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3 years ago

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A hollow cylindrical iron pipe with external and internal radii 8cm and 6cm respectively and length 35cm is melted and recast in

to a solid wure of thickness 2.8cm find the length of wire

1 answer:

marusya05 [52]3 years ago

3 0

Answer: length of wire = 500cm

Step-by-step explanation:

We would determine the volume of material used in making the hollow cylindrical iron pipe by subtracting the volume of the inner pipe from the volume of the outer pipe.

The formula for determining the volume of a cylinder is expressed as

Volume = πr²h

Volume of inner pipe = π × 6² × 35 = 1260π

Volume of outer pipe = π × 8² × 35 = 2240π

Volume of material used in making the hollow cylindrical iron pipe is

2240π - 1260π = 980πcm³

The solid wire made is cylindrical and has a thickness of 2.8cm. This is the diameter. It means that the radius of the wire is 2.8/2 = 1.4cm

Volume of wire = 980πcm³

Applying the volume of cylinder formula, it becomes

980π = π × 1.4² × h

where h = length of wire

h = 980/1.4² = 500cm

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3 years ago

Which of the following is true about the right triangle?

katrin2010 [14]

Option 1. $cos (a) = \frac{3}{5}$ , option 3. $sin (a) = \frac{4}{5}$ , and option 5. $tan (a) = \frac{4}{3}$ are true about the right triangle.

Step-by-step explanation:

Step 1:

According to the Pythagorean theorem, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Here the hypotenuse measures 15 units. Assume the other side measures x.

$15^{2} = x^{2} +9^{2} , x^{2} = 15^{2} -9^{2} .$

$x^{2} = 144, x= 12.$

Step 2:

$sin \theta = \frac{oppositeside}{hypotenuse} ,cos \theta = \frac{adjacentside}{hypotenuse} , tan \theta = \frac{oppositeside}{adjacentside}.$

For angle A, the opposite side is 12 units long, the adjacent side is 9 units long and the hypotenuse is 15 units long.

$sin (a) = \frac{12}{15} ,cos(a) = \frac{9}{15} , tan(a) = \frac{12}{9}.$

$sin (a) = \frac{4}{5} ,cos (a) = \frac{3}{5} , tan (a) = \frac{4}{3}.$

So option 1. $cos (a) = \frac{3}{5}$ , option 3. $sin (a) = \frac{4}{5}$ , and option 5. $tan (a) = \frac{4}{3}$ are true about the right triangle.

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3 years ago

What is 37/10 as a mixed number?

Dafna11 [192]

First, we need to find how many times 10 can fit into 37.

10*3 = 30

3 is the whole number.

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3 years ago

Read 2 more answers

A data set consists of the values 2, 6, 3, and 1. If we consider this a population (all the values available), the variance isA.

Hoochie [10]

Answer: E. none of the above

Step-by-step explanation:

The given data values that represents the population:

2, 6, 3, and 1.

Number of values : n=4

Mean of the data values = $\dfrac{\text{Sum of values}}{\text{No. of values}}$

$\dfrac{2+6+3+1}{4}=\dfrac{12}{4}=3$

Sum of the squares of the difference between each values and the mean =

$(2-3)^2+(6-3)^2+(3-3)^2+(1-3)^2$

$=-1^2+3^2+0^2+(-2)^2$

$=1+9+0+4=14$

Now , Variance = (Sum of the squares of the difference between each values and the mean ) ÷ (n)

= (14) ÷ (4)= 3.5

Hence, the variance is 3.5.

Therefore , the correct answer is "E. none of the above".

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3 years ago

sweet [91]

To model and solve our situation we are going to use the equation: $s= \frac{d}{t}$
where
$s$ is speed
$d$ is distance
$t$ is time

1. We know that the distance between the cities is 2400 miles, so $d=2400$ . We also know that the speed of the plane is 450 mi/h. Since we don't know the speed of the air, $S_{a}=?$ . We don't know how much the westward trip takes, so $t_{w}=?$ , and we also don't know how much the eastward trip takes, so $t_{e}=?$ .

Going westward. Here the plane is flying against the air, so we need to subtract the speed of the air from the speed of the plane:
$450-S_{a}= \frac{2400}{t_{w} }$
Going eastward. Here the plane is flying with the the air, so we need to add the speed of the air to the speed of the plane:
$450+S_{a}= \frac{2400}{t_{e} }$

2. We know for our problem that the round trip takes 11 hours; so the total time of the trip is 11, $t_{t}=11$ . Notice that we also know that the total time of the trip equals time of the tip going westward plus time of the trip going eastward, so $t_{t}=t_{w}+t_{e}$ . Since we know that the total trip takes 11 hours, we can replace that value in our total time equation and solve for $t_{w}$ :
$11=t_{w}+t_{e}$
$t_{w}=11-t_{e}$

Now we can replace $t_{w}$ in our going westward equation to model our round trip with a system of equations:
$450-S_{a}= \frac{2400}{t_{w}}$
$450-S_{a}= \frac{2400}{11-t_{e} }$ equation (1)
$450+S_{a}= \frac{2400}{t_{e}}$ equation (2)

3. To solve our system of equations, we are going to solve for $t_{e}$ in equations (1) (2):

From equation (1)
$450-S_{a}= \frac{2400}{11-t_{e} }$
$11-t_{e}= \frac{2400}{450-S_{a} }$
$-t_{e}= \frac{2400}{450-S_{a} } -11$
$t_{e}=11- \frac{2400}{450-S_{a} }$
$t_{e}= \frac{4950-11S_{a} -2400}{450-S_{a} }$
$t_{e}= \frac{2550-11S_{a} }{450-S_{a} }$ equation (3)

From equation (2):
$450+S_{a}= \frac{2400}{t_{e} }$
$t_{e}= \frac{2400}{450+S_{a} }$ equation (4)

Replacing (4) in (3)
$\frac{2400}{450+S_{a}} = \frac{2550-11S_{a}}{450-S_{a} }$
Now, we can solve for $S_{a}$ to find the speed of the wind:
$2400(450-S_{a})=(450+S_{a})(2550-11S_{a})$
$1080000-2400S_{a}=1147500-4950S_{a}+2550S_{a}-11(S_{a})^{2}$
$11(S_{a})^{2}-67500=0$
$11(S_{a})^{2}=67500$
$(S_{a})^{2}= \frac{67500}{11}$
$S_{a}=+/- \sqrt{ \frac{67500}{11} }$
Since speed cannot be negative, the solution of our equation is:
$S_{a}= \sqrt{ \frac{67500}{11} }$
$S_{a}=78.33$

We can conclude that the speed of the wind is 78 mph.

3 0

4 years ago