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VladimirAG [237]

1 year ago

15

A train travels 8.81 m/s in a -51.0° direction. The train accelerated for 2.23 s, changing its velocity to 9.66 m/s in a 37.0° d

irection. What is (delta)x?

1 answer:

Evgen [1.6K]1 year ago

7 0

The value of the $\rm \triangle x$ will be 29.14 m. Delta represent the change.

<h3>What is velocity?</h3>

The change of distance with respect to time is defined as speed. Speed is a scalar quantity. It is a time-based component. Its unit is m/sec.

The velocity in the part 1;

V₁ = 8.81

The velocity in the part 2;

V₂ = 9.66 m/sec

The resultant of the two vectors of velocity gives the velocity for the connecting path.

$\rm V = \sqrt{8.81 ^2 + 9.66^2 } \\\\ V= 13.07 \ m/sec$

The time period for the train accelerates;

$\rm \triangle x = v \times t \\\\ \triangle x = 13.07 \ m/sec \times 2.23 \ sec \\\\ \triangle x = 29.14 \ m$

Hence the value of the $\rm \triangle x$ will be 29.14 m.

To learn more about the velocity, refer to the link: brainly.com/question/862972.

#SPJ1

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On moon time period will become 2.45 times of the time period on earth

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Time period of simple pendulum is equal to $T=2\pi \sqrt{\frac{l}{g}}$ ....eqn 1 here l is length of the pendulum and g is acceleration due to gravity on earth

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2 years ago

melvin pulls a sled across level snow with a force of 317 N along a rope that is 33 degrees above the horizont?

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Answer: work Melvin did=9000J

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Given to complete the question: If the sled moved 33.9m,how much work did Melvin do? Answer in unit of J and round to the nearest thousandth.

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If a ball is thrown vertically upward from the roof of 32 foot building with a velocity of 64 ft/sec, its height after t seconds

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Answer:

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Explanation:

a) The maximum height is obtained with the help of the First and Second Derivative Tests:

First Derivative

$v(t) = 64 - 32\cdot t$

$64 - 32\cdot t = 0$

$t = 2\,s$

Second Derivative

$a(t) = -32$ (absolute maximum)

The maximum height reached by the ball is:

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$s_{max} = 96\,ft$

b) The time required by the ball to hit the ground is:

$32+64\cdot t - 16\cdot t^{2} = 0$

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$t^{2}-4\cdot t - 2 = 0$

$(t -4.449)\cdot (t+0.449)\approx 0$

Just one root offers a solution that is physically reasonable:

$t = 4.449\,s$

The velocity of the ball when it hits the ground is:

$v(4.449\,s) = 64 - 32\cdot (4.449\,s)$

$v(4.449\,s) = -78.368\,\frac{ft}{s}$

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