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Dima020 [189]
3 years ago
11

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

You might be interested in
The rate at which a force displaces a mass a horizontal distance is measured in?
kodGreya [7K]

Answer:

c. joules

Explanation:

The rate at which a force displaces a mass a horizontal distance is measured in joules. In science, this phenomenon is known as work done.

Work done can be defined as the rate at which a force acting on an object or a body causes it to experience a displacement. The work done is a scalar quantity and is measured in joules (J).

Mathematically, work done is given by the formula;

Work done = force * distance

W = F * d

Where,

  • W is the work done
  • F represents the force acting on a body.
  • d represents the distance covered by the body.

<em>For example, a bull pulling a plough through a farm, a girl pushing a shopping cart down the aisle of a supermarket etc. </em>

6 0
3 years ago
A spherical shell of radius 3.59 cm and a cylinder of radius 7.22 cm are rolling without slipping along the same floor. The two
vampirchik [111]

Answer:

(ω₁ / ω₂) = 1.9079

Explanation:

Given

R₁ = 3.59 cm

R₂ = 7.22 cm

m₁ = m₂ = m

K₁ = K₂

We know that

K₁ = Kt₁ + Kr₁ = 0.5*m₁*v₁²+0.5*I₁*ω₁²

if

v₁ = ω₁*R₁

and

I₁ = (2/3)*m₁*R₁² = (2/3)*m*R₁²

∴    K₁ = 0.5*m*ω₁²*R₁²+0.5*(2/3)*m*R₁²*ω₁²   <em>(I)</em>

then

K₂ = Kt₂ + Kr₂ = 0.5*m₂*v₂²+0.5*I₂*ω₂²

if

v₂ = ω₂*R₂

and

I₂ = 0.5*m₂*R₂² = 0.5*m*R₂²

∴    K₂ = 0.5*m*ω₂²*R₂²+0.5*(0.5*m*R₂²)*ω₂²   <em>(II)</em>

<em>∵   </em>K₁ = K₂    

⇒   0.5*m*ω₁²*R₁²+0.5*(2/3)*m*R₁²*ω₁² = 0.5*m*ω₂²*R₂²+0.5*(0.5*m*R₂²)*ω₂²

⇒  ω₁²*R₁²+(2/3)*R₁²*ω₁² = ω₂²*R₂²+0.5*R₂²*ω₂²

⇒  (5/3)*ω₁²*R₁² = (3/2)*ω₂²*R₂²

⇒  (ω₁ / ω₂)² = (3/2)*R₂² / ((5/3)*R₁²)

⇒  (ω₁ / ω₂)² = (9/10)*(7.22/ 3.59)²

⇒  (ω₁ / ω₂) = (7.22/ 3.59)√(9/10)

⇒  (ω₁ / ω₂) = 1.9079

8 0
3 years ago
If a sample emits 2000 counts per second when the detector is 1 meter from the sample, how many counts per second would be obser
Alona [7]

Answer:

<h2><em>6000 counts per second</em></h2>

Explanation:

If a sample emits 2000 counts per second when the detector is 1 meter from the sample, then;

2000 counts per second = 1 meter ... 1

In order to know the number of counts per second that would be observed when the detector is 3 meters from the sample, we will have;

x count per second = 3 meter ... 2

Solving the two expressions simultaneously for x we will have;

2000 counts per second = 1 meter

x counts per second = 3 meter

Cross multiply to get x

2000 * 3 = 1* x

6000 = x

<em></em>

<em>This shows that 6000 counts per second would be observed when the detector is 3 meters from the sample</em>

5 0
3 years ago
This picture represents the electric field diagram between two particles with static charges. Do the two particles have the same
dexar [7]

Answers:

No, They will attract each other, B, and neither direction

Explanation:

Since the two already presented particles in the diagram represent both opposing charges due to the direction of the arrows (the arrows facing away from the particle shows a positive charge and the particles facing towards the particle show a negative charge), not only because of this but as the arrows between the particles show an attracting magnetic field, then it can be concluded that the particles will attract to each other and if another particle was introduced into the diagram of a positive charge, then it would attract to the negatively charged particle. If you have any questions or need further explanation, please comment below. E2021, have a great day.

7 0
3 years ago
The radius of the aorta is about 1 cm and the blood flowing through it has a speed of about 30 cm/s. Calculate the average speed
puteri [66]

Answer:

The average speed of the blood in the capillaries is 0.047 cm/s.

Explanation:

Given;

radius of the aorta, r₁ = 1 cm

speed of blood, v₁ = 30 cm/s

Area of the aorta, A₁ = πr₁² = π(1)² = 3.142 cm²

Area of the capillaries, A₂ = 2000 cm²

let the average speed of the blood in the capillaries = v₂

Apply continuity equation to determine the average speed of the blood in the capillaries.

A₁v₁ = A₂v₂

v₂ = (A₁v₁) / (A₂)

v₂ = (3.142 x 30) / (2000)

v₂ = 0.047 cm/s

Therefore, the average speed of the blood in the capillaries is 0.047 cm/s.

4 0
2 years ago
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