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igor_vitrenko [27]
3 years ago
15

A hockey player hits a puck so that it comes to rest in 9 s after sliding 30 m on the ice. Determine (a) the acceleration in ter

ms of ,​(4 marks) (b) the initial velocity and acceleration of the puck.​(5 marks)
Physics
1 answer:
blondinia [14]3 years ago
4 0

Answer:

a) Acceleration, a = -u/9, where u is the initial velocity

b) Acceleration = 0.741 m/s², Initial velocity = 6.67 m/s

Explanation:

 We have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.

a)  We have v = 0 m/s, t = 9 s

 So,    0 = u + a x 9

      Acceleration, a = -u/9, where u is the initial velocity

b) We have equation of motion , s= ut+\frac{1}{2} at^2, s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

 s = 30 m , u = -9a, t =9 s

 So,  30= -9a*9+\frac{1}{2} *a*9^2\\ \\ 30=-40.5a\\ \\ a=-0.741m/s^2

u = -9a = 6.67 m/s

So acceleration = 0.741 m/s², Initial velocity = 6.67 m/s

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Salsk061 [2.6K]

Answer:

The resultant electric force is 14.8N to the right.

Explanation:

Since the three charges aren't in the same line, we have to break down the force in components. First, we need to know the distance from the third charge to the other ones. That is made using the Pythagorean Theorem. As the figure is symmetric with respect to the x-axis, the two distances are the same:

r=\sqrt{(0.50m)^{2}+(0.70m)^{2}}=0.86m

Now, we use the Coulomb's Law to obtain the magnitude of the individual forces caused by each charge on the third charge:

|F_{13}|=k\frac{q_1q_3}{r^{2}} \\\\|F_{13}|=(9*10^{9}Nm^{2}/C^{2})\frac{(2.5*10^{-9}C)(3.0*10^{-9}C)}{(0.86m)^{2}}\\\\|F_{13}|=9.1N

For the same reason the distances are the same, the magnitude of the forces are the same:

|F_{23}|=|F_{13}|=9.1N

So, to get the resultant force, we have to break down this forces in components. To do this, we need their angles with respect to the x-axis. Let θ₁ and θ₂ be these angles, respectively. Then, we calculate them using trigonometry:

\theta_1=\arctan(\frac{-0.50m}{0.70m})=-35.5\°\\\\\theta_2=\arctan(\frac{0.50m}{0.70m})=35.5\°

Now, we calculate the components of the forces:

F_{13}_x=F_{13}\cos\theta_1=9.1N\cos(-35.5\°)=7.4N\\\\F_{13}_y=F_{13}\sin\theta_1=9.1N\sin(-35.5\°)=-5.3N\\\\F_{23}_x=F_{23}\cos\theta_2=9.1N\cos(35.5\°)=7.4N\\\\F_{23}_y=F_{23}\sin\theta_2=9.1N\sin(35.5\°)=5.3N

Evidently, the y-components cancel out, and the resultant electric force on the third charge is 7.4N+7.4N=14.8N along the x-axis (to the right, because it's positive).

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Answer:

When force is applied to an object it causes to to accelerate.

Explanation:

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svet-max [94.6K]

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y=y_0+v_{0y}t+\dfrac12a_yt^2

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Meanwhile, the horizontal component of the ball's position vector is

x=x_0+v_{0x}t+\dfrac12a_xt^2

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8 0
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Answer:

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