What does it mean that a material transmits waves?

What single property was the most important in jesseca's material

jek_recluse [69]

Answer:

Jesseca wanted to create a material that reflected most of the light that fell on it.

Explanation: The Graphite was the material in the passage that had reflected most of the light.

6 0

3 years ago

Una partícula se mueve en el plano XY efectúa un desplazamiento mientras actúa sobre ella una fuerza constante. X= (4i + 3j) m,

dsp73

Answer:

a) La magnitud del desplazamiento es de 5 m

La magnitud de la fuerza es 20 N

b) El trabajo realizado por la fuerza es de 100 J

c) El ángulo entre la fuerza y el plano es 0 °

Explanation:

a) La magnitud del desplazamiento se encuentra por la relación;

$\left | X \right | = \sqrt{X_{x}^{2}+X_{y}^{2}}$

Lo que da;

$\left | X \right | = \sqrt{4^{2}+3^{2}} = 5 \ m$

De manera similar, la magnitud de la fuerza, F, se encuentra como sigue;

$\left | F \right | = \sqrt{F_{x}^{2}+F_{y}^{2}}$

Lo que da;

$\left | F \right | = \sqrt{16^{2}+12^{2}} = 20 \ N$

b) El trabajo, W, realizado por la fuerza = Fuerza, F × Distancia, X

∴ Ancho = 20 N × 5 m = 100 N · m = 100 J

c) La dirección de la fuerza viene dada por la siguiente fórmula;

$tan^{-1} \left (\dfrac{F_y}{F_x} \right ) = tan^{-1} \left (\dfrac{12}{16} \right ) = 38.9^{\circ}$

La dirección del plano viene dada por la siguiente fórmula;

$tan^{-1} \left (\dfrac{X_y}{X_x} \right ) = tan^{-1} \left (\dfrac{3}{4} \right ) = 38.9^{\circ}$

Por tanto, el ángulo entre la fuerza y el plano = 0 °

La fuerza actúa a lo largo del plano.

6 0

2 years ago

A swimmer of mass 64.38 kg is initially standing still at one end of a log of mass 237 kg which is floating at rest in water. He

nikklg [1K]

Answer:

0.9432 m/s

Explanation:

We are given;

Mass of swimmer;m_s = 64.38 kg

Mass of log; m_l = 237 kg

Velocity of swimmer; v_s = 3.472 m/s

Now, if we consider the first log and the swimmer as our system, then the force between the swimmer and the log and the log and the swimmer are internal forces. Thus, there are no external forces and therefore momentum must be conserved.

So;

Initial momentum = final momentum

m_l × v_l = m_s × v_s

Where v_l is speed of the log relative to water

Making v_l the subject, we have;

v_l = (m_s × v_s)/m_l

Plugging in the relevant values, we have;

v_l = (64.38 × 3.472)/237

v_l = 0.9432 m/s

8 0

2 years ago

The maximum allowable resistance for an underwater cable is one hundredth of an ohm per

Studentka2010 [4]

Answer:

A = 1.54 x 10⁻⁵ m² = 15.4 mm²

Explanation:

The resistance of a wire can be given by the following formula:

$R = \frac{\rho L}{A}\\\\A = \frac{\rho L}{R} = \frac{\rho}{\frac{R}{L} }$

where,

A = smallest cross-sectional area = ?

ρ = resistivity of copper = 1.54 x 10⁻⁸ Ωm

$\frac{R}{L}$ = resistance per unit length of wire = 0.001 Ω/m

Therefore,

$A = \frac{1.54\ x\ 10^{-8}\ \Omega m}{0.001\ \Omega/m}$

6 0

2 years ago

The standard molar heat of vaporization for water is 40.79 kj/mol. How much energy would be required to vaporize 4.72 x 10^10 mo

stellarik [79]

(5.81 x 10^10 molecules) / (6.022 x 10^23 molecules/mol) x (40.79 kj/mol) = 3.94 x 10^-12 kj

7 0

3 years ago