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2 years ago

How far could a rabbit run if it ran 36km/h for 5.0min?

Physics

2 answers:

krok68 [10]2 years ago

5 0

Distance = (speed) x (time)

The question GIVES us both the speed and the time. Sadly, though, in order to work with them, we need them to both have the same units, but the information we're given has two different units for time:

The speed is in (km per HOUR), but the time is in MINUTES. We'll have to change one or the other before we can multiply them.

Well, 5.0 minutes is (5/60) = 1/12 of an hour. We can use that.

Distance = (36 km/hr) x (1/12 hour)

Distance = (36/12) (km-hr/hr)

Distance = 3 km .

= = = = = = = = = =

There's a slightly different way to handle it. If the problem was more complicated, this way would actually turn out to be a better way, and you'd have LESS chance of making a mistake if you did it this way:

Distance = (36 km/hr) x (5.0 min)

Distance = (36 x 5) km - min / hr

Distance = 180 km · min / hr

This is a correct answer, but the unit is awkward. We can change the unit at THIS point in our calculation. All we have to do is multiply the answer by ' 1 ' !

We'll find a fraction that has the same number on top or bottom, but in different units, and we'll multiply our answer by this fraction. It won't CHANGE our answer because we're only multiplying it by ' 1 '; but If we're clever about how we build the fraction, it can solve the problem of the units for us.

The fraction to use is . . . (1 hour / 60 minutes) .

The top and the bottom are the same quantity, so the value of the fraction is ' 1 ', and we can multiply our answer by it without changing the answer. But watch what it does to our units:

Distance = (180 km · min / hr) · (1 hr / 60 min)

Distance = (180 · 1 / 60) (km · min · hr / hr · min)

(Do some canceling, and you get ...)

Distance = 3 km

Thank you. Thank you very much. Thank you. I'll be here until Thursday. don't forget to tip your waiter.

Korolek [52]2 years ago

4 0

3 kilometers, it is just 5/60 or 1/12 multiplied by 36.

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and also,

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2 years ago

A ladybug starts at the center of a 16.0 in .-diameter turntable and crawls in a straight radial line to the edge. While this is

mafiozo [28]

The problem is about magnitude of the displacement vector of the lady bug and its directions. so the magnitude of the displacement of the lady bug is 16 in / 2 because it started from the center so the magnitude is 8 in. and the direction is the rotation of the turn table which is 60 degrees

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2 years ago

Max and Jimmy want to jump on a trampoline. Max begins jumping in a steady pattern, making small waves in the trampoline. Jimmy

mylen [45]

Answer:

x_total = (A + B) cos (wt + Ф)

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Explanation:

In this case we can see that the first boy Max when he enters the trampoline and jumps creates a harmonic movement, with a given frequency. When the second boy Jimmy enters the trampoline and begins to jump he also creates a harmonic movement. If the frequency of the two movements is the same and they are in phase we have a resonant process, where the amplitude of the movement increases significantly.

Max

x₁ = A cos (wt + Ф)

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x₂ = B cos (wt + Ф)

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2 years ago

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Answer:

a.) F = 3515 N

b.) F = 140600 N

Explanation: given that the

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Time t = 0.16 s

Force F = change in momentum ÷ time

F = (74×7.6)/0.16

F = 3515 N

b.) If Logan had hit the concrete wall moving at the same speed, his momentum would have been reduced to zero in 0.0080 seconds

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6 0

2 years ago

An electron is in motion at 4.0 × 106 m/s horizontally when it enters a region of space between two parallel plates, as shown, s

max2010maxim [7]

Answer:

xmax = 9.5cm

Explanation:

In this case, the trajectory described by the electron, when it enters in the region between the parallel plates, is a semi parabolic trajectory.

In order to find the horizontal distance traveled by the electron you first calculate the vertical acceleration of the electron.

You use the Newton second law and the electric force on the electron:

$F_e=qE=ma$ (1)

q: charge of the electron = 1.6*10^-19 C

m: mass of the electron = 9.1*10-31 kg

E: magnitude of the electric field = 4.0*10^2N/C

You solve the equation (1) for a:

$a=\frac{qE}{m}=\frac{(1.6*10^{-19}C)(4.0*10^2N/C)}{9.1*10^{-31}kg}=7.03*10^{13}\frac{m}{s^2}$

Next, you use the following formula for the maximum horizontal distance reached by an object, with semi parabolic motion at a height of d:

$x_{max}=v_o\sqrt{\frac{2d}{a}}$ (2)

Here, the height d is the distance between the plates d = 2.0cm = 0.02m

vo: initial velocity of the electron = 4.0*10^6m/s

You replace the values of the parameters in the equation (2):

$x_{max}=(4.0*10^6m/s)\sqrt{\frac{2(0.02m)}{7.03*10^{13}m/s^2}}\\\\x_{max}=0.095m=9.5cm$

The horizontal distance traveled by the electron is 9.5cm

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