All categories

3 years ago

The edge of a cube whose volume is 1728 cm 3

Mathematics

1 answer:

mamaluj [8]3 years ago

3 0

Answer:

12 cm

Step-by-step explanation:

This would be the cubic root of 1728

= 12 cm

( 1729 is known as Ramanajan's number. Look it up - it's interesting)

You might be interested in

Find the value of both variables.

ella [17]

$\cos(45) = \frac{5 \sqrt{2} }{x} \\ \frac{1}{ \sqrt{2} } = \frac{5 \sqrt{2} }{x} \\ x = 10 \\ \\ \tan(45) = \frac{y}{5 \sqrt{2} } \\ 1 = \frac{y}{5 \sqrt{2} } \\ y = 5 \sqrt{2}$

I hope I helped you ^_^

8 0

3 years ago

A 1/17th scale model of a new hybrid car is tested in a wind tunnel at the same Reynolds number as that of the full-scale protot

Olegator [25]

Answer:

The ratio of the drag coefficients $\dfrac{F_m}{F_p}$ is approximately 0.0002

Step-by-step explanation:

The given Reynolds number of the model = The Reynolds number of the prototype

The drag coefficient of the model, $c_{m}$ = The drag coefficient of the prototype, $c_{p}$

The medium of the test for the model, $\rho_m$ = The medium of the test for the prototype, $\rho_p$

The drag force is given as follows;

$F_D = C_D \times A \times \dfrac{\rho \cdot V^2}{2}$

We have;

$L_p = \dfrac{\rho _p}{\rho _m} \times \left(\dfrac{V_p}{V_m} \right)^2 \times \left(\dfrac{c_p}{c_m} \right)^2 \times L_m$

Therefore;

$\dfrac{L_p}{L_m} = \dfrac{\rho _p}{\rho _m} \times \left(\dfrac{V_p}{V_m} \right)^2 \times \left(\dfrac{c_p}{c_m} \right)^2$

$\dfrac{L_p}{L_m} =\dfrac{17}{1}$

$\therefore \dfrac{L_p}{L_m} = \dfrac{17}{1} =\dfrac{\rho _p}{\rho _p} \times \left(\dfrac{V_p}{V_m} \right)^2 \times \left(\dfrac{c_p}{c_p} \right)^2 = \left(\dfrac{V_p}{V_m} \right)^2$

$\dfrac{17}{1} = \left(\dfrac{V_p}{V_m} \right)^2$

$\dfrac{F_p}{F_m} = \dfrac{c_p \times A_p \times \dfrac{\rho_p \cdot V_p^2}{2}}{c_m \times A_m \times \dfrac{\rho_m \cdot V_m^2}{2}} = \dfrac{A_p}{A_m} \times \dfrac{V_p^2}{V_m^2}$

$\dfrac{A_m}{A_p} = \left( \dfrac{1}{17} \right)^2$

$\dfrac{F_p}{F_m} = \dfrac{A_p}{A_m} \times \dfrac{V_p^2}{V_m^2}= \left (\dfrac{17}{1} \right)^2 \times \left( \left\dfrac{17}{1} \right) = 17^3$

$\dfrac{F_m}{F_p} = \left( \left\dfrac{1}{17} \right)^3$ = (1/17)^3 ≈ 0.0002

The ratio of the drag coefficients $\dfrac{F_m}{F_p}$ ≈ 0.0002.

5 0

3 years ago

Can x and y intercepts be in decimals or fractions?

Alenkinab [10]

Yes it can be in fractions and in decimals

8 0

3 years ago

Read 2 more answers

The kinetic energy (K) that an object has varies jointly with its mass (m) and the square of its velocity (v). The equation that

lord [1]

The kinetic energy of the bowling ball with the mass and traveling at the given velocity is 10.14 Joules.

<h3>What is Kinetic Energy?</h3>

Kinetic energy is simply a form of energy a particle or object possesses due to its motion.

It is expressed as;

K = (1/2)mv²

Where m is mass of the object and v is its velocity.

Given that;

Mass of the bowling ball m = 3kg

Velocity of the bowling ball v = 2.6m/s

Kinetic energy K = ?

We substitute the given values into the above equation.

K = (1/2)mv²

K = 0.5 × 3kg × (2.6m/s)²

K = 0.5 × 3kg × 6.76m²/s²

K = 10.14kgm²/s²

K = 10.14J

Therefore, the kinetic energy of the bowling ball with the mass and traveling at the given velocity is 10.14 Joules.

Learn more about kinetic energy here: brainly.com/question/12669551

#SPJ1

6 0

2 years ago

What additional information could you use to show that ΔSTU ≅ ΔVTU using SAS? Check all that apply.

Ugo [173]

I believe the correct answers are:

<span>UV = 14 ft and m∠TUV = 45°</span>
<span>ST = 20 ft, UV = 14 ft, and m∠UST = 98°

Or, in other words, Options A and D.
</span>

6 0

4 years ago

Read 2 more answers