Which statements describe properties of stars check all that apply

Response:

Explanation:

Let T denote the tension.

By employing Newton's second law to analyze the bucket's downward motion, we have:

mg - T = ma

A torque, TR, acts on the drum, inducing an angular acceleration α in it. If I refers to the moment of inertia of the drum, then:

TR = Iα

Rearranging gives: TR = Ia/R

This leads to T =  Ia/R²

Substituting this expression for T back into the previous equation yields:

mg - T = ma

mg - Ia/R² = ma

Consequently, we find that mg =  Ia/R² + ma

Therefore, a (I/R² + m) = mg

This results in: a = mg / (I/R² + m)

Next, we aim to express T as:

mg - T = ma

which simplifies to mg - ma  = T

Rearranging gives mg - m²g / (I/R² + m) = T

Thus, we arrive at: mg - mg / (1 + I / m R²) = T

For part (b), T =  Ia/R²

and for part (c), the moment of inertia of a hollow cylinder calculates to:

I = 1/2  M (R² - (R² / 4))

This simplifies to 3/4 x 1/2 MR², yielding 3/8 MR²

Thus, I / R² = 3/8 M

When we substitute, we find a = mg / (3/8 M + m)

and subsequently T =  Ia/R²

= 3/8 MR² × mg / (3/8 M + m) × 1/R²

Results in:

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The force is calculated by multiplying mass and gravitational acceleration (F= mg). To find the solution, the mass of the elephant (5600 kg) is multiplied by gravity (9.8 m/s²). The result is 55,880 N, representing the upward gravitational force the elephant exerts on the Earth.

Part a) The package's speed matches the helicopter's speed in the horizontal direction. Thus, after a time "t", the horizontal velocity remains constant, while in the Y-direction, it begins to fall under gravity. Part b) The distance relative to the helicopter is equivalent to the distance it falls freely. Part c) If the helicopter is ascending uniformly, the package's final speed after time t can be described in terms of its initial speed and gravity.

Answer:

a) W =400 kJ

b) W = 0 kJ

c) W =-160.944 KJ

Explanation:

Given

Process 1 ---> 2

Pressure at point (1) P1 = 10 bar = P2

Volume at point (1) V1 = 1 m^3

Volume at point (2) V2 =4 m^3

For Process 2 ---> 3, where V = constant

Pressure at point (3) P3 = 10 bar

Volume at point (3) V3 = 4 m^3

Process 3 ---> 1 defined as PV = constant.

Required

Sketch the processes on the PV coordinates To calculate the work in kJ

Work is calculated by W=

a=V2

b=V1

x=Pdx=dV

For Process 1 ---> 2 where P3 = P4 = 5 bar  

W=

a=V3

b=V2

x=4dx=dV

substituting the values here into the integral gives

W=400 kJFor Process 2 ---> 3

As V = constant in this case, the volume remains unchanged, resulting in W = 0 kJ  

For Process 3 ---> 1  By applying point (1) --> 5 x.2 = C ---> C = 1 P = 5V^-1  

 W=

a=V1

b=V3

x=1V^-1dx=dVsubstituting values into this integral results in

W=| ln V | limits a and b

  = -160.944 KJ