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Juli2301 [7.4K]

3 years ago

8

professor Xavier boarded a plane and flew East for 5558.4 km from houston to new orleans. he then boarded a second plane and fle

w 39 degrees east of north from new orleans to nashville for a distance of 758 km. what was professor xavier total displacement in km

1 answer:

Dovator [93]3 years ago

3 0

4761.4 km

I hope this helps

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Why do we warm up when we wear warm clothes? Explain the hiccup from the point of view of physics.

bazaltina [42]

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5 0

4 years ago

URGENT! Please help me with 7-9 and if possible can you please explain why :)!

faust18 [17]

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8 0

3 years ago

A person can survive a feet-first impact at a speed of about 12 m/s (27 mi/h) on concrete, 15 m/s (34 mi/h) on soil, and 34 m/s

PilotLPTM [1.2K]

Answer:

Different surfaces have different impact force during collision which depends on the time it takes the person to come to rest after collision.

Explanation:

Given;

speed on concrete = 12 m/s (27 mi/h)

speed on soil = 15 m/s (34 mi/h)

speed on water = 34 m/s (76 mi/h)

The impact force on this person during collision is rate of change of momentum;

$F = \frac{\delta P}{\delta t}$

During collision, the force exerted on this person depends on how long the collision lasts; that is, how long it takes for this person to come to rest after collision with each of the surfaces.

The longer the time of collision, the smaller the force exerted by each.

It takes shorter time for the person to come to rest on concrete surface than on soil surface, also it takes shorter time for the person to come to rest on soil surface than on water surface.

As a result of the reason above, the force exerted on the person during collision by the concrete surface is greater than that of soil surface which is greater than that of water surface.

8 0

4 years ago

Two blocks of masses 20 kg and 8.0 kg are connected togetherby

Dmitry [639]

Answer:

The tension is 14 N

Explanation:

For this problem we have to use newton's law, so:

$F=m*a$

The second string is connected to the mass of 8 kg, but the mass of 8 kg is connected to the mass of 20kg, so we can say that the second string is handling the two masses. so:

$F=28kg*0.5\frac{m}{s^2}=14N$

3 0

3 years ago

A particle initially located at the origin has an acceleration of a=3jm/s^2 and an initial velocity of Vi=5im/s.

-BARSIC- [3]

Given the particle's acceleration is

$\vec a(t) = \left(3\dfrac{\rm m}{\mathrm s^2}\right)\vec\jmath$

with initial velocity

$\vec v(0) = \left(5\dfrac{\rm m}{\rm s}\right)\,\vec\imath$

and starting at the origin, so that

$\vec r(0) = \vec 0$

you can compute the velocity and position functions by applying the fundamental theorem of calculus:

$\vec v(t) = \vec v(0) + \displaystyle \int_0^t \vec a(u)\,\mathrm du$

$\vec r(t) = \vec r(0) + \displaystyle \int_0^t \vec v(u)\,\mathrm du$

We have

• velocity at time <em>t</em> :

$\vec v(t) = \left(5\dfrac{\rm m}{\rm s}\right)\,\vec\imath + \displaystyle \int_0^t \left(3\dfrac{\rm m}{\mathrm s^2}\right)\,\vec\jmath\,\mathrm du \\\\ \vec v(t) = \left(5\dfrac{\rm m}{\rm s}\right)\,\vec\imath + \left(3\dfrac{\rm m}{\mathrm s^2}\right)t\,\vec\jmath \\\\ \boxed{\vec v(t) = \left(5\dfrac{\rm m}{\rm s}\right)\,\vec\imath + \left(3\dfrac{\rm m}{\mathrm s^2}\right)t\,\vec\jmath}$

• position at time <em>t</em> :

$\vec r(t) = \displaystyle \int_0^t \left(\left(5\dfrac{\rm m}{\rm s}\right)\,\vec\imath + \left(3\dfrac{\rm m}{\mathrm s^2}\right)u\,\vec\jmath\right) \,\mathrm du \\\\ \boxed{\vec r(t) = \left(5\dfrac{\rm m}{\rm s}\right)t\,\vec\imath + \frac12 \left(3\frac{\rm m}{\mathrm s^2}\right)t^2\,\vec\jmath}$

8 0

3 years ago