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julsineya [31]
3 years ago
5

You are designing a pendulum clock to have a period of 1.0 s. The acceleration of gravity is 9.81 m/s2 . How long should the pen

dulum be? Answer in units of m.
Physics
1 answer:
tatuchka [14]3 years ago
5 0

Answer:

0.25 m

Explanation:

The period of a pendulum is given by

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

where  in this case

T = 1.0 s is the period of the pendulum

g = 9.81 m/s^2 is the acceleration due to gravity

L is the length of the pendulum

Re-arranging the equation and solivng for L, we find the length of the pendulum:

L=g (\frac{T}{2 \pi})^2=(9.81 m/s^2)(\frac{1.0 s}{2\pi})^2=0.25 m

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3. A 142 g baseball is thrown at a speed of 42.9 m/s. What is the kinetic energy of the baseball at this moment?
krok68 [10]
Kinetic energy=1/2mv^2
=1/2(142*10^-3)(42.9)^2=130.6=131J
8 0
3 years ago
A particle executes simple harmonic motion with an amplitude of 2.18 cm.
Bogdan [553]

Answer:

The positive displacement from the midpoint of its motion at the speed equal one half of its maximum speed is 3.56 cm.

Explanation:

Maximum speed is  :

                          v (max) = Aω

Speed v at any displacement y is given by  

v^{2} = w^{2} (A^{2} - y^{2})   ........................................................  i

And,

               v = \frac{1}{2} v (max)  

          or,  2 × v = Aω     ....................................................   ii

Eliminating  ω from equations i and ii,

                       \frac{1}{4} A^{2}  w^{2}  =  w^{2}  ( A^{2}  - y^{2})

                     or, y^{2} =  (\frac{3}{4}) A^{2}  =(\frac{3}{4}) 2.18^{2}

                    or,  y =  3.56 cm.

3 0
3 years ago
If you push a crate across a factory floor at constant speed in a constant direction, what is the magnitude of the force of fric
poizon [28]

Answer:

The magnitude of the force of friction equals the magnitude of my push

Explanation:

Since the crate moves at a constant speed, there is no net acceleration and thus, my push is balanced by the frictional force on the crate. So, the magnitude of the force of friction equals the magnitude of my push.

Let F = push and f = frictional force and f' = net force

F - f = f' since the crate moves at constant speed, acceleration is zero and thus f' = ma = m (0) = 0

So, F - f = 0

Thus, F = f

So, the magnitude of the force of friction equals the magnitude of my push.

3 0
3 years ago
Please Answer the question in the picture ASAP PLEASE
attashe74 [19]

Answer:

HERE IS YOUR ANSWER

Explanation:

PLEASE MARK MY ANSWER AS BRAINLIEST IF THE ANSWERS ARE CORRECT .

Beacuse of the loose connection of the wire .

Straight

5 0
3 years ago
A cube 6.0 cm on each side is made of a metal alloy. After you drill a cylindrical hole 2.0 cm in diameter all the way through a
Crank

To solve this problem it is necessary to apply the concepts related to Newton's second law, the definition of density and the geometric relationships that allow us to find the volume of the figures presented.

For the particular case of the Cube with equal sides its volume is determined by

V_c = l^3

V_c = 6^3 = 216cm^3

In the case of perforated material we have that its volume is given according to the cylindrical geometry, that is to say

V_d = \pi r^2*l

V_d = \pi (\frac{2}{2})^2*6

V_d = 6\pi cm^3

In this way the net volume would be

\Delta V = V_c-V_d

\Delta V = 216cm^3-6\pi cm^3

\Delta V = 197.15cm^3 = 197.15*10^{-6}m^3

We need to find the mass, but we have the Weight and Gravity so from Newton's second Law

F= mg

m = \frac{F}{g}

m = \frac{6.6}{9.8}

m = 0.673kg

PART A) From the relation of density as a unit of mass and volume we have to

\rho = \frac{m}{V}

\rho = \frac{0.673}{197.15*10^{-6}}

\rho = 3413.64kg/m^3

PART B) To find the weight of the cube then we apply the ratio of

W = mg

W = V\rho g

W = (216*10^{-6})(3413.64)(9.8)

W = 7.22N

3 0
4 years ago
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