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geniusboy [140]
3 years ago
5

Two students walk in the same direction along a straight path, at a constant speedâone at 0.90 m/s and the other at 1.90 m/s.

Physics
1 answer:
mixer [17]3 years ago
3 0

Answer:

456.143684211 seconds

564.3 m

Explanation:

s = Distance

v = Velocity

Time is given by

t=\dfrac{s}{v}\\\Rightarrow t=\dfrac{780}{0.9}\\\Rightarrow t=866.67\ s

t=\dfrac{s}{v}\\\Rightarrow t=\dfrac{780}{1.9}\\\Rightarrow t=410.526315789\ s

Difference in time = 866.67-410.526315789 = 456.143684211 seconds

According to the question

\dfrac{x}{0.9}-\dfrac{x}{1.9}=5.5\times 60\\\Rightarrow x(\dfrac{1}{0.9}-\dfrac{1}{1.9})=330\\\Rightarrow x=\dfrac{330}{\dfrac{1}{0.9}-\dfrac{1}{1.9}}\\\Rightarrow x=564.3\ m

The students would have to walk 564.3 m

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Answer:

<h2>206.67N</h2>

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To get the resultant, we need to get the sum of the forces along each components. But first lets get the acceleration along the components first.

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a_x = \frac{d}{dt}(\frac{dx}{dt} )\\ \\a_x = \frac{d}{dt}(6t^{2}-4  )\\\\a_x = \frac{d}{dt}(12t  )\\\\a_x = 12m/s^{2}

Similarly,

a_y = \frac{d}{dt}(\frac{dy}{dt} )\\ \\a_y = \frac{d}{dt}(5t^{3} +6 )\\\\a_y = \frac{d}{dt}(15t^{2}   )\\\\a_y = 30t\\a_y \ at \ t= 2.15s; a_y = 30(2.15)\\a_y = 64.5m/s^2

\sum F_x = 3.15 * 12 = 37.8N\\\sum F_y = 3.15 * 64.5 = 203.18N

R = \sqrt{37.8^2+203.18^2}\\ \\R = \sqrt{1428.84+41,282.11}\\ \\R = \sqrt{42.710.95}\\ \\R = 206.67N

Hence, the magnitude of the net force acting on this object at t = 2.15 s is approximately 206.67N

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3 years ago
Write fulk form of FPS?<br>​
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Two trumpet players are trying to tune their instruments. When they are in tune, they will both be playing the same note and no
marusya05 [52]

Answer:

The second trumpeter will be playing at frequency = 515 Hz

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Since the note sound lower, the second trumpeter will be playing at 515 Hz frequency

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