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4 years ago

7

A foot is 12 inches and a mile is 5280 ft, exactly. A centimeter is exactly 0.01m or mm. Sammy is 5 feet and 5.3 inches tall. Wh

at is Sammy's Height in inches?

1 answer:

bixtya [17]4 years ago

4 0

The answer is 65.3 inches tall

Explanation:

To know the heigh of Sammy in inches it is necessary to convert the 5 feet to inches and add this number to 5.3 inches as the statement mentions "Sammy is 5 feet and 5.3 inches tall". Additionally, it is known each foot is equal to 12 inches ( 1 foot = 12 inches). According to this, the simplest method to convert feet to inches is to multiply the feet given by 12. The process is shown below:

1 foot = 12 inches

feet to inches = number of feet x 12

5 feet x 12 = 60 inches

This means 5 feet is equal to 60 inches. Now, 60 inches + 5.3 inches = 65.3 inches (total height of Sammy in inches)

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Read 2 more answers

A horizontal clothesline is tied between 2 poles, 12 meters apart. When a mass of 1 kilograms is tied to the middle of the cloth

nadya68 [22]

Answer:

The tension on the clotheslines is $T = 8.83 \ N$

Explanation:

The diagram illustrating this question is shown on the first uploaded image

From the question we are told that

The distance between the two poles is $d = 12 \ m$

The mass tie to the middle of the clotheslines $m = 1 \ kg$

The length at which the clotheslines sags is $l = 4 \ m$

Generally the weight due to gravity at the middle of the clotheslines is mathematically represented as

$W = mg$

let the angle which the tension on the clotheslines makes with the horizontal be $\theta$ which mathematically evaluated using the SOHCAHTOA as follows

$Tan \theta = \frac{ 4}{6}$

=> $\theta = tan^{-1}[\frac{4}{6} ]$

=> $\theta = 33.70^o$

So the vertical component of this tension is mathematically represented a

$T_y = 2* Tsin \theta$

Now at equilibrium the net horizontal force is zero which implies that

$T_y - mg = 0$

=> $T sin \theta - mg = 0$

substituting values

$T = \frac{m*g}{sin (\theta )}$

substituting values

$T = \frac{1 *9.8}{2 * sin (33.70 )}$

$T = 8.83 \ N$

6 0

3 years ago

You hang a heavy ball with a mass of 10 kg from a gold wire 2.6 m long that is 1.6 mm in diameter. You measure the stretch of th

PolarNik [594]

<u>Answer:</u> The Young's modulus for the wire is $6.378\times 10^{10}N/m^2$

<u>Explanation:</u>

Young's Modulus is defined as the ratio of stress acting on a substance to the amount of strain produced.

The equation representing Young's Modulus is:

$Y=\frac{F/A}{\Delta l/l}=\frac{Fl}{A\Delta l}$

where,

Y = Young's Modulus

F = force exerted by the weight = $m\times g$

m = mass of the ball = 10 kg

g = acceleration due to gravity = $9.81m/s^2$

l = length of wire = 2.6 m

A = area of cross section = $\pi r^2$

r = radius of the wire = $\frac{d}{2}=\frac{1.6mm}{2}=0.8mm=8\times 10^{-4}m$ (Conversion factor: 1 m = 1000 mm)

$\Delta l$ = change in length = 1.99 mm = $1.99\times 10^{-3}m$

Putting values in above equation, we get:

$Y=\frac{10\times 9.81\times 2.6}{(3.14\times (8\times 10^{-4})^2)\times 1.99\times 10^{-3}}\\\\Y=6.378\times 10^{10}N/m^2$

Hence, the Young's modulus for the wire is $6.378\times 10^{10}N/m^2$

3 0

4 years ago

An iron wire has length 8.0m and a diameter 0.50mm. The sir has a resistance R.

Rudik [331]

The re<span>sistance of the second wire is 16 R.
where R is the resistance of the first wire.

R = </span>ρ $\frac{l}{A}$
where l = length of the wire
A = area of the wire
A = $\pi r^{2}$ where, r = $\frac{diameter of wire}{2}$

Thus, on finding the ratio of resistance of the two wires, we get,

$\frac{R1}{R2} = \frac{l1A2}{l2A1}$

here, R1 = R
l1 = 8m
l2 = 2m
A1=π $0.25^{2}$
A1=π $0.50^{2}$

we get. R2 = 16R

7 0

4 years ago