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3 years ago

What affects how an object falls toward earth?

Physics

2 answers:

Ad libitum [116K]3 years ago

5 0

And also the gravitational acceleration varies with respect to distance from the surface.

ioda3 years ago

4 0

1. Air resistance (therefore, the air density, shape (aerodynamic or not), and surface area affected
2. The gravitational acceleration constant (g = 9.8 m/s^2 on Earth)

You might be interested in

A 50kg box is pulled on by a rope. The rope applies a horizontal force of 180 N to the box. What is the normal force exerted on

Over [174]

The normal force would be the opposing and equal force to gravity ... that is why the box isn't floating or going through the ground ... this is Newton's third law of motion

4 0

4 years ago

Answer the following questions.<br><br> show your full steps please

aleksandr82 [10.1K]

Answer:

Explanation:

If we let our reference frame travel at 30 m/s with the constant speed car, The accelerating car increases its velocity by 10 m/s in 3 seconds.

The average velocity of the accelerating car is (0 + 10) / 2 = 5 m/s

It will advance its position 5 m/s(3 s) = 15 m in the accelerating period.

It takes 5 + 3 = 8 m for the two cars to become side by side.

It would take another 5 + 3 = 8 m for the accelerating car to leave a gap of 3 m between.

The car requires 8 + 8 = 16 m to pass the other safely but the acceleration period only gets him to 15 m.

So despite your saying this is not a YES / NO question, the answer is NO the acceleration is too low or not long enough to meet the required clearances.

Input needed is 10000 J/s / 0.30 = 333333 = J/s

three hours requires 333333(3)(3600) = 360 MJ of energy

360 MJ / 34 MJ/liter = 10.6 liters.

8 0

3 years ago

A. What is the basic building block of matter?

Nikitich [7]

Answer:

a. atoms are the basic building blocks of matter. The atom is made of electrons and protons. It contains the chemical identity of an element.

b. Three types of particles that make up an atom are:

electrons - these are negatively charged particles and lightest constituents of an atom.

protons are positively charges and is 1873 times heavier than an electron.

Neutrons are neutral particles having mass equivalent to mass of proton.

Protons and neutrons exist in the center of an atom in nucleus and electrons revolve around the nucleus.

c. An atom is a neutral entity where as an ion is charged. An ion has excess of electrons or lacks electrons due to which it has negative or positive charge. An atom has an equal number of electrons and protons.

d. In a negative ion, the number electrons is greater than number of protons. The atom when gains electrons becomes negative ion.

e. John Dalton proposed Solid sphere model in which tiny invisible particles constitute matter. These invisible particles are atoms and these are indivisible.

4 0

3 years ago

Read 2 more answers

Air enters the compressor of an ideal air-standard Braytoncycle at 100 kPa, 300 K, with a volumetric flow rate of 5 m3/s.The tur

Harrizon [31]

Answer:

Explanation:

Given that

Air Inlet Pressure, P1 = 100 KPa

Air Inlet temperature, T1 = 300 K

Volume flow rate, Q = 5 m³/s

Turbine inlet temperature, T₃ = 1400 K

Compressor pressure ratio, r = 6, 8, 12

Heat capacity ratio or air = 1.4

γ= 1.4

Specific heat constant pressure of air, cp = 1.005 KJ/kg.k

At r = 6,

For Brayton cycle,

T2/T1 = r ^ (γ - 1)/γ

T3/T4 = r ^ (γ - 1)/γ

Now by putting the values

T2/300 = 6 ^ (1.4 - 1)/1.4

1400/T4 = 6 ^ (1.4 - 1)/1.4

T₂ = 1.67 × 300

= 500 K

T₄ = 1400/1.67

= 839.07 K

Efficiency, η = 1 - ((T4 - T1)/(T3 - T2)

Inputting values,

= 1 - ((839.07 - 300)/(1400 - 500))

= 0.40

= 40%

Bwr = Wcomp/Wturb

Where,

Wcomp = workdone by compressor

Wturb = workdone by turbine

= ((T2 - T1)/(T3 - T4))

= ((500 - 300)/(1400 - 839.07))

= 0.36

Net work = Net heat

Net heat = Qa - Qr

Qr = Cp ( T₄-T₁)

Qa = Cp ( T₃-T₂)

Imputting values,

Net heat, Qnet = 1.005 (1400 - 500 - 839.07 + 300)

= 1.005 × 360.93

= 362.74 kJ/kg

Net heat, Qnet = 362.74 kJ/kg

Using the ideal gas equation,

P V = n R T

But n = mass/molar mass,

P = ρ R T

By putting the values

P = ρ R T

Inputting values,

100 = ρ x 0.287 x 300

ρ = 1.16 kg/m³

mass flow rate, m = ρ × Q

= 1.16 × 5

= 5.80 kg/s

Net power, Pnet = ms × Net heat, Qnet

= 5.8 × 362.74

= 2103.9 kW.

At r = 8,

For Brayton cycle,

T2/T1 = r ^ (γ - 1)/γ

T3/T4 = r ^ (γ - 1)/γ

Now by putting the values

T2/300 = 8 ^ (1.4 - 1)/1.4

1400/T4 = 8 ^ (1.4 - 1)/1.4

T₂ = 1.81 × 300

= 543.4 K

T₄ = 1400/1.81

= 772.9 K

Efficiency, η = 1 - ((T4 - T1)/(T3 - T2)

Inputting values,

= 1 - ((772.9 - 300)/(1400 - 543.4))

= 0.448

= 45%

Bwr = Wcomp/Wturb

Where,

Wcomp = workdone by compressor

Wturb = workdone by turbine

= ((T2 - T1)/(T3 - T4))

= ((543.4 - 300)/(1400 - 772.9))

= 0.39

Net work = Net heat

Net heat = Qa - Qr

Qr = Cp ( T₄-T₁)

Qa = Cp ( T₃-T₂)

Imputting values,

Net heat, Qnet = 1.005 (1400 - 543.4 - 772.9 + 300)

= 1.005 × 383.7

= 385.62 kJ/kg

Net heat, Qnet = 385.62 kJ/kg

Using the ideal gas equation,

P V = n R T

But n = mass/molar mass,

P = ρ R T

By putting the values

P = ρ R T

Inputting values,

100 = ρ x 0.287 x 300

ρ = 1.16 kg/m³

mass flow rate, m = ρ × Q

= 1.16 × 5

= 5.80 kg/s

Net power, Pnet = ms × Net heat, Qnet

= 5.8 × 385.62

= 2236.59 kW.

At r = 12,

For Brayton cycle,

T2/T1 = r ^ (γ - 1)/γ

T3/T4 = r ^ (γ - 1)/γ

Now by putting the values

T2/300 = 12 ^ (1.4 - 1)/1.4

1400/T4 = 12 ^ (1.4 - 1)/1.4

T₂ = 2.03 × 300

= 610 K

T₄ = 1400/2.03

= 688.32 K

Efficiency, η = 1 - ((T4 - T1)/(T3 - T2)

Inputting values,

= 1 - ((688.32 - 300)/(1400 - 610))

= 0.509

= 51%

Bwr = Wcomp/Wturb

Where,

Wcomp = workdone by compressor

Wturb = workdone by turbine

= ((T2 - T1)/(T3 - T4))

= ((610 - 300)/(1400 - 688.32))

= 0.44

Net work = Net heat

Net heat = Qa - Qr

Qr = Cp ( T₄-T₁)

Qa = Cp ( T₃-T₂)

Imputting values,

Net heat, Qnet = 1.005 (1400 - 610 - 688.32 + 300)

= 1.005 × 401.68

= 403.7 kJ/kg

Net heat, Qnet = 403.7 kJ/kg

Using the ideal gas equation,

P V = n R T

But n = mass/molar mass,

P = ρ R T

By putting the values

P = ρ R T

Inputting values,

100 = ρ x 0.287 x 300

ρ = 1.16 kg/m³

mass flow rate, m = ρ × Q

= 1.16 × 5

= 5.80 kg/s

Net power, Pnet = ms × Net heat, Qnet

= 5.8 × 403.7

= 2341.39 kW.

3 0

3 years ago

Read 2 more answers

On flat ground, a 70-kg person requires about 300 W of metabolic power to walk at a steady pace of 5.0 km/h (1.4 m/s). Using the

Darina [25.2K]

Answer:

C 350W

Explanation:

Given power output to walk on a flat ground to be 300W, h = 0.05x, v = 1.4m/s

m = 70kg and g =9.8m/s².

x = horizontal distance covered

Total energy used = potential energy used in climbing and the energy used in a walking the horizontal distance.

E = mgh + 300t

Where t is the time taken to cover the distance

x = vt and h = 0.05vt

E = mg×0.05×vt + 300t

Substituting respective values

E = 70×9.8×0.05×1.4t +300t = 348t

P = E/t = 348W ≈ 350W.

4 0

4 years ago