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2 years ago

9

A honeybee with a mass of 0.150 g lands on one end of a floating 4.75-g popsicle stick, as shown in (Figure 1) . After sitting a

t rest for a moment, it runs toward the other end with a velocity v⃗ b relative to the still water. The stick moves in the opposite direction with a speed of 0.140 cm/s.
What is the velocity of the bee? (Let the direction of the bee's motion be the positive x direction.)

Express your answer to three significant figures and include appropriate units.

1 answer:

Alekssandra [29.7K]2 years ago

5 0

Answer:

0.0443 m/s

Explanation:

$m_1$ = Mass of honeybee = 0.15 g

$m_2$ = Mass of popsicle stick = 4.75 g

$v_1$ = Velocity of honeybee

$v_2$ = Velocity of stick = 0.14 cm/s

In this system the linear momentum is conserved

$m_1v_1=m_2v_2\\\Rightarrow v_1=\dfrac{m_2v_2}{m_1}\\\Rightarrow v_1=\dfrac{4.75\times 0.14}{0.15}\\\Rightarrow v_1=4.43\ m/s$

The velocity of the bee is 4.43 cm/s or 0.0443 m/s

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Associations Clusters

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2 years ago

Two astronauts, each with a mass of 50 kg, are connected by a 7 m massless rope. Initially they are rotating around their center

kiruha [24]

Answer:

The angular velocity is $w_f = 1.531 \ rad/ s$

Explanation:

From the question we are told that

The mass of each astronauts is $m = 50 \ kg$

The initial distance between the two astronauts $d_i = 7 \ m$

Generally the radius is mathematically represented as $r_i = \frac{d_i}{2} = \frac{7}{2} = 3.5 \ m$

The initial angular velocity is $w_1 = 0.5 \ rad /s$

The distance between the two astronauts after the rope is pulled is $d_f = 4 \ m$

Generally the radius is mathematically represented as $r_f = \frac{d_f}{2} = \frac{4}{2} = 2\ m$

Generally from the law of angular momentum conservation we have that

$I_{k_1} w_{k_1}+ I_{p_1} w_{p_1} = I_{k_2} w_{k_2}+ I_{p_2} w_{p_2}$

Here $I_{k_1 }$ is the initial moment of inertia of the first astronauts which is equal to $I_{p_1}$ the initial moment of inertia of the second astronauts So

$I_{k_1} = I_{p_1 } = m * r_i^2$

Also $w_{k_1 }$ is the initial angular velocity of the first astronauts which is equal to $w_{p_1}$ the initial angular velocity of the second astronauts So

$w_{k_1} =w_{p_1 } = w_1$

Here $I_{k_2 }$ is the final moment of inertia of the first astronauts which is equal to $I_{p_2}$ the final moment of inertia of the second astronauts So

$I_{k_2} = I_{p_2} = m * r_f^2$

Also $w_{k_2 }$ is the final angular velocity of the first astronauts which is equal to $w_{p_2}$ the final angular velocity of the second astronauts So

$w_{k_2} =w_{p_2 } = w_2$

So

$mr_i^2 w_1 + mr_i^2 w_1 = mr_f^2 w_2 + mr_f^2 w_2$

=> $2 mr_i^2 w_1 = 2 mr_f^2 w_2$

=> $w_f = \frac{2 * m * r_i^2 w_1}{2 * m * r_f^2 }$

=> $w_f = \frac{3.5^2 * 0.5}{ 2^2 }$

=> $w_f = 1.531 \ rad/ s$

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2 years ago

Air rushing against an airplane is an example of _____friction.

slava [35]

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3 years ago

Suppose you lived in a pre-industrial society and needed to lift a heavy (20 kg) block a height of 5 m and had two choices for h

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Let's break the question into two parts:

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2) The effort force needed in the lever scenario.

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Since the height is 5m and the length of the ramp is 8m, $sin \alpha$ would be 5/8 or 0.625. Now that you have $sin \alpha$ , mutiple it with mg.

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Just apply "moment action" in this case, which is:
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$F_{e}$ = ?

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Plug-in the values in the above equation:
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3 years ago