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30.11.2020 •
Mathematics
A hockey puck with a mass of 0.18 kg is at rest on the horizontal frictionless surface of the rink. A player applies a horizontal force of 0.5 N to the puck. Find the speed and the traveled distance 5 s later.
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Ответ:
1) The speed after 5 seconds is 13.
m/s
2) The distance traveled after 5 seconds is 34.7
meters
Step-by-step explanation:
1) The given parameters are;
The mass of the hockey puck, m = 0.18 kg
The horizontal force applied by the player, m = 0.5 N
The time at which the speed is calculated, t = 5 seconds after the force is applied
Therefore;
From the equation for the applied force, F = m × a. where a represents the acceleration, by substituting the known values get;
0.5 = 0.18 × a
∴ a = 0.5/0.18 = 50/18 = 25/9 = 2.![\bar 7](/tpl/images/0935/5230/93299.png)
a = 2.
m/s²
The speed after 5 seconds is given from the kinematic equation, v = u + a·t
Where;
v = The final velocity
u = The initial velocity = 0 m/s
t = The time = 5 seconds
By substitution of the known values, we get;
v = 0 + 2.
× 5 = 13.![\bar 8](/tpl/images/0935/5230/e7094.png)
The speed after 5 seconds = 13.
m/s
2) The distance traveled, s, after 5 seconds is given by the equation, s = u·t + 1/2·a·t²
From which we have;
s = 0 × 5 + 1/2 × 2.
× 5² = 34.7![\bar 2](/tpl/images/0935/5230/2cf26.png)
The distance traveled after 5 seconds = 34.7
meters
Ответ:
5
Step-by-step explanation: