The length of a 2 sec pendulum is 1 m.
Given that, initial length of the simple pendulum L₁ = 1 m
Initial time period T₁ = 2 sec
We need to find the length of the pendulum whose time period is 2 sec
T₂ = 2 sec
L₂ = ?
We know that the time period of the simple pendulum is given by the formula,
T = 2π√(L/g)
From the above relation, we can write T ∝ √L
T₁ / T₂ = √(L₁/L₂)
Making L₂ from the above relation, we have,
L₂ = (T₂² * L₁)/ T₁² = 2² * 1/ 2² = 1 m
Thus, the length of a 2 sec pendulum is 1 m.
To know more about time period:
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I see the word "when..." kind of fading out at the end of the first line.
Whatever comes after it may be important.
If you're just supposed to copy the expression into the box,
then the problem is that you left the 'e' out of it.
I'm guessing that you're supposed to enter whatever the expression becomes
when either N₀ or ' t ' has some special value that's in the first line.
Just taking a wild guess here . . . . .
If it's "Enter the expression ..... , when t=0 ." ,
then the correct answer in the box is N₀ .
But that's just a wild guess. As I pointed out, you cut off
the picture in the middle of the word 'when', and I've got
a hunch that there's something important after it.
1) 4°C : It has the highest density as shown on the graph.
2) Water expands when it freezes, making it less dense than just water.
3) The ice would sink to the bottom, then the rest of the water would freeze as well, the entire lake/river/whatever will freeze eliminating the organisms that live there.
Answer:
619.8 N
Explanation:
The tension in the string provides the centripetal force that keeps the rock in circular motion, so we can write:
where
T is the tension
m is the mass of the rock
v is the speed
r is the radius of the circular path
At the beginning,
T = 50.4 N
v = 21.1 m/s
r = 2.51 m
So we can use the equation to find the mass of the rock:
Later, the radius of the string is decreased to
r' = 1.22 m
While the speed is increased to
v' = 51.6 m/s
Substituting these new data into the equation, we find the tension at which the string breaks:
Answer:
1) In a concave mirror parallel rays falling on it converges at F and 2F.
Explanation:
Spherical mirrors can be used for magnification of images. There are basically two types of spherical mirrors and they are converging mirror and diverging mirrors. The converging mirrors are also termed as concave mirrors and its basic work is to converge or combine light rays coming from a larger distance to a single point. Mostly the light beams falling parallel to the principle axis of the concave mirror will be acting as parallel rays. And when these parallel rays fall on the mirror, the converging point can be the focal point of the mirror.
Thus the location of converging point in concave mirrors will be based on the position or distance of object from the mirror. If the object distance is very far from the twice the focal length distance of mirror, then the converging point will be the focal point or F. And if the object is placed slightly greater than twice the distance of focal point, then the image will be obtained at 2F. But the parallel beams will be converging at F and 2F.
