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3 years ago

As Luke rides his bike down a hill, potential energy is converted to kinetic energy. What is this an example of?

Physics

1 answer:

Musya8 [376]3 years ago

5 0

Answer:

If I am correct, the answer is D. Law of conservation of energy

Explanation:

the potential energy "converts" to kinetic once Luke is in motion.

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3 years ago

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the cross section area of a hole is 725cm^2. Given that the area of a circle is A=3.14r^2 , find the radius of the hole.

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2 years ago

Optimal cardiorespiratory fitness requires a BMI of __________.

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An unstable particle at rest breaks up into two fragments of unequal mass. The mass of the lighter fragment is equal to 2.90 ✕ 1

motikmotik

Answer:

The speed of the heavier fragment is 0.335c.

Explanation:

Given that,

Mass of the lighter fragment $M_{l}=2.90\times10^{-28}\ kg$

Mass of the heavier fragment $M_{h}=1.62\times10^{-27}\ Kg$

Speed of lighter fragment = 0.893c

We need to calculate the speed of the heavier fragment

Let v is the speed of the second fragment after decay

Using conservation of relativistic momentum

$0=\drac{m_{1}v_{1}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}-\drac{m_{2}v_{2}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}$

$\drac{m_{1}v_{1}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}=\drac{m_{2}v_{2}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}$

$\dfrac{2.90\times10^{-28}\times0.893c}{\sqrt{1-(0.893)^2}}=\dfrac{1.62\times10^{-27}v_{2}}{\sqrt{1-\dfrac{v_{2}^2}{c^2}}}$

$\dfrac{v_{2}}{\sqrt{1-\dfrac{v_{2}^2}{c^2}}}=\dfrac{2.90\times10^{-28}\times0.893c}{1.62\times10^{-27}\times0.45}$

$\dfrac{v_{2}}{\sqrt{1-\dfrac{v_{2}^2}{c^2}}}=0.355c$

$\dfrac{v_{2}}{1-\dfrac{v_{2}^{2}}{c^2}}=(0.355c)^2$

$\dfrac{1-\dfrac{v_{2}^2}{c^2}}{v_{2}^2}=\dfrac{1}{(0.355c)}$

$\dfrac{1}{v_{2}^2}-\dfrac{1}{c^2}=\dfrac{1}{(0.355c)^2}$

$\dfrac{1}{v_{2}^2}=\dfrac{1}{c^2}+\dfrac{1}{0.126c^2}$

$\dfrac{1}{v_{2}^2}=\dfrac{1}{c^2}(1+\dfrac{1}{0.126})$

$\dfrac{1}{v_{2}^2}=\dfrac{8.93}{c^2}$

$v_{2}^2=\dfrac{c^2}{8.93}$

$v_{2}=0.335c$

Hence, The speed of the heavier fragment is 0.335c.

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3 years ago

William WangHomework #33 Regents Review (3)0989Assignment Mode : Open (Time On Task) 9 of 25 ListenDuring a collision, an 84-kil

Lemur [1.5K]

Answer:

$1.7\cdot 10^3 N$

Explanation:

The impulse theorem states that the product between the force and the time interval of the collision is equal to the change in momentum:

$F \Delta t = m \Delta v$

where

F is the force

$\Delta t$ is the time interval

m is the mass

$\Delta v$ is the change in velocity

Here we have

m = 84 kg

$\Delta t = 1.2 s$

$\Delta v = 24 m/s$

So we can solve the equation to find the force:

$F= \frac{m \Delta v}{\Delta t }=\frac{(84 kg)(24 m/s)}{1.2 s}=1680 N \sim 1.7\cdot 10^3 N$

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2 years ago