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3 years ago

A ball is dropped from the top of a building. It initially

Physics

1 answer:

AysviL [449]3 years ago

8 0

Answer:

Explanation:

I'm assuming you missed the decimal point in the initial velocity and that it should be 4.0 m/s. If after a half of a second it is moving a tiny bit slower, it would be because of air resistance. You can only neglect air resistance if the problems you are doing tell you neglect it.

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How can you make an item that is made of magnetic material become a permanent magnet instead of a temporary magnet?

koban [17]

Answer:

Explanation:

You can heat it and then let it cool in a very strong magnetic field.

4 0

3 years ago

Read 2 more answers

A device for acclimating military pilots to the high accelerations they must experience consists of a horizontal beam that rotat

vladimir2022 [97]

Answer:

f = 0.4 Hz

Explanation:

The frequency of rotation of an object in order to achieve required centripetal or radial acceleration, can be found out by using the following formula:

f = (1/2π)√(ac/r)

where,

f = frequency of rotation = ?

ac = radial acceleration = 34.1 m/s²

r = radius = length of beam = 5.55 m

Therefore,

f = (1/2π)√[(34.1 m/s²)/(5.55 m)]

6 0

3 years ago

Please help

cupoosta [38]

I definitely agree with choise that you consider to be as a correct one. I am pretty sure that prganism is correct because this is a unit which ebrases all the mentioned points. And in simple words, the rest of options represent collective objects but you have to answer with sole one. Hope you find it helpful.

8 0

4 years ago

Read 2 more answers

Explain how the distance to a lightning bolt (Fig. CQ17.5) can be determined by counting the seconds between the flash and the s

Sonbull [250]

The distance to a lightning bolt can be determined by counting the seconds between the flash and the sound of thunder due to the speed of light is greater than the speed of sound

We must consider that the speed of light (3.00× 10⁸ m/s) is greater than the speed of sound (340 m/s), this is why we can first see the lightning and after seeing it we can hear it.

In order to calculate the distance we must count the time since the lightning bolt and use the speed of sound, applying the following formula:

x = v * t

Where:

x = distance
t = time
v = velocity

<h3>What is velocity?</h3>

It is a physical quantity that indicates the displacement of a mobile per unit of time, it is expressed in units of distance per time, for example (miles/h, km/h).

Learn more about velocity at: brainly.com/question/80295?source=archive

#SPJ4

7 0

2 years ago

A boat crossing a 153.0 m wide river is directed so that it will cross the river as quickly as possible. The boat has a speed of

Lynna [10]

We have the relation

$\vec v_{B \mid E} = \vec v_{B \mid R} + \vec v_{R \mid E}$

where $v_{A \mid B}$ denotes the velocity of a body A relative to another body B; here I use B for boat, E for Earth, and R for river.

We're given speeds

$v_{B \mid R} = 5.10 \dfrac{\rm m}{\rm s}$

$v_{R \mid E} = 3.70 \dfrac{\rm m}{\rm s}$

Let's assume the river flows South-to-North, so that

$\vec v_{R \mid E} = v_{R \mid E} \, \vec\jmath$

and let $-90^\circ < \theta < 90^\circ$ be the angle made by the boat relative to East (i.e. -90° corresponds to due South, 0° to due East, and +90° to due North), so that

$\vec v_{B \mid R} = v_{B \mid R} \left(\cos(\theta) \,\vec\imath + \sin(\theta) \, \vec\jmath\right)$

Then the velocity of the boat relative to the Earth is

$\vec v_{B\mid E} = v_{B \mid R} \cos(\theta) \, \vec\imath + \left(v_{B \mid R} \sin(\theta) + v_{R \mid E}\right) \,\vec\jmath$

The crossing is 153.0 m wide, so that for some time $t$ we have

$153.0\,\mathrm m = v_{B\mid R} \cos(\theta) t \implies t = \dfrac{153.0\,\rm m}{\left(5.10\frac{\rm m}{\rm s}\right) \cos(\theta)} = 30.0 \sec(\theta) \, \mathrm s$

which is minimized when $\theta=0^\circ$ so the crossing takes the minimum 30.0 s when the boat is pointing due East.

It follows that

$\vec v_{B \mid E} = v_{B \mid R} \,\vec\imath + \vec v_{R \mid E} \,\vec\jmath \\\\ \implies v_{B \mid E} = \sqrt{\left(5.10\dfrac{\rm m}{\rm s}\right)^2 + \left(3.70\dfrac{\rm m}{\rm s}\right)^2} \approx 6.30 \dfrac{\rm m}{\rm s}$

The boat's position $\vec x$ at time $t$ is

$\vec x = \vec v_{B\mid E} t$

so that after 30.0 s, the boat's final position on the other side of the river is

$\vec x(30.0\,\mathrm s) = (153\,\mathrm m) \,\vec\imath + (111\,\mathrm m)\,\vec\jmath$

and the boat would have traveled a total distance of

$\|\vec x(30.0\,\mathrm s)\| = \sqrt{(153\,\mathrm m)^2 + (111\,\mathrm m)^2} \approx \boxed{189\,\mathrm m}$

3 0

2 years ago