As we know that spring force is given as
here we know that
F = 4 N
x = 2 cm = 0.02 m
now from the above equation we will have
so the elastic constant of the spring will be 200 N/m
.Ryan boiled a liter of water and then stirred sugar into it, adding more sugar until no more would dissolve in the water, creat
1 answer:
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Answer:
61440 peaks
Explanation:
A hertz represents one cycle for every second:
So if a wave have frequency of 2Hz for example, this means the wave does two cycles per second.
Normally the waves are represented by cosine or sine waves. These kind of waves have two peaks in each cycle, one positive and one negative. With this in mind, let's calculate how many peaks of the wave pass each minute.
A minute has 60 seconds, hence:
And we know already that every cycle has two peaks, so:
Answer:
Impulse = 88 kg m/s
Mass = 8.8 kg
Explanation:
<u>We are given a graph of Force vs. Time. Looking at the graph we can see that the Force acts approximately between the time interval from 1sec to 4sec. </u>
Newton's Second Law relates an object's acceleration as a function of both the object's mass and the applied net force on the object. It is expressed as:
Eqn. (1)
where
: is the Net Force in Newtons (
)
: is the mass (
)
: is the acceleration (
)
We also know that the acceleration is denoted by the velocity ( ) of an object as a function of time (
) with
Eqn. (2)
Now substituting Eqn. (2) into Eqn. (1) we have
Eqn. (3)
However since in Eqn. (3) the time-variable is present, as a result the left hand side (i.e. is in fact the Impulse
of the cart ), whilst the right hand side denotes the change in momentum of the cart, which by definition gives as the impulse. Also from the graph we can say that the Net Force is approximately ≈
and
(thus just before the cut-off time of the force acting).
Thus to find the Impulse we have:
So the impulse of the cart is
Then, we know that the cart is moving at . Plugging in the values in Eqn. (3) we have:
So the mass of the cart is .
Answer:
a) Please find attached the required drawing of light passing through the lens
By the use of similar triangles;
The image distance from the lens = 26 cm
The height of the image = 14 cm
c) The image distance from the lens = 26 cm
The height of the image = 14 cm
Explanation:
Question;
a) Determine the image distance and the height of the image
b) Calculate the image position and height
The given parameters are;
The height of the object, h = 14 cm
The distance of the object from the mirror, u = 26 cm
The focal length of the mirror, f = 13 cm
The location of the object = 2 × The focal length
Therefore, given that the center of curvature ≈ 2 × The focal length, we have;
The location of the object ≈ The center of curvature of the lens
The diagram of the object, lens and image created with MS Visio is attached
From the diagram, it can be observed, using similar triangles, that the image distance from the lens = The object distance from the lens = 26 m
The height of the image = The height of the object - 14 cm
b) The lens equation is used for finding the image distance from the lens as follows;
Where;
v = The image distance from the lens
We get;
Therefore;
The distance of the image from the lens, v = 26 cm
The magnification, M =v/u
∴ M = 26/26 = 1, therefore, the object and the image are the same size
Therefore;
The height of the image = The height of the object = 14 cm.
