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3 years ago

What is mass formula?

Physics

2 answers:

kherson [118]3 years ago

4 0

Answer:

The mass formula is given as

Mass = ρ × v

Where,

ρ = density and

v = the volume

The weight mass formula is given as

m = w / g

Where,

w= weight

m = mass

g = gravity

The mass formula is also given as

m = F / a

If acceleration itself is the gravity, then

M = F / g

Where,

F = force

G = gravity

According to Einstein’s mass-energy relation

m = (E / c2)

Where,

m = mass

E = energy

c = speed of light (3×108 m/s)

The kinetic energy mass formula is given as

K.E = ½ mv2

Where,

m = mass,

v = velocity.

Explanation:

pls mark me as brainliest

Aleksandr [31]3 years ago

4 0

Featured snippet from the web
The molar mass (M) is a physical property and it is defined as the mass of one mole of the chemical substance or it is a ratio of the mass of a chemical compound to its amount of chemical substance. The unit of molar mass is kg/mol. The mass formula is given as. Mass = ρ × v.

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An oceanographer is studying how the ion concentration in seawater depends on depth. She makes a measurement by lowering into th

Black_prince [1.1K]

Answer:

a) R = ρ₀ L /π(r_b² - R_a²) , b) ρ₀ = V / I π (r_b² - R_a²) / L

Explanation:

a) The resistance of a material is given by

R = ρ l / A

where ρ is the resistivity, l is the length and A is the area

the length is l = L and the resistivity is ρ = ρ₀

the area is the area of the cylindrical shell

A = π r_b² - π r_a²

A = π (r_b² - r_a²)

we substitute

R = ρ₀ L /π(r_b² - R_a²)

b) The potential difference is related to current and resistance by ohm's law

V = i R

we subsist the expression of resistance

V = I ρ₀ L /π (r_b² - R_a²)

ρ₀ = V / I π (r_b² - R_a²) / L

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3 years ago

What are the two major characteristics that change when air is heated and cooled ? How do these characteristics change the movem

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Two major characteristics that change when air is heated and cooled is the density and the direction of the air flow. They change the movement of the air by heated air rises and cooled air doesn't.

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3 years ago

Find the gravitational force between the

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Answer:

The process is given in the pic.

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3 years ago

A block of mass 10.0 kg is pulled to the right along a rough horizontal surface with a constant horizontal force of 20.0 N. The

Yakvenalex [24]

Answer:

the magnitude of acceleration will be 1.50m/s^2

Explanation:

To calculate your acceleration, you can use your formula that states that the net force on an object is equal to the mass of the object multiplied by the acceleration of the object. Fnet=ma

if you draw out this situation and label the forces you will have your vector towards the right with a magnitude of 20.0N and then your friction vector will be pointing to the left (in other words, in the negative direction) (opposing the direction of movement) with a magnitude of 5.00N, with the 10.0 kg box in the middle.

The net force will be calculated using F1+F2=Fnet where your F1=20.0N and F2= -5.00N (since it is towards the negative direction).

you will find that Fnet=15.0N

With that, plug in the values you know to calculate the acceleration of the block:

Fnet=ma

(15.0N)=(10.0kg)a from her you can divide both sides by 10 to isolate a:

1.50=a (and now make sure to label the units of your answer)

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2 years ago

The y-position of a damped oscillator as a function of time is shown in the figure.

NISA [10]

(1) The period of the oscillator is 1 second.

(2) The damping coefficient is 0.93.

<h3>What is period of oscillation?</h3>

The period of oscillation is the time taken to make one complete cycle.

From the graph, the time taken to make one complete oscillation is 1 second.

<h3>Damping coefficient</h3>

equation of the wave is given as;

y(t) = Ae^(-btx) cos(ωt)

3.5 = Ae^(-0) cos(0)

3.5 = A x 1

A = 3.5 cm

-3 = 3.5e^(-bx) cos(ω)

-3/3.5 = e^(-bx) cos(ω)

-0.857 = e^(-bx) cos(ω)

-0.857 / cos(ω) = e^(-bx)

ln[-0.857 / cos(ω)] = -bx

ln[-0.857 / cos(ω)] / b = - x ---- (1)

-2 = 3.5e^(-2bx) cos(2ω)

-0.57 = e^(-2bx) cos(2ω)

ln[-0.57 / cos(2ω)] = -2bx

ln[-0.57 / cos(2ω)] /2b = - x ------(2)

solve (1) and (2)

ln[-0.57 / cos(2ω)]/2b = ln[-0.857 / cos(ω)] /b

-0.57 / cos(ω) = 2(-0.857 / cos(ω))

2(-0.857/cosω) = -0.57/cos2ω

-(2 x 0.857) / (-0.57) = cosω/cos 2ω

3 = cosω/cos 2ω

3(cos 2ω) = cosω

3(2cos²ω - 1) = cos ω

6cos²ω - 6 = cosω

6cos²ω - cosω - 6 = 0

let cosω = y

6y² - y - 6 = 0

solve the quadratic equation;

y = 1.1 or -0.92

cosω = -0.92

ω = arc cos(-0.92)

ω = 2.74 rad/s

From equation (1)

ln[-0.857 / cos(ω)] / x = -b ---- (1)

let x = 1

ln(-0.857/cos(2.74) = -b

-0.93 = -b

b = 0.93

Thus, the damping coefficient is 0.93.

Learn more about damping coefficient here: brainly.com/question/14058210

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4 0

2 years ago