Ask question

Ask question

All categories

3 years ago

14

1. An airplane flies 800 kilometers east from Denver to Wichita in one hour. Then the plane flies west back to Denver in the sam

e amount of time. Which of the following best relates the speed and velocity of the two flights?
A.
The speed and velocity of the flights are not related for either flight.
B.
The velocity of the second flight is negative compared to the speed.
C.
The speed of the second flight is negative compared to the velocity.
D.
The velocity and speed of both flights are equal.

1 answer:

Bumek [7]3 years ago

3 0

B. The velocity of the second flight is negative compared to the speed.

You might be interested in

Speed is a scalar quantity that describes only the

Anna [14]

Answer:

magnitude of the velocity

Explanation:

4 0

3 years ago

A good train took 7 hours to complete its journey. For the first 3 hours, it travelled an average speed of 186km/h. What was the

Sliva [168]

I really hope this helps

4 0

2 years ago

A swimming pool has the shape of a right circular cylinder with radius 21 feet and height 10 feet. Suppose that the pool is full

AysviL [449]

Answer:

The water required to pump all the water to a platform 2 feet above the top of the pool is is 6061310.32 foot-pound.

Explanation:

Given that,

Radius = 21 feet

Height = 10 feet

Weighing = 62.5 pounds/cubic

Work = 4329507.37572

Height = 2 feet

Let's look at a horizontal slice of water at a height of h from bottom of pool

We need to calculate the area of slice

Using formula of area

$A=\pi r^2$

Put the value into the formula

$A=\pi\times21^2$

$A=441\pi\ feet^2$

Thickness of slice $t=\Delta h\ ft$

The volume is,

$V=(441\pi\times\Delta h)\ ft^3$

We need to calculate the force

Using formula of force

$F=W\times V$

Where, W = water weight

V = volume

Put the value into the formula

$F=62.5\times(441\pi\times\Delta h)$

$F=27562.5\pi\times\Delta h\ lbs$

We need to calculate the work done

Using formula of work done

$W=F\times d$

Put the value into the formula

$W=27562.5\pi\times\Delta h\times(10-h)\ ft\ lbs$

We do this by integrating from h = 0 to h = 10

We need to find the total work,

Using formula of work done

$W=\int_{0}^{h}{W}$

Put the value into the formula

$W=\int_{0}^{10}{27562.5\pi\\times(10-h)}dh$

$W=27562.5\pi(10h-\dfrac{h^2}{2})_{0}^{10}$

$W=27562.5\pi(10\times10-\dfrac{100}{2}-0)$

$W=4329507.37572$

To pump 2 feet above platform, then each slice has to be lifted an extra 2 feet,

So, the total distance to lift slice is (12-h) instead of of 10-h

We need to calculate the water required to pump all the water to a platform 2 feet above the top of the pool

Using formula of work done

$W=\int_{0}^{h}{W}$

Put the value into the formula

$W=\int_{0}^{10}{27562.5\pi\\times(12-h)}dh$

$W=27562.5\pi(12h-\dfrac{h^2}{2})_{0}^{10}$

$W=27562.5\pi(12\times10-\dfrac{100}{2}-0)$

$W=1929375\pi$

$W=6061310.32\ foot- pound$

Hence, The water required to pump all the water to a platform 2 feet above the top of the pool is is 6061310.32 foot-pound.

8 0

3 years ago

Which is not a common extrusive igneous rock

KonstantinChe [14]

Granite is the answer i think

8 0

3 years ago

*A car is going through a dip in the road whose curvature approximates a circle of radius 150m. At what velocity will the occupa

Valentin [98]

Answer:

v= 14.85 m/s

Explanation:

When at the bottom of the dip, the only force that keeps the car in the circular trajectory, is the centripetal force.
This force is not a new force, is just the net force aiming to the center of the circle.
In this case, is just the difference between the normal force (always perpendicular to the surface, pointing upward) and the force that gravity exerts on the car (which is known as the weight), pointing downward.
So, we can write the following expression:

$F_{cent} = F_{n} - F_{g} (1)$

It can be showed that the centripetal force is related to the speed by the following expression:
$F_{cent} = m*\frac{v^{2}}{r} (2)$

The normal force, it is called the apparent weight, because it would be the weight as measured by a scale.
Replacing (2) in (1), and solving for Fn, we get:

$F_{n} = m*\frac{v^{2} }{r} + m*g (3)$

Now, we need to find the value of v that makes Fn, exactly 15% more than the weight m*g, so we can write the following equation:

$F_{n} = 1.15*F_{g} = m*\frac{v^{2}}{r} +F_{g} (4)$

Replacing Fg by its value, simplifying, and solving for v, we get:

$v = \sqrt{0.15*g*r} = \sqrt{9.8 m/s2*0.15*150m} = 14.85 m/s (5)$

3 0

2 years ago