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3 years ago

When passing through a lock, which light means âapproach the lock under full control?â?

Physics

1 answer:

vekshin13 years ago

6 0

The locks referred to here are the elevators that are used to transport boats safely from one water level to another in dams. These two varying water depths allow river traffic to operate The attached picture shows how boats enter locks in dam sites.

To regulate traffic, there are traffic lights that signal boatmen to adjust their speed when approaching the lock. The red light means to stop and to steer clear away from the lock to allows the boats inside to exit. The green light signals to enter the lock. Lastly, the amber light means approach the lock at a safe speed and under full control.

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The magnitude of a negative of a vector is negative. O True O False

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Answer:

The given statement is false.

Explanation:

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$\overrightarrow{r}=-x\widehat{i}-y\widehat{j}$

The magnitude of the vector is given by

$|r|=\sqrt{(-x)^{2}+(-y)^{2}}\\\\|r|=\sqrt{x^2+y^2}$

As we know that square root of any quantity cannot be negative thus we conclude that the right hand term in the above expression cannot be negative hence we conclude that magnitude of any vector cannot be negative.

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den301095 [7]

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3 years ago

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A 45.0 kg girl is standing on a 159 kg plank. The plank, originally at rest, is free to slide on a frozen lake, which is a flat,

Ganezh [65]

Answer:

(a) $v_g_i=1.08\frac{m}{s}$

(b) $v_p_i=-0.3\frac{m}{s}$

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$\Delta p=0\\p_i=p_f$

Initially, the girl and the plank are at rest. So, relative to the ice, we have:

$0=m_gv_g_i+m_pv_p_i\\v_p_i=-\frac{m_gv_g_i}{m_p_i}(1)$

(a) The velocity of the girl relative to the ice is:

$v_g_i=v_g_p+v_p_i(2)$

Here, $v_g_p$ is the velocity of the girl relative to the plank and $v_p_i$ is the velocity of the plank relative to the ice.

Replacing (1) in (2):

$v_g_i=v_g_p-\frac{m_gv_g_i}{m_p_i}\\v_g_i+\frac{m_gv_g_i}{m_p_i}=v_g_p\\v_g_i(1+\frac{m_g}{m_p})=v_g_p\\v_g_i=\frac{v_g_p}{1+\frac{m_g}{m_p}}\\v_g_i=\frac{1.38\frac{m}{s}}{1+\frac{45kg}{159kg}}\\v_g_i=1.08\frac{m}{s}$

(b) According to (2), the velocity of the plank relative to the surface of ice is:

$v_p_i=v_g_i-v_g_p\\v_p_i=1.08\frac{m}{s}-1.38\frac{m}{s}\\v_p_i=-0.3\frac{m}{s}$

The negative sing indicates that the plank is moving to the left.

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3 years ago

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