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OLEGan [10]
3 years ago
13

Converting 15 miles to kilometer

Physics
2 answers:
Dmitry [639]3 years ago
8 0
15 miles to kilometers would be: 24.14 kilometers
dimaraw [331]3 years ago
5 0

1 mile = 1609.344... meters

1,000 meters = 1 km

(15 miles) x (1609 m/mile) x (1 km/1000 m) =

(15 x 1609 / 1000) (mile-meter-km / mile-meter) =  <em>24.14 km</em>


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A pole vaulter takes 1.3 seconds to fall from peak height to the landing mat. What is his vertical velocity at contact with the
julia-pushkina [17]

12.75 \frac{m}{s}

Assuming vertical acceleration of 9.81 \frac{m}{s^2}, the speed after x seconds of falling is 9.81x \to 9.81(1.3) = 12.75

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A golf ball is hit off a tee at the edge of a cliff. Its x and y coordinates as functions of time are given by x = 18.0t and y =
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Position:                 x = 18t    y = 4t - 4.9t²

First derivative:        x' = 18      y' = 4 - 9.8t

Second derivative:    x'' = 0        y'' = - 9.8


Position vector:      P  =  (18t) i  +  (4t - 4.9t²) j

Velocity vector:      V  =  (18) i  +  (4 - 9.8t) j

Acceleration vector  A  =              (- 9.8) j

6 0
3 years ago
Read 2 more answers
Given a particle that has the velocity v(t) = 3 cos(mt) = 3 cos (0.5t) meters, a. Find the acceleration at 3 seconds. b. Find th
DiKsa [7]

Answer:

a.\rm -1.49\ m/s^2.

b. \rm 50.49\ m.

Explanation:

<u>Given:</u>

  • Velocity of the particle, v(t) = 3 cos(mt) = 3 cos (0.5t) .

<h2>(a):</h2>

The acceleration of the particle at a time is defined as the rate of change of velocity of the particle at that time.

\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(3\cos(0.5\ t ))\\=3(-0.5\sin(0.5\ t.))\\=-1.5\sin(0.5\ t).

At time t = 3 seconds,

\rm a=-1.5\sin(0.5\times 3)=-1.49\ m/s^2.

<u>Note</u>:<em> The arguments of the sine is calculated in unit of radian and not in degree.</em>

<h2>(b):</h2>

The velocity of the particle at some is defined as the rate of change of the position of the particle.

\rm v = \dfrac{dr}{dt}.\\\therefore dr = vdt\Rightarrow \int dr=\int v\ dt.

For the time interval of 2 seconds,

\rm \int\limits^2_0 dr=\int\limits^2_0 v\ dt\\r(t=2)-r(t=0)=\int\limits^2_0 3\cos(0.5\ t)\ dt

The term of the left is the displacement of the particle in time interval of 2 seconds, therefore,

\Delta r=3\ \left (\dfrac{\sin(0.5\ t)}{0.05} \right )\limits^2_0\\=3\ \left (\dfrac{\sin(0.5\times 2)-sin(0.5\times 0)}{0.05} \right )\\=3\ \left (\dfrac{\sin(1.0)}{0.05} \right )\\=50.49\ m.

It is the displacement of the particle in 2 seconds.

7 0
3 years ago
According to the cooling curve, what is the approximate freezing point of the substance?
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Answer:

I do not know which substance you are referring to, but the freezing point of water is 32°F, or 0°C.

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