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3 years ago

7

While goofing off at the ice skating rink, a student takes off her shoes and places each of them on the ice. Her friend, a hocke

y player, then shoots a hockey puck at each shoe. The first puck immediately comes to rest after it collides with the left shoe. The second puck rebounds after it collides with the right shoe. If each hockey puck has the same incoming speed, which shoe has greater speed after the collision?

1 answer:

sukhopar [10]3 years ago

6 0

Answer:

Right shoe

Explanation:

Let the mass and velocity of incoming puck be m and v respectively.

Momentum of the colliding puck will be mv

In case of first case , the momentum of puck becomes zero so change in momentum after collision with left shoe

= mv - 0 = mv

If time duration of collision be t

rate of change of momentum

= mv / t

This is the force exerted by puck on the left shoe .

Now let us consider collision with right shoe

momentum after collision with right shoe

- mv

change in momentum

= mv - ( - mv ) = 2mv

If time duration of collision be t

rate of change of momentum

= 2mv / t

This is the force exerted by puck on the right shoe .

Since the force on the right shoe is more , this shoe will have greater speed

after collision.

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Kay [80]

Answer:

The true course: $40.29^\circ$ north of east

The ground speed of the plane: 96.68 m/s

Explanation:

Given:

$V_w$ = velocity of wind = $30\ km/h\ S45^\circ E = (30\cos 45^\circ\ \hat{i}-30\sin 45^\circ\ \hat{j})\ km/h = (21.21\ \hat{i}-21.21\ \hat{j})\ km/h$

$V_p$ = velocity of plane in still air = $100\ km/h\ N60^\circ E = (100\cos 60^\circ\ \hat{i}+100\sin 60^\circ\ \hat{j})\ km/h = (50\ \hat{i}+86.60\ \hat{j})\ km/h$

Assume:

$V_r$ = resultant velocity of the plane

$\theta$ = direction of the plane with the east

Since the resultant is the vector addition of all the vectors. So, the resultant velocity of the plane will be the vector sum of the wind velocity and the plane velocity in still air.

$\therefore V_r = V_p+V_w\\\Rightarrow V_r = (50\ \hat{i}+86.60\ \hat{j})\ km/h+(21.21\ \hat{i}-21.21\ \hat{j})\ km/h\\\Rightarrow V_r = (71.21\ \hat{i}+65.39\ \hat{j})\ km/h$

Let us find the direction of this resultant velocity with respect to east direction:

$\theta = \tan^{-1}(\dfrac{65.39}{71.21})\\\Rightarrow \theta = 40.29^\circ$

This means the the true course of the plane is in the direction of $40.29^\circ$ north of east.

The ground speed will be the magnitude of the resultant velocity of the plane.

$\therefore Magnitude = \sqrt{71.21^2+65.39^2} = 96.68\ km/h$

Hence, the ground speed of the plane is 96.68 km/h.

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The motion of an object undergoing constant acceleration can be modeled by the kinematic equations. One such equation is xf=xi+v

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Answer:

a = 1.72 m/s²

Explanation:

The given kinematic equation is the 2nd equation of motion. The equation is as follows:

xf = xi + (Vi)(t) + (1/2)(a)t²

where,

xf = the final position = 5000 m

xi = the initial position = 1000 m

Vi = the initial velocity = 15 m/s

t = the time taken = 60 s

a = acceleration = ?

Therefore,

5000 m = 1000 m + (15 m/s)(60 s) + (1/2)(a)(60 s)²

5000 m = 1000 m + 900 m + a(1800 s²)

5000 m = 1900 m + a(1800 s²)

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a(1800 s²) = 3100 m

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Part (a):
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2- Since the two resistors are in series, therefore, the current flowing in both is the same. We will use ohm's law to get the current as follows:
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Therefore:
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Part (b):
To get the voltage across the second resistor, we will again use Ohm's law as follows:
V = I*R
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I is the current in the second resistor = 2.59 ampere
R is the value of the second resistor = 6.15 ohm
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