If the skater coasts only around 65 degrees of the circle, find the magnitude of his displacement vector.

1 answer:

4 0

We are given a skater who coasts only around 65 degrees of the circle. The magnitude of the vector is determined by converting 65 degrees to radians:

65//180 = 0.36 radians

The displacement of the vector is 65 degrees east.

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A 5 kg fish swimming at 1 m/s swallows an absentminded 500 g fish swimming toward it at a velocity that brings both fish to a ha

AlexFokin [52]

To solve this problem we will apply the concepts related to the conservation of momentum. Momentum is defined as the product between mass and velocity of each body. And its conservation as the equality between the initial and final momentum. Mathematically described as

$m_1u_1+m_2u_2 = (m_1+m_2)v_f$

Here

$m_1$ = Mass of big fish

$m_2$ = Mass of small fish

$v_1$ = Velocity of big fish

$v_2$ = Velocity of small fish

$v_F$ = Final Velocity

The big fish eats small fish and the final velocity is zero. Rearrange the equation for the initial velocity of small fish we have

$m_1u_1=-m_2u_2$

$u_2 = -\frac{m_1u_1}{m_2}$

Replacing we have,

$u_2 = -\frac{(5kg)(1m/s)}{0.5kg}$

$u_2 = -10m/s$

The negative sign indicates that the small fish is swimming in the direction opposite to that of the big fish.

Therefore the speed of the small fish is 10m/s

8 0

3 years ago

Harry and ron set up this experiment with a glider, whose mass they have measured to be 565 g, and seven washers hanging from th

svetlana [45]

Let's call $m=565~g=0.565~kg$ the mass of the glider and $m_w=7\cdot12~g =84~g=0.084~kg$ the total mass of the seven washers hanging from the string.
The net force on the system is given by the weight of the hanging washers:
$F_{net} = m_w g$
For Newton's second law, this net force is equal to the product between the total mass of the system (which is $m+m_w$ ) and the acceleration a:
$F_{net}=(m+m_w)a$
So, if we equalize the two equations, we get
$m_w g = (m+m_w)a$
and from this we can find the acceleration:
$a= \frac{m_w g}{(m+m_w)} = \frac{0.084~kg \cdot 9.81~m/s^2}{(0.565~kg+0.084~kg)}=1.27~m/s^2$

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3 years ago

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Yakvenalex [24]

Answer:

Acceleration will be $-1.620m/sec^2$