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3 years ago

A furnace is rated 4000 BTU per hour. How many foot pounds of energy are released in one hour?

1 answer:

4 0

To solve this we are going to use a conversion factor. We know that 1 BTU per hour is approximately 778 foot-pounds of energy per hour, so our conversion factor will be: $\frac{778foot-pound}{1BTU}$ .

Now we just need to multiply 4000 BTU by our conversion factor:
 $4000BTU*\frac{778foot-pound}{1BTU}=3112000foot-pound$

We can conclude that the furnace released 3,112,000 foot pounds of energy in one hour.

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Write a linear function f with f(0) = 7 and f(3) =1.

Kipish [7]

(0,7) and (3,1) are points on the line.

Slope of line passing through (x₁, y₁) and (x₂, y₂) = (y₂-y₁)/(x₂-x₁)

Slope of line passing through (0, 7) and (3, 1) = (1-7)/(3-0) = -2

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3 years ago

ANOTHER FILL IN THE BLANKS QUESTION, MARKING BRAINLIEST FOR CORRECT ANSWER

Mamont248 [21]

Hi, again!
First blank: multiplication property of equality
Both sides of the equation were multiplied by (1/2).

Second blank: m<E
Angle E is an inscribed angle intercepting arc FGD, so by the inscribed angle theorem, its measure is half the measure of the arc.

Third blank: m<G
Same explanation as the prior blank

Fourth blank: substitution property of equality
Since angle E is equal to half the measure of arc FGD, we can replace that term. Likewise with angle G.

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3 years ago

A spherical balloon currently has a radius of 19cm. If the radius is still growing at a rate of 5cm or minute, at what rate is a

miv72 [106K]

Answer:

22670.8 cm³/min

Step-by-step explanation:

Given:

Radius of the balloon at a certain time (r) = 19 cm

Rate of growth of radius is, $\frac{dr}{dt}=5\ cm/min$

The rate at which the air is pumped in the balloon can be calculated by finding the rate of increase in the volume of the balloon.

So, first we find the volume of the sphere in terms of 'r'. As the balloon is spherical in shape, the volume of the balloon is equal to the volume of a sphere. Therefore,

Volume of balloon is given as:

$V=\frac{4}{3}\pi r^3$

Now, rate of increase of volume is obtained by differentiating both sides of the equation with respect to time 't'.

Differentiating both sides with respect to time 't', we get:

$\frac{dV}{dt}=\frac{d}{dt}(\frac{4}{3}\pi r^3)\\\\\frac{dV}{dt}=\frac{4\pi}{3}(3r^2)(\frac{dr}{dt})\\\\\frac{dV}{dt}=4\pi r^2(\frac{dr}{dt})$

Now, plug in 19 cm for 'r', 5 cm per minute for $\frac{dr}{dt}$ and solve for $\frac{dV}{dt}$ . This gives,

$\frac{dV}{dt}=4\pi (19 cm)^2(5\ cm/min)\\\\\frac{dV}{dt}=4\times 3.14\times 361\times 5\ cm^3/min\\\\\frac{dV}{dt}=22670.8\ cm^3/min$

Therefore, the rate at which the air is being pumped into the balloon is 22670.8 cm³/min.

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4 years ago

Josiah skateboarded from his house to school 4 miles away. It takes him 20 minutes. a. If Josiah's speed, in miles per minute, i

blondinia [14]

Answer: 35 miles

Step-by-step explanation:

If 4 miles is 20 minutes that mean 1 mile is 5 minutes so just add 4+3 which is 7 then multiply that by 5 which is 35 so it takes him 35 minutes to go to his friends house

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3 years ago

Please help me! Wil give brainliest!

Mashcka [7]

Answer:

Help With what?

Step-by-step explanation:

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3 years ago