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2 years ago

Every winter I fly home to Chicago. It takes 3 hours. What is my average speed?

Physics

1 answer:

Tanya [424]2 years ago

5 0

It depends on where you live when you're not visiting Chicago. We need to know the distance of the trip.

Your average speed on the trip is . . .

(total distance in miles) / (3 hours)

miles per hour

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A balloon drifts 140m toward the west in 45s ; then the wind suddenly changes and the balloon flies 90m toward the east in the n

Bogdan [553]

Answer: 140 m

Explanation:

Let's begin by stating clear that motiont is the change of position of a body at a certain time. So, during this motion, the balloon will have a trajectory and a displacement, being both different:

The trajectory is the path followed by the body, the distance it travelled (is a scalar quantity).

The displacement is the distance in a straight line between the initial and final position (is a vector quantity).

So, according to this, the distance the balloon traveled during the first 45 s (its trajectory) is 140 m.

But, if we talk about displacement, we have to draw a straight line between the initial position of the balloon (point 0) to its final position (point 90 m). Being its displacement 95 m.

8 0

1 year ago

A shift in one fringe in the Michelson-Morley experiment corresponds to a change in the round-trip travel time along one arm of

olya-2409 [2.1K]

Explanation:

When Michelson-Morley apparatus is turned through $90^{o}$ then position of two mirrors will be changed. The resultant path difference will be as follows.

$\frac{lv^{2}}{\lambda c^{2}} - (-\frac{lv^{2}}{\lambda c^{2}}) = \frac{2lv^{2}}{\lambda c^{2}}$

Formula for change in fringe shift is as follows.

n = $\frac{2lv^{2}}{\lambda c^{2}}$

$v^{2} = \frac{n \lambda c^{2}}{2l}$

v = $\sqrt{\frac{n \lambda c^{2}}{2l}}$

According to the given data change in fringe is n = 1. The data is Michelson and Morley experiment is as follows.

l = 11 m

$\lambda = 5.9 \times 10^{-7} m$

c = $3.0 \times 10^{8}$ m/s

Hence, putting the given values into the above formula as follows.

v = $\sqrt{\frac{n \lambda c^{2}}{2l}}$

= $\sqrt{\frac{1 \times (5.9 \times 10^{-7} m) \times (3.0 \times 10^{8})^{2}}{2 \times 11 m}}$

= $2.41363 \times 10^{9} m/s$

Thus, we can conclude that velocity deduced is $2.41363 \times 10^{9} m/s$ .

3 0

2 years ago

A rock is thrown straight up with an initial velocity of 19.6 m/s. What time interval elapses between the rock’s being thrown an

inna [77]

Answer:

It will take 4 sec rock to comes its original point

Explanation:

It is given that the rock comes to its original point

So displacement S = 0 m

Initial velocity u = 19.6 m/sec

Acceleration due to gravity $g=9.8m/sec^2$

According to second equation of motion $h=ut+\frac{1}{2}gt^2$

$0=19.6\times t+\frac{1}{2}\times 9.8t^2$

$19.6=4.9t$

t = 4 sec

3 0

2 years ago

Read 2 more answers

The burj Khalifa in Dubai is the worlds tallest building. It rises to an amazing 828M above the ground and if you were to get to

antoniya [11.8K]

I don't understand what you mean by "depth" of the steps. The flat part of the step has a front-to-back dimension, and the 'riser' has a height. I don't care about the horizontal dimension of the step because it doesn't add anything to the climber's potential energy. And if the riser of each step is 20cm high, then 3,234 of them only take him (3,234 x 0.2) = 646.8 meters up off the ground. So something is definitely fishy about the steps.

Fortunately, we don't need to worry at all about the steps in order to derive a first approximation to the answer ... one that's certainly good enough for high school Physics.

In order to lift his bulk 828 meters from the street to the top of the Burj, the climber has to provide a force of 800 newtons, and maintain it through a distance of 828 meters. The work [s]he does is (force) x (distance) = 662,400 joules. 

6 0

2 years ago

A (1.25+A) kg bowling ball is hung on a (2.50+B) m long rope. It is then pulled back until the rope makes an angle of (12.0+C)o

daser333 [38]

Answer:

F = 0.535 N

Explanation:

Let's use the concepts of energy, at the highest and lowest point of the trajectory

Higher

Em₀ = U = mg y

Lower

$Em_{f}$ = K = ½ m v²

Emo = $Em_{f}$

mg y = ½ m v2

v = √ 2gy

y = L - L cos θ

v = √ (2g L (1-cos θ))

Now let's use Newton's second law n at the lowest point where the acceleration is centripetal

F = ma

a = v² / r

In turning radius is the cable length r = L

F = m 2g (1-cos θ)

Let's calculate

F = 2 1.25 9.8 (1 - cos 12)

F = 0.535 N

7 0

2 years ago