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Luba_88 [7]
3 years ago
13

Pure water is a(n) _____.

Physics
1 answer:
Mashutka [201]3 years ago
6 0
The correct answer would be compound
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Newton's Second Law tells us that the more mass an object has, the more force is needed to move it. A larger rocket will need stronger forces (eg. more fuel) to make it accelerate. The space shuttles required seven pounds of fuel for every pound of payload they carry.

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Planet A takes 1 year to go around its star at an average of 1 A.U. distance. Planet B is 4 A.U. from the star. Calculate how lo
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Read 2 more answers
A 0.6 kg block attached to a spring of force constant 13.6 N/m oscillates with an amplitude of 9 cm. Find the maximum speed of t
mash [69]

Answer:

1) 0.43 meters per second

2) 0.21 meters per second

3) 1.02 \frac{m}{s^{2}}

4) 0.66 seconds

Explanation:

part 1

By conservation of energy, the maximum kinetic energy (K) of the block is at equilibrium point where the potential energy is zero. So, at the equilibrium kinetic energy is equal to maximum potential energy (U):

K=U

\frac{mv^2}{2}=\frac{kx_{max}^2}{2}

With m the mass, v the speed, k the spring constant and xmax the maximum position respect equilibrium position. Solving for v

v=\sqrt{\frac{kx_{max}^2}{m}}=\sqrt{\frac{(13.6)(0.09m)^2}{0.6}}=0.43\frac{m}{s}

part 2

Again by conservation of energy we have kinetic energy equal potential energy:

\frac{mv^2}{2}=\frac{kx_{max}^2}{2}=

v=\sqrt{\frac{kx_{max}^2}{m}}=\sqrt{\frac{(13.6)(0.045m)^2}{0.6}}=0.21\frac{m}{s}

part 3

Acceleration can be find using Newton's second law:

F=ma

with F the force, m the mass and a the acceleration, but elastic force is -kx, so:

-kx=ma

a= -\frac{kx}{m}=-\frac{(13.6)(0.045)}{0.6}=-1.02\frac{m}{s^{2}}

part 4

The period of an oscillator is the time it takes going from one extreme to the other one, that is going form 4.5 cm to -4.5 cm respect the equilibrium position. That period is:

T=2\pi\sqrt{\frac{m}{k}}=T=2\pi\sqrt{\frac{0.6}{13.6}}=1.32s

So between 0 and 4.5 cm we have half a period:

t=\frac{T}{2}=0.66s

7 0
3 years ago
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