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2 years ago

Which object has more momentum?

Physics

1 answer:

lakkis [162]2 years ago

8 0

Answer:

3. Large butterfly in flight, flying through the air?

Explanation:

Momentum is simply defined as the quantity of motion a body possess. It is mathematically given as;

Momentum = mass x velocity

The larger the mass, the larger the momentum and also the velocity

Since the large butterfly is in flight, it has the largest velocity.

A sleeping bear and resting caterpillar have no momentum because their velocity is 0

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2. If you were trying to determine if cars in your neighborhood were travelling at or below the posted speed limit, how would yo

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Speed is defined as the distance over time. So in measuring the speed of a car, the most manual thing that we can do besides using a speedometer is to measure a certain distance then measure the time at which the car passes that distance then divide the distance over the time. Then determine the speed limit.

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2 years ago

A sprinter starting to run from rest, reaches a speed of 8m/s in 2 second, what is the average power needed to accelerate like t

larisa86 [58]

Answer:

okay

Explanation:

power is rate of doing work

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1 year ago

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Yuliya22 [10]

As wavelength increase, frequency decrease

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2 years ago

Two objects each moving with speed v travel in opposite directions along a straight line passing through both their centers. The

Tpy6a [65]

Answer:

$\dfrac{1}{16}$

$\dfrac{5}{3}$

Explanation:

$m_1$ = Mass of first object

$m_2$ = Mass of second object

v = Speed of both objects

$\dfrac{v}{4}$ = Combined velocity

The ratio of final kinetic energy to initial kinetic energy will be

$\dfrac{K_f}{K_i}=\dfrac{\dfrac{1}{2}(m_1+m_2)(\dfrac{v}{4})^2}{\dfrac{1}{2}(m_1v^2+m_2v^2)} \\\Rightarrow \dfrac{K_f}{K_i}=\dfrac{(m_1+m_2)\dfrac{v^2}{16}}{m_1v^2+m_2v^2}\\\Rightarrow \dfrac{K_f}{K_i}=\dfrac{(m_1+m_2)\dfrac{1}{16}}{m_1+m_2}\\\Rightarrow \dfrac{K_f}{K_i}=\dfrac{1}{16}$

The ratio is $\dfrac{1}{16}$

As the linear momentum is conserved

$m_1v-m_2v=(m_1+m_2)\dfrac{v}{4}\\\Rightarrow m_1-m_2=(m_1+m_2)\dfrac{1}{4}$

Divide by $m_2$ on both sides

$\dfrac{m_1}{m_2}-1=\dfrac{m_1}{4m_2}+\dfrac{1}{4}\\\Rightarrow \dfrac{m_1}{m_2}-\dfrac{m_1}{4m_2}=\dfrac{1}{4}+1\\\Rightarrow \dfrac{3m_1}{4m_2}=\dfrac{5}{4}\\\Rightarrow \dfrac{m_1}{m_2}=\dfrac{5\times 4}{3\times 4}\\\Rightarrow \dfrac{m_1}{m_2}=\dfrac{5}{3}$

The ratio of mass is $\dfrac{5}{3}$

6 0

2 years ago

A playground merry-go-round has a mass of 120 kg and a radius of 1.80 m and it is rotating with an angular velocity of 0.400 rev

saw5 [17]

Answer:

The final angular velocity is rev/s is 0.293 rev/s.

Explanation:

Given;

mass of the merry-go-round, m₁ = 120 kg

radius of the merry-go-round, r = 1.8 m

initial angular velocity, ω = 0.4 rev/s

mass of the child, m₂ = 22 kg

Apply the principle of conservation angular momentum to determine the final angular velocity;

$I_i= I_f\\\\\frac{1}{2} m_1r^2 \omega _i = \frac{1}{2} m_1r^2 \omega _f + m_2r^2 \omega _f\\\\ \frac{1}{2} m_1r^2 \omega _i =( \frac{1}{2} m_1r^2 + m_2r^2 )\omega _f\\\\\omega _f = \frac{ \frac{1}{2} m_1r^2 \omega _i}{\frac{1}{2} m_1r^2 + m_2r^2} \\\\\omega _f = \frac{ \frac{1}{2} m_1 \omega _i}{\frac{1}{2} m_1 + m_2}\\\\\omega _f = \frac{0.5 \ \times \ 120\ kg \ \times \ 0.4\ rev/s}{0.5 \ \times 120\ kg \ \ + \ \ 22 \ kg} \\\\\omega _f = 0.293 \ rev/s\\$

Therefore, the final angular velocity is rev/s is 0.293 rev/s.

3 0

2 years ago