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4 years ago

When all parts of a circuit are composed of conducting materials, the circuit is said to be

1 answer:

4 0

When all parts of a circuit ... all the way from one terminal of the
battery or power supply to the other terminal ... are composed of
conducting materials, then current can flow in the circuit, and the
circuit is said to be 'closed'.

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do clouds and wind appear to follow the same patterns? Can you find any patterns in the direction that they move?

aniked [119]

What scientists use to make models of the Earth's water cycle so they can<span> see how it is ... Where </span>does<span> the water that we use to meet our everyday needs come from? .... what </span>you<span> notice about the </span>patterns<span> the</span>winds<span> and </span>clouds follow<span>: </span>Do clouds and<span> ... </span>same patterns<span>? </span>Can you find any patterns in the direction that they move? Precipitation is a vital component of how water moves through Earth’s water cycle, connecting the ocean, land and atmosphere. Water evaporates from the surface of the land and oceans, rises and cools, condenses into rain or snow, and falls again to the surface as precipitation. The water falling on land collects in rivers and lakes, soil, and porous layers of rock, and much of it flows back into the oceans. The cycling of water in and out of the atmosphere is a significant aspect of the weather patterns on Earth. so that will be probs the best i can do

6 0

4 years ago

Drops of rain fall perpendicular to the roof of a parked car during a rainstorm. The drops strike the roof with a speed of 15 m/

IrinaK [193]

Answer: (a) 1.065 N (b) 2.13 N

Explanation:

<h2>(a) average force exerted by the rain on the roof</h2><h2 />

According Newton's 2nd Law of Motion the force $F$ is defined as <u>the variation of linear momentum</u> $p$ <u>in time:</u>

$F=\frac{dp}{dt}$ (1)

Where the linear momentum is:

$p=mV$ (2) Being $m$ the mass and $V$ the velocity.

In the case of the rain drops, which initial velocity is $V_{i}=15m/s$ and final velocity is $V_{f}=0$ (we are told the drops come to rest after striking the roof). The momentum of the drops $p_{drops}$ is:

$p_{drops}=mV_{i}+mV_{f}$ (3)

If $V_{f}=0$ , then:

$p_{drops}=mV_{i}$ (4)

Now the force $F_{drops}$ exerted by the drops is:

$F_{drops}=\frac{dp_{drops}}{dt}=\frac{d}{dt}mV_{i}$ (5)

$F_{drops}=\frac{dm}{dt}V_{i}+m\frac{dV_{i}}{dt}$ (6)

At this point we know the mass of rain per second (mass rate) $\frac{dm}{dt}=0.071 kg/s$ and we also know the initial velocity does not change with time, because that is the velocity at that exact moment (instantaneous velocity). Therefore is a constant, and the derivation of a constant is zero.

This means (6) must be rewritten as:

$F_{drops}=\frac{dm}{dt}V_{i}$ (7)

$F_{drops}=(0.071 kg/s)(15m/s)$ (8)

$F_{drops}=1.065kg.m/s^{2}=1.065N$ (9) This is the force exerted by the rain drops on the roof of the car.

<h2>(b) average force exerted by hailstones on the roof </h2><h2 />

Now let's assume that instead of rain drops, hailstones fall on the roof of the car, and let's also assume these hailstones bounce back up off after striking the roof (this means they do not come to rest as the rain drops).

In addition, we know the hailstones fall with the same velocity as the rain drops and have the same mass rate.

So, in this case the linear momentum $p_{hailstones}$ is:

$p_{hailstones}=mV_{i}+mV_{f}$ (9) Being $V_{i}=V_{f}$

$p_{hailstones}=mV+mV=2mV$ (10)

Deriving with respect to time to find the force $F_{hailstones}$ exerted by the hailstones:

$F_{hailstones}=\frac{d}{dt}p_{hailstones}=\frac{d}{dt}(2mV)$ (10)

$F_{hailstones}=2\frac{d}{dt}(mV)=2(\frac{dm}{dt}V+m\frac{dV}{dt})$ (11)

Assuming $\frac{dV}{dt}=0$ :

$F_{hailstones}=2(\frac{dm}{dt}V)$ (12)

$F_{hailstones}=2(0.071 kg/s)(15m/s)$ (13)

Finally:

$F_{hailstones}=2.13kg.m/s^{2}=2.13N$ (14) This is the force exerted by the hailstones

Comparing (9) and (14) we can conclude the force exerted by the hailstones is two times greater than the force exerted by the raindrops.

5 0

3 years ago

El Sol está en promedio a 93 millones de millas de la Tierra. ¿A cuántos metros equivale esto? Expréselo usando potencias de die

mars1129 [50]

Hola!

93 millones de millas equivalen a 1,50 x 10¹¹ metros.

Para calcular este valor, necesitamos saber la equivalencia entre millas y metros (1 milla=1609,34m), y aplicar el siguiente factor de conversión. Para expresar el resultado en potencias de 10, se debe correr la coma a la izquierda hasta obtener un valor de una sola unidad y multiplicar este valor por 10 elevado a la cantidad de espacios que la coma se tuvo que correr a la izquierda:

$93000000millas*\frac{1609,34 metros}{1milla}=150000000000,0 metros=1,50* 10^{11} metros$

Saludos!

5 0

3 years ago

Are the Sun's rays like mechanical or electromagnetic waves

Dmitry_Shevchenko [17]

If the sun generates any mechanical waves, none of them
have ever reached us, because only electromagnetic waves
can travel through empty space.

5 0

3 years ago

Read 2 more answers

Work is the product of

Mkey [24]

You have to separate out Choice 2 from the top and the last one down. The other 2 are incorrect.

The formula is Work = Force * distance.
Distance in physics is well defined. It is a change in location. But if it is spinning on an axis, the axis at least is not changing location. I'm pick the second one down.

8 0

4 years ago