All categories

3 years ago

What is the definition of Cubit

Physics

1 answer:

disa [49]3 years ago

7 0

Answer:

See below.

Explanation:

Cubit is a unit of length based on the length of the forearm from the elbow to the tip of the middle finger and usually equal to about 18 inches (46 centimeters).

It is an ancient unit of length used in ancient Egypt and is also known as "ancient Egyptian royal cubit."

Other similar units of measurements are displayed in the image below. Thanks!

You might be interested in

What is gibbs paradox?

mr_godi [17]

This leads to a paradox known as the Gibbs paradox, after Josiah Willard Gibbs. The paradox allows for the entropy of closed systems to decrease, violating the second law of thermodynamics. A related paradox is the "mixing paradox".

7 0

3 years ago

(a) At what height above Earth’s surface is the energy required to lift a satellite to that height equal to the kinetic energy r

Nikitich [7]

Answer:

Explanation:

Gravitational Potential Energy at earth surface $U_1=\frac{GM_em}{R_e}$

Gravitational Potential Energy at height h is $U_2=\frac{GM_em}{R_e+h}$

Energy required to lift the satellite $E_1=U_1-U_2$

$E_1=\frac{GM_em}{R_e}-\frac{GM_em}{R_e+h}$

Now Energy required to orbit around the earth

$E_2=\frac{1}{2}mv_{orbit}^2=\frac{GM_2m}{2(R_e+h)}$

$\Delta E=E_1-E_2$

$\Delta E=\frac{GM_em}{R_e}-\frac{GM_em}{R_e+h}-\frac{GM_2m}{2(R_e+h)}$

$E_1=E_2$ (given)

$\frac{GM_em}{R_e}-\frac{GM_em}{R_e+h}-\frac{GM_2m}{2(R_e+h)}=0$

$\frac{1}{R_e}-\frac{3}{2(R_e+h)}=0$

$h=\frac{R_e}{2}$

$h=3.19\times 10^6\ m$

(b)For greater height $E_1$ is greater than $E_2$

thus energy to lift the satellite is more than orbiting around earth

4 0

3 years ago

Sapting Leng You have a string with a mass of 13.7 q. You stretch the string with a force of 8.39 N, giving it a length of 1.87

marshall27 [118]

Answer:

a) $\lambda=0.935\ \textup{m}$

b) $f=36.19\approx 36\ \textup{Hz}$

Explanation:

Given:

String vibrates transversely fourth dynamic, thus n = 4

mass of the string, m = 13.7 g = 13.7 × 10⁻¹³ kg

Tension in the string, T = 8.39 N

Length of the string, L = 1.87 m

a) we know

$L= n\frac{\lambda}{2}$

where,

$\lambda$ = wavelength

on substituting the values, we get

$1.87= 4\times \frac{\lambda}{2}$

$\lambda=0.935\ \textup{m}$

b) Speed of the wave (v) in the string is given as:

$v =f\lambda$

also,

$v=\sqrt\frac{T}{(\frac{m}{L})}$

equating both the formula for 'v' we get,

$f\lambda=\sqrt\frac{T}{(\frac{m}{L})}$

on substituting the values, we get

$f\times 0.935=\sqrt\frac{8.39}{(\frac{13.7\times 10^{3}}{1.87})}$

$f=\frac{33.84}{0.935}$

$f=36.19\approx 36\ \textup{Hz}$

5 0

3 years ago

A truck initially traveling at a speed of 22 meters per second increases speed at a constant rate of 2.4 meters per second^2 for

Usimov [2.4K]

Thank you for posting your question here. The total distance traveled by the truck during the 3.2 seconds interval is 83 m. Below is the solution:

d = vit + 1/2 at^2
d = (22m/ s) (3.2s) + 1/2 (2.4m/ s^2) (3.2s)^2
d = 83 m
Hope the answer helps.

8 0

3 years ago

In an experiment, a variable, position-dependent force FC) is exerted on a block of mass 1.0 kg that is moving on a horizontal s

marshall27 [118]

Answer:

C) The function F(x) for 0 < x < 5, the block's initial velocity, and the value of Fr.

Explanation:

Yo want to prove the following equation:

$W_N=\Delta K\\\\$

That is, the net force exerted on an object is equal to the change in the kinetic energy of the object.

The previous equation is also equal to:

$F(x)x-F_f=\frac{1}{2}m(v_f^2-v_o^2)$ (1)

m: mass of the block

vf: final velocity

v_o: initial velocity

Ff: friction force

F(x): Force

x: distance

You know the values of vf, m and x.

In order to prove the equation (1) it is necessary that you have C The function F(x) for 0 < x < 5, the block's initial velocity, and the value of F. Thus you can calculate experimentally both sides of the equation.

8 0

3 years ago