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2 years ago

An object is thrown from the ground with an initial velocity of 30 m/s. What is the velocity at the point 25 m above the ground?

Physics

2 answers:

Dimas [21]2 years ago

8 0

Answer:

Explanation:

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dsp732 years ago

3 0

Answer:

It's a pretty simple suvat linear projectile motion question, using the following equation and plugging in your values it's a pretty trivial calculation.

V^2=U^2+2*a*x

V=0 (as it is at max height)

U=30ms^-1 (initial speed)

a=-g /-9.8ms^-2 (as it is moving against gravity)

x is the variable you want to calculate (height)

0=30^2+2*(-9.8)*x

x=-30^2/2*-9.8

x=45.92m

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3 years ago

Four traveling waves are described by the following equations, where all quantities are measured in SI units and y represents th

agasfer [191]

Answer:

$T_1=T_3=\dfrac{2\pi}{21}$

$T_2=T_4=\dfrac{2\pi}{42}$

Explanation:

Wave 1, $y_1=0.12\ cos(3x-21t)$

Wave 2, $y_2=0.15\ sin(6x+42t)$

Wave 3, $y_3=0.13\ cos(6x+21t)$

Wave 4, $y_4=-0.27\ sin(3x-42t)$

The general equation of travelling wave is given by :

$y=A\ cos(kx\pm \omega t)$

The value of $\omega$ will remain the same if we take phase difference into account.

For first wave,

$\omega_1=21$

$\dfrac{2\pi }{T_1}=21$

$T_1=\dfrac{2\pi}{21}$

For second wave,

$\omega_2=42$

$\dfrac{2\pi }{T_2}=42$

$T_2=\dfrac{2\pi}{42}$

For the third wave,

$\omega_3=21$

$\dfrac{2\pi }{T_3}=21$

$T_3=\dfrac{2\pi}{21}$

For the fourth wave,

$\omega_4=42$

$\dfrac{2\pi }{T_4}=42$

$T_4=\dfrac{2\pi}{42}$

It is clear from above calculations that waves 1 and 3 have same time period. Also, wave 2 and 4 have same time period. Hence, this is the required solution.

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3 years ago