For an object to be seen, light must leave _______ and enter ______

How do Ohm's Law relate current, voltage difference, and resistance?

Snowcat [4.5K]

The relationship between voltage, current, and resistance is described by Ohm's law. The equation, i = v/r, tells us that the current, i, flowing through a circuit is directly proportional to the voltage, v, and inversely proportional to the resistance, r.

6 0

3 years ago

When must batteries in an emergency locator transmitter (ELT) be replaced or recharged, if re-chargeable?

tatiyna

Answer:

After one half of the battery's useful life.

Explanation:

Batteries of the emergency locator transmitter (ELT) must be replaced or recharged after one half of the battery's useful life because if it is exposed to the high temperature for a long period of time such as the air plane parked in the sun will result in the deterioration of battery which may makes the transmitter out of order before the expiry date of the battery. So it will be safe to do that after the use of one half of the battery's life.

6 0

3 years ago

What happens when a body of cold air approaches a body of warm air

inn [45]

The warm air gets pushed up, this also could cause rain, and possibly high winds that can cause a tornado to form

7 0

3 years ago

Se deja caer un objeto desde la cornisa de una casa la cual tarda en llegar al suelo 645/500 segundos. ¿Cúal es la altura de la

salantis [7]

8.15m. La altura de la casa de la cual se deja caer un objeto que tarda en caer $\frac{645}{500}s$ es de 8.15m.

La clave para resolver este problema es mediante caída libre, la velocidad inicial del objeto es 0 por lo que podemos calcular la altura h de la casa mediante la relación $h= \frac{gt^{2}}{2}$ .

Conocemos la gravedad $g=9.8\frac{m}{s^{2}}$ y el tiempo en que tarda en caer el objeto $t=\frac{645}{500}s$ , sustituyendo los valores tenemos que:

$h=\frac{(9.8\frac{m}{s^{2}})(\frac{645}{500}s)^{2}}{2}=8.15m$

4 0

3 years ago

A book rest on a table which it's face having a sides 30cm by 25cm . if it exerts apressure of 200pa then determine the mass of

Simora [160]

Answer:

Approximately $1.5\; \rm kg$ . (Assuming that this table is level, and that the gravitational field strength is $g = 9.8\; \rm N \cdot kg^{-1}$ .

Explanation:

Convert the dimensions of this book to standard units:

$\displaystyle 30\; \rm cm = 30\; \rm cm \times \frac{1\; \rm m}{100\; \rm cm} = 0.30\; \rm m$ .

$\displaystyle 25\; \rm cm = 25\; \rm cm \times \frac{1\; \rm m}{100\; \rm cm} = 0.25\; \rm m$ .

Calculate the surface area of this book:

$0.30\; \rm m \times 0.25\; \rm m = 0.075\; \rm m^{2}$ .

Pressure is the ratio between normal force and the area over which this force is applied.

$\displaystyle \text{Pressure} = \frac{\text{normal Force}}{\text{contact Area}}$ .

Equivalently:

$(\text{normal Force}) = \text{Pressure} \cdot (\text{contact Area})$ .

In this question, $\text{Pressure} = 200\; \rm Pa = 200\; \rm N \cdot m^{-2}$ .

It was found that $(\text{contact Area}) = 0.075\; \rm m^{2}$ (assuming that the entire side of this book is in contact with the table.

Hence:

$\begin{aligned}& (\text{normal Force}) \\ &= \text{Pressure} \cdot (\text{contact Area}) \\ &= 200\; \rm N \cdot m^{-2} \times 0.075\; \rm m^{2} \\ &= 15\; \rm N \end{aligned}$ .

If that the table is level, this normal force would be equal to the weight of this book:

$\text{weight} = 15\; \rm N$ .

Assuming that the gravitational field strength is $g = 9.8\; \rm N \cdot kg^{-1}$ . The mass of this book would be:

$\begin{aligned}\text{mass} &= \frac{\text{weight}}{g} \\ &= \frac{15\; \rm N}{9.8\; \rm N \cdot kg^{-1}}\approx 1.5\; \rm kg\end{aligned}$ .

4 0

3 years ago

For an object to be seen, light must leave _______ and enter _______.

For an object to be seen, light must leave _ and enter _.