In the following reaction, identify the missing particle:

A car travels a distance of 100 km. For the first 30 minutes it is driven at a constant speed of 80 km/hr. The motor begins to v

gregori [183]

Explanation:

First, we need to determine the distance traveled by the car in the first 30 minutes, $d_{\frac{1}{2}}$ .

Notice that the unit measurement for speed, in this case, is km/hr. Thus, a unit conversion of from minutes into hours is required before proceeding with the calculation, as shown below

$d_{\frac{1}{2}\text{h}} \ = \ \text{speed} \ \times \ \text{time taken} \\ \\ \\ d_{\frac{1}{2}\text{h}} \ = \ 80 \ \text{km h}^{-1} \ \times \ \left(\displaystyle\frac{30}{60} \ \text{h}\right) \\ \\ \\ d_{\frac{1}{2}\text{h}} \ = \ 80 \ \text{km h}^{-1} \ \times \ 0.5 \ \text{h} \\ \\ \\ d_{\frac{1}{2}\text{h}} \ = \ 40 \ \text{km}$

Now, it is known that the car traveled 40 km for the first 30 minutes. Hence, the remaining distance, $d_{\text{remain}}$ , in which the driver reduces the speed to 40km/hr is

$d_{\text{remain}} \ = \ 100 \ \text{km} \ - \ 40 \ \text{km} \\ \\ \\ d_{\text{remain}} \ = \ 60 \ \text{km}$ .

Subsequently, we would also like to know the time taken for the car to reach its destination, denoted by $t_{\text{remian}}$ .

$t_{\text{remain}} \ = \ \displaystyle\frac{\text{distance}}{\text{speed}} \\ \\ \\ t_{\text{remain}} \ = \ \displaystyle\frac{60 \ \text{km}}{40 \ \text{km hr}^{-1}} \\ \\ \\ t_{\text{remain}} \ = \ 1.5 \ \text{hours}$ .

Finally, with all the required values at hand, the average speed of the car for the entire trip is calculated as the ratio of the change in distance over the change in time.

$\text{speed} \ = \ \displaystyle\frac{\Delta d}{\Delta t} \\ \\ \\ \text{speed} \ = \ \displaystyle\frac{100 \ \text{km}}{(0.5 \ \text{hr} \ + \ 1.5 \ \text{hr})} \\ \\ \\ \text{speed} \ = \ \displaystyle\frac{100 \ \text{km}}{2 \ \text{hr}} \\ \\ \\ \text{speed} \ = \ 50 \ \text{km hr}^{-1}$

Therefore, the average speed of the car is 50 km/hr.

8 0

2 years ago

A meteoroid is in a circular orbit 600 km above the surface of a distant planet. The planet has the same mass as Earth but has a

AVprozaik [17]

<h2>Answer:</h2><h2>The acceleration of the meteoroid due to the gravitational force exerted by the planet = 12.12 m/ $s^{2}$ </h2>

Explanation:

A meteoroid is in a circular orbit 600 km above the surface of a distant planet.

Mass of the planet = mass of earth = 5.972 x $10^{24}$ Kg

Radius of the earth = 90% of earth radius = 90% 6370 = 5733 km

The acceleration of the meteoroid due to the gravitational force exerted by the planet = ?

By formula, g = $\frac{GM}{r^{2} }$

where g is the acceleration due to the gravity

G is the universal gravitational constant = 6.67 x $10^{-11}$ $m^{3} kg^{-1} s^{-2}$

M is the mass of the planet

r is the radius of the planet

Substituting the values, we get

g = $\frac{(6.67 * 10^{-11}) (5.972 * 10^{24}) }{5733^{2} }$

g = 12.12 m/ $s^{2}$

The acceleration of the meteoroid due to the gravitational force exerted by the planet = 12.12 m/ $s^{2}$

6 0

3 years ago

Johnson made a hole at the bottom of a plastic bottle containing water. However, he noticed that the water did not flow out from

Sphinxa [80]

Answer:

a) See explanation below

b) At X.

Explanation:

Please, see the picture attached with the image of the plastic bottle for this question.

<u>(a) Explain why the water could not flow out of the bottle.</u>

What makes the water flow out of the botlle is the force of gravity, whic attracts the water towards the Earth.

When Johnson made a small hole at the bottom of the plastic bottle containing water, the air outside the bottle, which surrounds it and exerts a pressure all over the outer walls of the bottle, exerted a force against the small area of water "over" the hole that is in contact with the air.

Thus, this force of the air pushing upward through the wall opposed the force of gravity pulling downward making the net force zero and the water cannot fall.

<u>(b) To make the water flow out more easily, his teacher suggested making another hole. At which position - X, Y or Z, should he make the 2nd hole in order for the water to flow out the fastest?</u>

You must open the hole at a place where there is not water but air, such that the outer air can enter in the bottle.

That will make that the pressure in the space over the water inside the bottle be equal to the pressure outside.

The pressure of the air above the water will push it downward. Now, the force from the pressure of air inside the water, which is downward, opposes the upward force from the pressure of air around the first hole, and the net force will be downward, making the water flow out more easily.

Thus, the position where he should make the second hole in order for the water to flow fastest is at X.