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Arada [10]
3 years ago
9

A simple pendulum is used to determine the acceleration due to gravity at the surface of a planet. The pendulum has a length of

2 m and its period is measured to be 2 s. The value of g obtained in this investigation is most nearly __________.A. 1 m/s²B. 2 m/s²C. 5 m/s²D. 10 m/s²E. 20 m/s²
Physics
1 answer:
SVEN [57.7K]3 years ago
8 0

Answer:

Acceleration due to gravity is 20 m/sec^2

So option (E) will be correct answer

Explanation:

We have given length of the pendulum l = 2 m

Time period of the pendulum T = 2 sec

We have to find acceleration due to gravity g

We know that time period of pendulum is given by

T=2\pi \sqrt{\frac{l}{g}}

2=2\times 3.14 \sqrt{\frac{2}{g}}

0.3184= \sqrt{\frac{2}{g}}

Squaring both side

0.1014= {\frac{2}{g}}

g=19.71=20m/sec^2

So acceleration due to gravity is 20 m/sec^2

So option (E) will be correct answer.

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worty [1.4K]
The White colored star is likely to be the coldest
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3 years ago
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Ms. Kasper is in a panic. Her cat, Penny, is stuck in a tree and about to jump out. In order to save her cat, Ms. Kasper needs t
Misha Larkins [42]

Answer:

Speed greater than 4 m/s

Explanation:

Given that Ms. Kasper is in a panic. Her cat, Penny, is stuck in a tree and about to jump out. In order to save her cat, Ms. Kasper needs to run to the tree, 12 meters away. If it takes her cat, 3 seconds to fall, how fast would Ms. Kasper have to run to save her cat?

The distance = 12 m

Time = 3s

Speed = distance/time

Speed = 12/3

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Ms Kasper must run at speed more than 4m/s for her to save the cat.

3 0
3 years ago
©
Andru [333]

Answer:

Two positively charged particles

Explanation:

I said two positively charged particles because if I say c or d what ever it is for you guy it can be wrong so just pick the one that says Two positively charged particles

8 0
3 years ago
Sound waves have the ability to cause objects to vibrate. If a paperback book is placed near a speaker and the volume of the spe
erik [133]

Answer:

yes with a lot of time it eventualy could be, but in short term no

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4 0
2 years ago
A cutting tool several forces acting on it. One force is F=-axy^2 j , a force in the negative y-direction whose magnitude depend
liq [111]
The force on the tool is entirely in the negative-y direction.
So no work is done during any moves in the x-direction.

The work will be completely defined by

                     (Force) x (distance in the y-direction),

and it won't matter what route the tool follows to get anywhere. 
Only the initial and final y-coordinates matter.

We know that    F = - 2.85 y².  (I have no idea what that ' j ' is doing there.)
Remember that 'F' is pointing down.

From  y=0  to  y=2.40  is a distance of  2.40  upward.

Sadly, since the force is not linear over the distance, I don't think
we can use the usual formula for  Work = (force) x (distance).
I think instead we'll need to integrate the force over the distance,
and I can't wait to see whether I still know how to do that.

        Work  =  integral of (F·dy) evaluated from  0  to  2.40

                  =  integral of (-2.85 y² dy) evaluated from  0  to  2.40

                 =  (-2.85) · integral of  (y² dy)  evaluated from  0  to  2.40 .


Now, integral of (y² dy)  =  1/3  y³ .

Evaluated from  0  to  2.40 , it's  (1/3 · 2.40³) - (1/3 · 0³)

                                            =  1/3 · 13.824  =  4.608 .

And the work  =  (-2.85) · the integral

                     =  (-2.85) · (4.608)

                     =      - 13.133  .

-- There are no units in the question (except for that mysterious ' j ' after the 'F',
which totally doesn't make any sense at all).
If the ' F ' is newtons and the 2.40 is meters, then the  -13.133 is joules.

-- The work done by the force is negative, because the force points
DOWN but we lifted the tool UP to 2.40.  Somebody had to provide
13.133 of positive work to lift the tool up against the force, and the force
itself did 13.133 of negative work to 'allow' the tool to move up. 

-- It doesn't matter whether the tool goes there along the line  x=y , or
by some other route.  WHATEVER the route is, the work done by ' F ' 
is going to total up to be  -13.133 joules at the end of the day.

As I hinted earlier, the last time I actually studied integration was in 1972,
and I haven't really used it too much since then.  But that's my answer
and I'm stickin to it.  If I'm wrong, then I'm wrong, and I hope somebody
will show me where I'm wrong.
3 0
3 years ago
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