Physics, 19.08.2019 01:30 Fsamrai

Ahobo hops a freight train traveling at 2.5 m/s due east. the man boards the train from the rear of the boxcar and walks from the southwest corner to the north east corner of the car at a rate of 5.0 m/s. the boxcar is 15 meters long and 3 meters wide. what is the velocity of the man relative to the train tracks?

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Answer from: akeemsimpson15p2034o

Since it isn't specified, I have to assume that when he's walking inside, along
the diagonal of the car, at 5m/s, that speed is relative to the floor of the car.
The reason I have to assume that is because I think it makes the problem
easier. I could be wrong. And it's still troubling, because 5 m/s is a hefty
11.18 mph, which is a pretty energetic walk. (In fact, it's a 5min 22sec mile,
which I for one never accomplished, even when I was running.) But so be it.

-- The length of the car's diagonal is √(3² + 15²) = √(9 + 225) = √334

-- The angle of his walk along the diagonal is the angle whose tangent is 3/15.

-- His velocity consists of the components [ 5 cos(angle) east ] and
[ 5 sin(angle) north ].

That's [ 5 x 15/√334 east ] and [ 5 x 3/√334 ] north .

The train's motion adds to the easterly component of his velocity,
and that becomes [ (2.5) + (5 x 15/√334) ] . The train's motion has
no effect on the northerly component of his velocity.

So now we're ready to put the components together and find his velocity
relative to the tracks. I think it'll be easier to go ahead and get the numerical
value of each component, and then combine them.

Easterly component: (2.5) + (5 x 15/√334) = 6.6038 m/s

Northerly component: 5 x 3/√334 = 0.8208 m/s

Just before finding the magnitude, we note that the direction of his velocity
is (the angle whose tangent is 0.8208/6.6038) north of east. That's about
7.085 degrees north of east ...the compass bearing of 82.92 degrees.

Now for the magnitude. It's the square root of the sum of the squares of
the easterly component and the northerly component.

√ (6.6038² + 0.8208²) = √44.2841 = 6.655 m/s

(All numbers are rounded.)

That's my story, and I'm sticking with it.

Answer from: Quest

answer: the correct answer is 1.

explanation: there are various process in which a radioactive nuclei decays:

1) alpha decay: in this process, a heavier nuclei decays into lighter nuclei by releasing alpha particle. the mass number is reduced by 4 units.

$_a^z\textrm{x}\rightarrow _{a-2}^{z-4}+_2^4\alpha$

2) electron capture: in this decay process, a parent nuclei absorbs an electron and gets converted into a neutron. simply, a proton and an electron combines together to form a neutron. mass number does not change in this process.

$_a^z\textrm{x}+_{-1}^0\textrm{e}\rightarrow _{z-1}^a\textrm{y}$

3) gamma ray emission: it is a decay process in which an unstable nuclei gives excess energy by a spontaneous electromagnetic process. this decay releases $\gamma -radiations$ . this process does not change the mass number.

$_a^z\textrm{x}^*\rightarrow _a^z\textrm{x}+_0^0\gamma$

4) positron emission: it is a type of decay process, in which a proton gets converted to neutron and an electron neutrino. this is also known as $\beta ^+$ -decay. in this the mass number remains same.

$_a^z\textrm{x}\rightarrow _{z-1}^a\textrm{y}+_{+1}^0e$

from the above information, we see that only alpha-decay is producing a change in the mass number of the nucleus.

Answer from: Quest

speed

ratio of distance traveled to time interval

net force

vector sum of forces on object

object

source of diverging light rays either luminous or illuminated

don't remove my answer im trying my best to answer them

explanation:

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Ahobo hops a freight train traveling at 2.5 m/s due east. the man boards the train from the rear of...

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