Ask question

Ask question

All categories

4 years ago

12

If a resident accidentally causes a chunk of ice to fall from the roof, how fast would the ice be moving when it hits the ground

? Neglect air resistance.

1 answer:

Elena L [17]4 years ago

5 0

Answer:

v = √2gh

v = 7 m / s

Explanation:

For this exercise we can use the law of conservation of energy

starting point. In the roof

Em₀ = U = m g h

final point. On the floor

Em_f = K = ½ m v²

as there is no friction, the energy is conserved

Em₀ = Em_f

m g h = ½ m v²

v = √2gh

In the exercise the height of the ceiling is not indicated, but in general the height of the ceiling of a house is h = 2.5 m

v = √ (2 9.8 2.5)

v = 7 m / s

You might be interested in

The temperature of a smelting furnace is found to be 2000 ℃. Find the

IRISSAK [1]

Answer:

a) The equivalent temperature of 2000 ºC is 3632 ºF.

b) The equivalent temperature of 2000 ºC is 4091.67 R.

c) The equivalent temperature of 2000 ºC is 2273.15 K.

d) The equivalent temperature of 2000 ºC is 1600 ºRe.

Explanation:

a) The equivalent temperature on the Fahrenheit scale is defined by the following formula:

$T_{F} = \frac{9}{5}\cdot T_{C}+32$ (Eq. 1)

Where:

$T_{C}$ - Temperature, measured in degrees Celsius.

$T_{F}$ - Temperature, measured in degrees Fahrenheit.

If we know that $T_{C} = 2000\,^{\circ}C$ , then the temperature is:

$T_{F} = \frac{9}{5}\cdot (2000\,^{\circ}C)+32\,^{\circ}F$

$T_{F} = 3632\,^{\circ}F$

The equivalent temperature of 2000 ºC is 3632 ºF.

b) From result in a) we determine the equivalent temperature on the Rankine scale by using the following formula:

$T_{R} = T_{F}+459.67$ (Eq. 2)

Where:

$T_{F}$ - Temperature, measured in degrees Fahrenheit.

$T_{R}$ - Temperature, measured in Rankine.

If we know that $T_{F} = 3632\,^{\circ}F$ , then the temperature is:

$T_{R} = 3632\,^{\circ}F+459.67\,R$

$T_{R} = 4091.67\,R$

The equivalent temperature of 2000 ºC is 4091.67 R.

c) The equivalent temperature on the absolute scale is calculated by using this expression:

$T_{K} = T_{C}+273.15$ (Eq. 3)

Where:

$T_{C}$ - Temperature, measured in degrees Celsius.

$T_{K}$ - Temperature, measured in Kelvin.

If we know that $T_{C} = 2000\,^{\circ}C$ , then the temperature is:

$T_{K} = 2000\,^{\circ}C+273.15\,K$

$T_{K} = 2273.15\,K$

The equivalent temperature of 2000 ºC is 2273.15 K.

d) The equivalent temperature on the Réaumur scale is calculated by applying this expression:

$T_{Re} = \frac{4}{5}\cdot T_{C}$ (Eq. 4)

Where:

$T_{C}$ - Temperature, measured in degrees Celsius.

$T_{Re}$ - Temperature, measured in degrees Réaumur.

If we know that $T_{C} = 2000\,^{\circ}C$ , then the temperature is:

$T_{Re} = \frac{4}{5}\cdot (2000\,^{\circ}C)$

$T_{Re} = 1600\,^{\circ}Re$

The equivalent temperature of 2000 ºC is 1600 ºRe.

5 0

3 years ago

A diode has a power rating of 5W. If the diode voltage is 3.6V and the diode current is 1.75A, what is the power dissipation?

son4ous [18]

Answer:

Explanation:

Power = current * voltage

Power = 3.6 * 1.75

Power = 6.3 watts.

I hope this diode does not have to do this for very long. The power dissipation is above the diode's rated power -- a condition that won't last forever.

The power dissipation is 6.3 watts.

4 0

3 years ago

Here is the image again

Komok [63]

I matched the colors, but make sure you double check.

3 0

3 years ago

Explain the difference between the synodic period of a planet and the sidereal period. why is the synodic period often much less

Ilia_Sergeevich [38]

-Synodic period is the period of celestial bodies observed on the moving planet(mostly earth)
Sideral period is the period comparing to the fixed stars without motion of the earth involved.

(I will explain the second question with an example, so it's easier to understand)
-For Sideral month for example of the moon it cactually complete one revolution in around 27.3 days.
However, since the earth moves, for us it took some more time to see the moon the same as before (fullmoon to fullmoon) again. That make synodic month of the moon to be around 29.5 days.

4 0

4 years ago

A hair dryer draws 14.5 a when plugged into a 120-v line. assume direct current.

AVprozaik [17]

Missing part in the exercise. Complete text:
"A hair dryer draws 14.5 A when plugged into a 120-V line. Assume direct current.

(a) What is its resistance? Express your answer to two significant figures and include the appropriate units."

(b) How much charge passes through it in 11 min ? Express your answer to two significant figures and include the appropriate units.

Solution

(a) We can calculate the resistance by using Ohm's law:

$\Delta V= I R$

Since we know that $\Delta V=120 V$ and $I=14.5 A$ , we can find R:

$R= \frac{\Delta V}{I} = \frac{120 V}{14.5 A}=8.3 \Omega$

(b) The current is equal to the quantity of charge by unit of time:

$I= \frac{Q}{t}$

In our problem, t=11min=660 s, so we can calculate the charge passed through the hair dryer:

$Q=It=14.5 A \cdot 660 s = 9570 C$

5 0

3 years ago