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3 years ago

Find the height of a cylinder of volume 200cm^3 and radius 4

Mathematics

1 answer:

liubo4ka [24]3 years ago

8 0

Answer:

3.98 cm

Step-by-step explanation:

V= πr²h

V= 200 cm³, r= 4 cm, h=?

h= V/(πr²)= 200/(3.14*4²)= 3.98 cm

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Convert 1 cal/(m^2 * sec * °C) into BTU/(ft^2 * hr * °F)

Crazy boy [7]

Answer:

$1\ \frac{\text{cal}}{m^2\times sec\times ^\circ C}=0.03926\frac{\text{BTU}}{ft^2\times hr\times ^\circ F}$

Step-by-step explanation:

To find : Convert $1\ \frac{\text{cal}}{m^2\times sec\times ^\circ C}$ into $\frac{\text{BTU}}{ft^2\times hr\times ^\circ F}$

Solution :

We convert units one by one,

$1\text{ m}^2=10.7639\text{ ft}^2$

$1\text{ sec}=\frac{1}{3600}\text{ hour}$

$1\text{ cal}=0.003968\text{ BTU}$

Converting temperature unit,

$^\circ C\times \frac{9}{5}+32=^\circ F$

$1^\circ C\times \frac{9}{5}+32=33.8^\circ F$

So, $1^\circ C=33.8^\circ F$

Substitute all the values in the unit conversion,

$1\ \frac{\text{cal}}{m^2\times sec\times ^\circ C}=\frac{0.003968}{10.7639\times \frac{1}{3600}\times 33.8}\frac{\text{BTU}}{ft^2\times hr\times ^\circ F}$

$1\ \frac{\text{cal}}{m^2\times sec\times ^\circ C}=\frac{0.003968}{0.101061}\frac{\text{BTU}}{ft^2\times hr\times ^\circ F}$

$1\ \frac{\text{cal}}{m^2\times sec\times ^\circ C}=0.03926\frac{\text{BTU}}{ft^2\times hr\times ^\circ F}$

Therefore, The conversion of unit is $1\ \frac{\text{cal}}{m^2\times sec\times ^\circ C}=0.03926\frac{\text{BTU}}{ft^2\times hr\times ^\circ F}$

3 0

3 years ago

Given rectangle QUAD, If DU = 5x - 4 and QP = 2x + 7, then find the length of DU.

ira [324]

Answer:

DU = 86

Step-by-step explanation:

Properties of a rectangle,

- Opposite sides are equal in measure.

- Diagonals of a rectangle are equal.

- Diagonals bisect each other.

In the given rectangle QUAD,

Diagonals AQ and DU bisect each other at P.

AQ = DU

2(QP) = DU [Since, AQ = 2(PQ)]

2(2x + 7) = 5x - 4

4x + 14 = 5x - 4

5x - 4x = 14 + 4

x = 18

Since, DU = 5x - 4

DU = 5(18) - 4

= 90 - 4

= 86

Therefore, measure of DU = 86 units

3 0

2 years ago

The experimental probability that Pedro will hit a hole in one at the golf course is 2/5. If Pedro goes golfing 40 times in a ye

Sati [7]

I'm pretty sure you would need to multiply 2/5 by 40.

So, if you multiply 2/5 by 40, you need to turn 40 into a fraction with 1 being the denominator.

2/5 x 40/1 = 80/5

Since the product is an improper fraction, you would simplify it to a whole number.

80/5 = 16.

He can expect to hit it 16 times to get a hole in one.

I hope I am right and have a great day.

6 0

3 years ago

Read 2 more answers

What is the least common multiple of 9 and 8

leonid [27]

Answer:

72 is the least common multiple

6 0

2 years ago

Read 2 more answers

If f(x) = integral of 1/ (sqrt of t^3 +2)dt, then f'1 = ?

mihalych1998 [28]

Differentiating an integral removes the integral.

f(x) = integral of dt/sqrt(t^3 + 2)
f'(x) = 1/sqrt(x^3 + 2)
f'(1) = 1/sqrt(1^3 + 2)
f'(1) = 1/sqrt(3) = sqrt(3)/3.

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!

4 0

3 years ago