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3 years ago

A typical home in the northern U.S. might require 120 MBtu of heat for the average winter.

Physics

1 answer:

eimsori [14]3 years ago

0 0

200 MBtu of heat

Explanation:

As provided in the question, heat supplying natural gas furnace operates at 60% efficiency. It means that out of 100 units of heat energy produced by the furnace, only 60 units of heat can be transferred.

Thereby as the demand of the heat energy is 120 MBtu, amount of gas that needs to be purchased equals-

∴ 60% of (cubic metres of gas that needs to be purchased) = 120 MBtu

⇒ (60/100) * gas = 120

⇒ Gas= 120*100/60

∴ Cubic metres of gas that needs to be purchased= 200MBtu

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A model train engine was moving at a constant speed on a straight horizontal track. As the engine moved along, a marble was fir

bagirrra123 [75]

Answer:

The marble was moving in a projectile and the speed of the engine was 2.716 m/s

Explanation:

The vertical component of the marble's flight path relative to the train

is given by the equation y(t) = v*t - (4.9)*t^2,

where, v is the initial upward velocity of the marble relative to the train.

So with y(1) = v - 4.9 = 0 we have

v = 4.9 m/s.

The marble will reach maximum height after 0.5 seconds, at which the

height will be y(0.5) = (4.9)*(0.5) - (4.9)*(0.5)^2 = (4.9)*(0.25) = 1.225 m.

Now, the marble has a vertical velocity component of 4.9 m/s and a horizontal velocity component

of V m/s such that tan(61) = 4.9 / V

V = 4.9 / tan(61) = 2.716 m/s

This horizontal velocity component of the marble is the same as the

speed of the train i.e. 2.716 m/s.

3 0

3 years ago

Which statement best describes the relationship between plants and oxygen?

saveliy_v [14]

A statement would be to the effect that plants use oxygen as part of photosynthesis

4 0

4 years ago

Read 2 more answers

Vector ????⃗ has a magnitude of 16.6 and is at an angle of 50.5∘ counterclockwise from the +x‑axis. Vector ????⃗ has a magnitude

natka813 [3]

Answer:

For vector u, x component = 10.558 and y component =12.808

unit vector = 0.636 i+ 0.7716 j

For vector v, x component = 23.6316 and y component = -6.464

unit vector = 0.9645 i-0.2638 j

Explanation:

Let the vector u has magnitude 16.6

u makes an angle of 50.5° from x axis

So $u_x=ucos\Theta =16.6\times cos50.5=10.558$

Vertical component $u_y=usin\Theta =16.6\times sin50.5=12.808$

So vector u will be u = 10.558 i+12.808 j

Unit vector $u=\frac{10.558i+12.808j}{\sqrt{10.558^2+12.808^2}}=0.636i+0.7716j$

Now in second case let vector v has a magnitude of 24.5

Making an angle with -15.3° from x axis

So horizontal component $v_x=vcos\Theta =24.5\times cos(-15.3)=23.6316$

Vertical component $v_y=vsin\Theta =24.5\times sin(-15.3)=-6.464$

So vector v will be 23.6316 i - 6.464 j

Unit vector of v $=\frac{23.6316i-6.464}{\sqrt{23.6316^2+6.464^2}}=0.9645i-0.2638j$

8 0

3 years ago

A normal human heart, beating about once per second, creates a maximum 4.00 mV potential across 0.400 m of a person's chest, pro

kvasek [131]

Answer:

The maximum electric field strength = 0.01 V/m

Explanation:

Given

ΔV(max) = 4.00 mV = 0.004 V

d = 0.400 m

f = 1.00 Hz

Maximum electric field = (maximum potential)/(length)

Maximum electric field = E(max)

Maximum potential = 4.00 mV = 0.004 V

Length = 0.400 m

E(max) = (0.004/0.4) = 0.01 V/m

Hope this Helps!!!

3 0

3 years ago

Read 2 more answers

A sample of monatomic ideal gas occupies 5.00 L at atmospheric pressure and 300 K (point A). It is warmed at constant volume to

leonid [27]

Answer:

(a) 0.203 moles

(b) 900 K

(d) 15 L

(e) A → B, W = 0, Q = Eint = 1,518.91596 J

B → C, W = Q ≈ 1668.69974 J Eint = 0 J

C → A, Q = -2,531.5266 J, W = -1,013.25 J, Eint = -1,518.91596 J

(g) ∑Q = 656.089 J, ∑W = 655.449 J, ∑Eint = 0 J

Explanation:

At point A

The volume of the gas, V₁ = 5.00 L

The pressure of the gas, P₁ = 1 atm

The temperature of the gas, T₁ = 300 K

At point B

The volume of the gas, V₂ = V₁ = 5.00 L

The pressure of the gas, P₂ = 3.00 atm

The temperature of the gas, T₂ = Not given

At point C

The volume of the gas, V₃ = Not given

The pressure of the gas, P₃ = 1 atm

The temperature of the gas, T₂ = T₃ = 300 K

(a) The ideal gas equation is given as follows;

P·V = n·R·T

Where;

P = The pressure of the gas

V = The volume of the gas

n = The number of moles present

R = The universal gas constant = 0.08205 L·atm·mol⁻¹·K⁻¹

n = PV/(R·T)

∴ The number of moles, n = 1 × 5/(0.08205 × 300) ≈ 0.203 moles

The number of moles in the sample, n ≈ 0.203 moles

(b) The process from points A to B is a constant volume process, therefore, we have, by Gay-Lussac's law;

P₁/T₁ = P₂/T₂

∴ T₂ = P₂·T₁/P₁

From which we get;

T₂ = 3.0 atm. × 300 K/(1.00 atm.) = 900 K

The temperature at point B, T₂ = 900 K

(d) For a constant temperature process, according to Boyle's law, we have;

P₂·V₂ = P₃·V₃

V₃ = P₂·V₂/P₃

∴ V₃ = 3.00 atm. × 5.00 L/(1.00 atm.) = 15 L

The volume at point C, V₃ = 15 L

(e) The process A → B, which is a constant volume process, can be carried out in a vessel with a fixed volume

The process B → C, which is a constant temperature process, can be carried out in an insulated adjustable vessel

The process C → A, which is a constant pressure process, can be carried out in an adjustable vessel with a fixed amount of force applied to the piston

(f) For A → B, W = 0,

Q = Eint = n·cv·(T₂ - T₁)

Cv for monoatomic gas = 3/2·R

∴ Q = 0.203 moles × 3/2×0.08205 L·atm·mol⁻¹·K⁻¹×(900 K - 300 K) = 1,518.91596 J

Q = Eint = 1,518.91596 J

For B → C, we have a constant temperature process

Q = n·R·T₂·㏑(V₃/V₂)

∴ Q = 0.203 moles × 0.08205 L·atm/(mol·K) × 900 K × ln(15 L/5.00 L) ≈ 1668.69974 J

Eint = 0

Q = W ≈ 1668.69974 J

For C → A, we have a constant pressure process

Q = n·Cp·(T₁ - T₃)

∴ Q = 0.203 moles × (5/2) × 0.08205 L·atm/(mol·K) × (300 K - 900 K) = -2,531.5266 J

Q = -2,531.5266 J

W = P·(V₂ - V₁)

∴ W = 1.00 atm × (5.00 L - 15.00 L) = -1,013.25 J

W = -1,013.25 J

Eint = n·Cv·(T₁ - T₃)

Eint = 0.203 moles × (3/2) × 0.08205 L·atm/(mol·K) × (300 K - 900 K) = -1,518.91596 J

Eint = -1,518.91596 J

(g) ∑Q = 1,518.91596 J + 1668.69974 J - 2,531.5266 J = 656.089 J

∑W = 0 + 1668.69974 J -1,013.25 J = 655.449 J

∑Eint = 1,518.91596 J + 0 -1,518.91596 J = 0 J

5 0

3 years ago