A 2.70 kg cat is sitting on a windowsill. The cat is sleeping peacefully until a dog barks at him. Startled, the cat falls from
1 answer:
Answer:
The speed of the cat when it hits the ground is approximately 7.586 meters per second.
Explanation:
By Principle of Energy Conservation and Work-Energy Theorem, we have that initial potential gravitational energy of the cat (), in joules, is equal to the sum of the final translational kinetic energy (
), in joules, and work losses due to air resistance (
), in joules:
(1)
By definition of potential gravitational energy, translational kinetic energy and work, we expand the equation presented above:
(2)
Where:
- Mass of the cat, in kilograms.
- Gravitational acceleration, in meters per square second.
- Initial height of the cat, in meters.
- Final speed of the cat, in meters per second.
If we know that ,
,
and
, then the final speed of the cat is:
The speed of the cat when it hits the ground is approximately 7.586 meters per second.
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Answer:
a)
b)
Step-by-step explanation.
In order to solve this problem, we mus start by plotting the given points and charges. That will help us visualize the problem better and determine the direction of the forces (see attached picture).
Once we drew the points, we can start calculating the forces:
which yields:
(I will assume the positions are in meters)
Next, we can make use of the force formula:
so we substitute the values:
which yields:
Now we can find its components:
And we can now write them together for the first force, so we get:
We continue with the next force. The procedure is the same so we get:
which yields:
Next, we can make use of the force formula:
which yields:
Now we can find its components:
And we can now write them together for the second, so we get:
We continue with the next force. The procedure is the same so we get:
which yields:
Next, we can make use of the force formula:
which yields:
Now we can find its components:
And we can now write them together for the third force, so we get:
So in order to find the resultant force, we need to add the forces together:
so we get:
So when adding the problem together we get that:
which is the answer to part a), now let's take a look at part b).
b)
Basically, we need to find the magnitude of the force and divide it into the test charge, so we get:
which yields:
and now we take the formula for the electric field which is:
so we go ahead and substitute:
Answer: The observing friend will the swimmer moving at a speed of 0.25 m/s.
Explanation:
- Let <em>S</em> be the speed of the swimmer, given as 1.25 m/s
- Let
be the speed of the river's current given as 1.00 m/s.
- Note that this speed is the magnitude of the velocity which is a vector quantity.
- The direction of the swimmer is upstream.
Hence the resultant velocity is given as, = S — S 0
= 1.25 — 1
= 0.25 m/s.
Therefore, the observing friend will see the swimmer moving at a speed of 0.25 m/s due to resistance produced by the current of the river.
The tension in the string B) It quadruples.
Explanation:
The ball is in uniform circular motion in a horizontal circle, so the tension in the string is providing the centripetal force that keeps the ball in circular motion. So we can write:
where:
T is the tension in the string
m is the mass of the ball
v is the speed of the ball
r is the radius of the circle (the lenght of the string)
In this problem, we are told that the speed of the ball is doubled, so
v' = 2v
Substituting into the previous equation, we find the new tension in the string:
Therefore, the tension in the string will quadruple.
Learn more about circular motion:
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