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3 years ago

Which formula describes Newton’s second law of motion?

Physics

2 answers:

dimulka [17.4K]3 years ago

7 0

Answer:

Force = Mass x Acceleration

Allushta [10]3 years ago

6 0

Newton's second law of motion pertains to the behavior of objects for which all existing forces are not balanced. The second law states that the acceleration of an object is dependent upon two variables - the net force acting upon the object and the mass of the object. The acceleration of an object depends directly upon the net force acting upon the object, and inversely upon the mass of the object. As the force acting upon an object is increased, the acceleration of the object is increased. As the mass of an object is increased, the acceleration of the object is decreased.

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Darwin was the first scientist to propose the concept of evolution. A True B False

vladimir2022 [97]

Answer:

B: False

Explanation:

In the early 19th century Jean-Baptiste Lamarck proposed the first fully formed theory of evolution. Darwin's theory would come a few decades later in 1858

6 0

3 years ago

Read 2 more answers

10 points A baseball game can end in a tie. True False Help???

Vinvika [58]

Answer:

False

Explanation:

The game will go into more innings until a team outscores another. So, a game cannot end in a tie.

3 0

3 years ago

A rocket engine has a chamber pressure 4 MPa and a chamber temperature of 2000 K. Assuming isentropic expansion through the nozz

gladu [14]

This question is incomplete, the complete question is;

A rocket engine has a chamber pressure 4 MPa and a chamber temperature of 2000 K. Assuming isentropic expansion through the nozzle, and an exit Mach number of 3.2, what are the stagnation pressure and temperature in the exit plane of the nozzle? Assume the specific heat ratio is 1.2.

Answer:

- stagnation pressure is 274.993 Mpa

- the stagnation temperature Tt is 4048 K

Explanation:

Given the data in the question;

To determine the stagnation pressure and temperature in the exit plane of the nozzle;

we us the expression;

Pt/P = (1 + (γ-1 / 2) M²)^(γ/γ -1) = ( Tt/T )^(γ/γ -1)

where Pt is stagnant pressure = ?

P is static pressure = 4 MPa = 4 × 10⁶ Pa

Tt is stagnation temperature = ?

T is the static temperature = 2000 K

γ is ratio of specific heats = 1.2

M is Mach number M = 3.2

we substitute

Pt/P = (1 + (γ-1 / 2) M²)^(γ/γ -1)

Pt = P(1 + (γ-1 / 2) M²)^(γ/γ -1)

Pt = 4 × 10⁶(1 + (1.2-1 / 2) 3.2²)^(1.2/1.2 -1)

Pt = 4 × 10⁶ × 68.7484

Pt = 274.993 × 10⁶ Pa

Pt = 274.993 Mpa

Therefore stagnation pressure is 274.993 Mpa

Now, to get our stagnation Temperature

Pt/P = ( Tt/T )^(γ/γ -1)

we substitute

274.993 × 10⁶ Pa / 4 × 10⁶ Pa = ( Tt / 2000 )^(1.2/1.2 -1)

68.7484 = Tt⁶ / 6.4 × 10¹⁹

Tt⁶ = 68.7484 × 6.4 × 10¹⁹

Tt⁶ = 4.3998976 × 10²¹

Tt = ⁶√(4.3998976 × 10²¹)

Tt = 4047.999 ≈ 4048 K

Therefore, the stagnation temperature Tt is 4048 K

6 0

3 years ago

How does an ocean wave transfer energy across the ocean?

Alborosie

The energy travels in a disturbance, in an ocean that disturbance is a wave, so the wave makes energy and moves it through the water

3 0

3 years ago

A piece of aluminum has a volume of 1.50 10-3 m3. the coefficient of volume expansion for aluminum is β = 69 ✕ 10-6 (°c)-1. the

Alex17521 [72]

Answer:

W = 3.12 J

Explanation:

Given the volume is 1.50*10^-3 m^3 and the coefficient of volume for aluminum is β = 69*10^-6 (°C)^-1. The temperature rises from 22°C to 320°C. The difference in temperature is 320 - 22 = 298°C, so ΔT = 298°C. To reiterate our known values we have:

β = 69*10^-6 (°C)^-1 V = 1.50*10^-3 m^3 ΔT = 298°C

So we can plug into the thermal expansion equation to find ΔV which is how much the volume expanded (I'll use d instead of Δ because of format):

$dV = \beta V_{0} dT\\dV = (69*10^{-6})( C)^{-1} * (1.50*10^{-3})m^{3} * (298)C\\dV = 3.0843*10^-5$

So ΔV = 3.0843*10^-5 m^3

Now we have ΔV, next we have to solve for the work done by thermal expansion. The air pressure is 1.01 * 10^5 Pa

To get work, multiply the air pressure and the volume change.

$W = P * dV = (1.01 * 10^5)Pa * (3.0843*10^{-5})m^3\\W = 3.115143J$

W = 3.12 J

Hope this helps!

4 0

3 years ago