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2 years ago

A train has a final velocity of 110 km/h. It accelerated for 36 s at 0.50 m/s2 . What was its initial velocity

Physics

1 answer:

Tems11 [23]2 years ago

3 0

The initial velocity of the train is 12.56 m/s.

<h3>Initial velocity</h3>

The initial velocity of the train can be determined by using the first kinematic equation as shown below;

v = u + at

u = v - at

where;

v is the final velocity = 110 km/h = 30.56 m/s
u is the initial velocity

u = 30.56 - (36 x 0.5)

u = 12.56 m/s

Thus, the initial velocity of the train is 12.56 m/s.

Learn more about initial velocity here: brainly.com/question/19365526

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A satellite of mass m is moving in a circular orbit around the earth at a constant speed v and at an altitude h above the earth'

qwelly [4]

The satellite executes a rotation motion around the earth, because Earth's force of attraction plays the role of centripetal force:

Fa=Fcp=>k*Mp*m/(Rp+r)²=mv²/(Rp+r)=>v=√(k*Mp/(Rp+r))=√(6.67*10⁻¹¹*5.98*10²⁴/(6371*10³+1000*10³))=√(39.88*10¹³/(7371*10³))=√(5.41*10⁷)=7355.53 m/s

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3 years ago

Jane is sliding down a slide. What kind of motion is she demonstrating?

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When Jane is sliding down a slide, she is demonstrating translational motion.

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3 years ago

A rock is projected upward from the surface of the moon, at time t = 0.0 s, w a velocity of 30 m/s. The acceleration due to grav

Vinvika [58]

<h2>Answer: 277.777 m</h2>

Explanation:

The situation described here is parabolic movement. However, as we are told that the rock was<u> projected upward from the surface</u>, we will only use the equations related to the Y axis.

In this sense, the movement equations in the Y axis are:

$y-y_{o}=V_{o}.t+\frac{1}{2}g.t^{2}$ (1)

$V=V_{o}-g.t$ (2)

Where:

$y$ is the rock's final position

$y_{o}=0$ is the rock's initial position

$V_{o}=30\frac{m}{s}$ is the rock's initial velocity

$V$ is the final velocity

$t$ is the time the parabolic movement lasts

$g=1.62\frac{m}{s^{2}}$ is the acceleration due to gravity at the surface of the moon

As we know $y_{o}=0$ , equation (2) is rewritten as:

$y=V_{o}.t+\frac{1}{2}g.t^{2}$ (3)

On the other hand, the maximum height is accomplished when $V=0$ :

$V=V_{o}-g.t=0$ (4)

$V_{o}-g.t=0$

$V_{o}=g.t$ (5)

Finding $t$ :

$t=\frac{V_{o}}{g}$ (6)

Substituting (6) in (3):

$y=V_{o}(\frac{V_{o}}{g})+\frac{1}{2}g(\frac{V_{o}}{g})^{2}$ (7)

$y_{max}=\frac{{V_{o}}^{2}}{2g}$ (8) Now we can calculate the maximum height of the rock

$y_{max}=\frac{{(30m/s)}^{2}}{(2)(1.62m/s^{2})}$ (9)

Finally:

$y_{max}=277.777m$

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3 years ago

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babunello [35]

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3 years ago