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4 years ago

If a particle's position is given by x=4-12t+3t^2, where t is in seconds and x is in meters, what is its velocity at t=1 second?

Physics

1 answer:

andreyandreev [35.5K]4 years ago

3 0

Answer:

v = -6m/s

Explanation:

$x=4-12t+3t^2$

$\frac{dx}{dt}=-12+6t$

For t = 1:

$\frac{dx}{dt}=-6$

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A 1.0-m-tall vertical tube is filled with 20°C water. A tuning fork vibrating at 580 Hz is held just over the top of the tube as

Mademuasel [1]

Answer:

water heights of the tube are 0.851 m , 0.553 m, 0.255 m

Explanation:

given data

frequency = 580 Hz

temperature = 20°C

tube = 1 m

to find out

water heights of the tube

solution

we will apply here formula for length that is

length L = v ( 2n -1 ) / 4f

here v is velocity o sound that is 343.2 m/s

so for n = 1

L = 343.2 ( 2(1) -1 ) / 4(580) = 0.147931 m

for n = 2

L = 343.2 ( 2(2) -1 ) / 4(580) = 0.443793 m

for n = 3

L = 343.2 ( 2(3) -1 ) / 4(580) = 0.739655 m

for n = 4

L = 343.2 ( 2(4) -1 ) / 4(580) = 1.035517 m is greater than 1

and so here height is measured less than 1 m

so water heights of the tube are 1 m - 0.147931 m , 1 m - 0.443793 m, 1 m - 0.739655 m

so water heights of the tube are 0.851 m , 0.553 m, 0.255 m

3 0

3 years ago

Jake has noticed his best friend Alison cheating at chess several times this week. If she gets caught, she will be kicked off th

tensa zangetsu [6.8K]

He should confront her about it and if after that point she continues report it to the chess team

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4 years ago

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A spaceship ferrying workers to moon Base i takes a straight-line pat from the earth to the moon, a distance of 384,000km, Suppo

GrogVix [38]

Answer:

95.78125%

18000 m/s

22233.33 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

$v=u+at\\\Rightarrow v=0+20\times 15\times 60\\\Rightarrow v=18000\ m/s$

The velocity of the rocket at the end of the first 15 minutes is 18000 m/s which is the maximum speed of the rocket in the complete journey.

Distance traveled while speeding up

$s=ut+\dfrac{1}{2}at^2\\\Rightarrow s=0\times t+\dfrac{1}{2}\times 20\times 900^2\\\Rightarrow s=8100000\ m$

Distance traveled while slowing down

$s=18000\times 900+\dfrac{1}{2}\times -20\times 900^2\\\Rightarrow s=8100000\ m$

Distance traveled during constant speed

$384000000-(2\times 8100000)=367800000\ m$

Fraction

$\dfrac{367800000}{384000000}\times 100=95.78125\ \%$

Fraction of the total distance is traveled at constant speed is 95.78125%

Time taken at constant speed

$t=\dfrac{367800000}{18000}=20433.33\ s$

Total time taken is $900+20433.33+900=22233.33\ s$

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