot\theta$

$v = \sqrt{2gh(1 - \mu_k cot\theta)}$

For cast iron we have

$4 = \sqrt{2(9.81)(h)(1 - 0.15cot52)}$

$h = 0.92 m$

For copper

$4 = \sqrt{2(9.81)(h)(1 - 0.29cot52)}$

$h = 1.05 m$

For Lead

$4 = \sqrt{2(9.81)(h)(1 - 0.43cot52)}$

$h = 1.23 m$

For Zinc

$4 = \sqrt{2(9.81)(h)(1 - 0.85cot52)}$

$h = 2.43 m$

4 0

3 years ago

A uniform plank 8.00 m in length with mass 50.0 kg is supported at two points located 1.00 m and 5.00 m, respectively, from the

andreev551 [17]

To solve the problem it is necessary to use Newton's second law and statistical equilibrium equations.

According to Newton's second law we have to

$F = mg$

where,

m= mass

g = gravitational acceleration

For the balance to break, there must be a mass M located at the right end.

We will define the mass m as the mass of the body, located in an equidistant center of the corners equal to 4m.

In this way, applying the static equilibrium equations, we have to sum up torques at point B,

$\sum \tau = 0$

Regarding the forces we have,

$3Mg-1mg=0$

Re-arrange to find M,

$M = \frac{m}{3}$

$M = \frac{50}{3}$

$M = 16.67Kg$

Therefore the maximum additional mass you could place on the right hand end of the plank and have the plank still be at rest is 16.67Kg

8 0

3 years ago

Why is force not on a scalar quantity??

snow_lady [41]

Because force always has a direction, it always works towards or against something.

you might know that force,
is rate of change of momentum i.e

force = m (v-u)/t
= (mv - mu )/ t

as we know momentum is a vector quantity so, the rate of change of momentum i.e Force would also be a vector quantity.

momentum = mass × velocity

velocity has a direction so,
momentum has also got a direction.
so, momentum is also a vector quantity.

3 0

4 years ago

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Can someone help me please​

Can someone help me please