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3 years ago

8

Neglecting friction, of a pendulum bob has 100 joules of kinetic energy at the bottom of its swing, how much potential energy do

es it have at the top of its swing

1 answer:

galben [10]3 years ago

7 0

100 joule of kinetic energy

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If the angular frequency of the motion of a simple harmonic oscillator is doubled, by what factor does the maximum acceleration

Nataly_w [17]

Answer:

When we double the angular velocity the maximum acceleration $(a_{max})$ will changes by a factor of 4.

Explanation:

Given the angular frequency $(\omega)$ of the simple harmonic oscillator is doubled.

We need to find the change in the maximum acceleration of the oscillator.

$a_{max}=A\omega^2$

Now, according to the problem, the angular frequency $(\omega)$ got doubled.

Let us plug $\omega=2\times \omega$ . Then the maximum acceleration will be $a_{max'}$

$a_{max}=A\omega^2$

$a_{max'}=A(2\times \omega)^2\\a_{max'}=A\times 4\omega\\a_{max'}=4A\omega$

$a_{max'}=4a_{max}$

We can see, when we double the angular velocity the maximum acceleration will changes by a factor of 4.

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2 years ago

A 90 kg ice skater moving at 12.0 m/s on the ice encounters a region of roughed up ice with a coefficient of kinetic friction of

balandron [24]

Answer:

The skater covers a distance of <u>15 m</u> before stopping.

Explanation:

Let the distance traveled before stopping be 'd' m.

Given:

Mass of the skater (m) = 90 kg

Initial velocity of the skater (u) = 12.0 m/s

Final velocity of the skater (v) = 0 m/s (Stops finally)

Coefficient of kinetic friction (μ) = 0.490

Acceleration due to gravity (g) = 9.8 m/s²

Now, we know that, from work-energy theorem, the work done by the net force on a body is equal to the change in its kinetic energy.

Here, the net force acting on the skater is only frictional force which acts in the direction opposite to motion.

Frictional force is given as:

$f=\mu N$

Where, 'N' is the normal force acting on the skater. As there is no vertical motion, $N=mg$

∴ $f=\mu mg=0.490\times 90\times 9.8=432.18\ N$

Now, work done by friction is a negative work as friction and displacement are in opposite direction and is given as:

$W=-fd=-432.18d$

Now, change in kinetic energy is given as:

$\Delta K=\frac{1}{2}m(v^2-u^2)\\\\\Delta K=\frac{1}{2}\times 90(0-12^2)\\\\\Delta K=45\times (-144)=-6480\ J$

Therefore, from work-energy theorem,

$W=\Delta K\\\\-432.18d=6480\\\\d=\frac{6480}{432.18}\\\\d=14.99\approx 15\ m$

Hence, the skater covers a distance of 15 m before stopping.

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3 years ago

2. What is the percent composition of sulfur in H2SO4?

allochka39001 [22]

Answer:

C. 32.7%

Explanation:

% composition = ( mass S / mass H2SO4 ) × 100 = 32.08/ 98.10 × 100 = 32.7 % pls mark brainliest

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