First you must calculate the acceleration of the puck using F=ma. The force F is negative .14 N, the frictional force, and the mass is .12 kg. Solving for a=F/m, we get a=-1.17 m/s^2. Now we use a kinematic equation. vf^2=vi^2 + 2ax. We need to solve for x. We know the final speed vf is 0 because we are finding the stopping distance. Initial velocity vi is given as 18.3 m/s. We just found acceleration a is -1.17 m/s^2. Now we need to solve for displacement x, which is the answer to the question. Plugging in the appropriate values to the kinematic equation gives us 0=18.3^2 - 2(1.17)x. 2(1.17)x=18.3^2. x=143 m
Answer:
She covers the distance is 12 km.
The magnitude of displacement is 8.6 km.
The direction of her displacement is north east.
Explanation:
Given that,
Christina drives his moped 7 kilometers North and stop for lunch and then drive 5 km east.
We need to calculate the total distance
Using formula of distance
Put the value into the formula
We need to calculate the magnitude of displacement
Using formula of displacement
The direction of her displacement is north east.
Hence, She covers the distance is 12 km.
The magnitude of displacement is 8.6 km.
The direction of her displacement is north east.
So, physical properties are what we can detect with our basic 5 senses or measuring tools, and the things that, when changed, dont actually change the chemical properties (like atoms and molecules). Lets take wood for an example: its brown, its solid, it can be big or small, it has a taste and smell, its boiling, freezing or melting point...
Chemical properties, on the other hand, are the things we can change with, for example, experiments and tools. Does it burn? Can it rust/oxidize? How does it react with other chemicals? Is it radioactive, or toxic? All of these are chemical properties you can probably answer.
At the speed of light wich is approx 3 x 108 meters/ second
Sphere A travels a horizontal distance of (10 m/s) <em>t</em> after time <em>t</em>, while sphere B travels a distance of (5 m/s) <em>t</em>. So sphere B lands on the floor some point between the table's edge and the point X (A).