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laiz [17]
3 years ago
6

A simple pendulum consists of a point mass suspended by a weightless, rigid wire in a uniform gravitation field. Which of the fo

llowing statements are true when the system undergoes small oscillations? Check all that apply. a. The period is independent of the length of the wire. b. The period is inversely proportional to the length of the wire. c. The period is independent of the suspended mass. d. The period is proportional to the suspended mass. e. The period is proportional to the square root of the length of the wire. f. The period is inversely proportional to the suspended mass.
Physics
1 answer:
madreJ [45]3 years ago
6 0

Answer:

Option (c) and (e)

Explanation:

In case of a simple pendulum having suspended mass 'M', if the length of the wire is 'L', and the acceleration due to gravity that acts on it is 'g', then the time period, 'T' of the pendulum for one complete oscillation is given by:

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

From the above equation, we can say that:

  • Time period of the pendulum does not depend on the suspended mass.
  • Time period is in direct proportion to the square root of the  wire length
  • Time period is depends inversely on the acceleration due to gravity.
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Collar P slides outward at a constant relative speed along rod AB, which rotates counterclockwise with a constant angular veloci
diamong [38]

Answer:

a= 23.65 ft/s²

Explanation:

given

r= 14.34m

ω=3.65rad/s

Ф=Ф₀ + ωt

t = Ф - Ф₀/ω

= (98-0)×\frac{\pi}{180}/3.65

98°= 1.71042 rad

1.7104/3.65

t= 0.47 s

r₁(not given)

assuming r₁ =20 in

r₁ = r₀ + ut(uniform motion)

u = r₁ - r₀/t

r₀ = 14.34 in= 1.195 ft

r₁ = 20 in = 1.67 ft

= (1.667 - 1.195)/0.47

0.472/0.47

u= 1.00ft/s

acceleration at collar p

a=rω²

= 1.67 × 3.65²

a = 22.25ft/s²

acceleration of collar p related to the rod = 0

coriolis acceleration = 2ωu

= 2× 3.65×1 = 7.3 ft/s²

acceleration of collar p

= 22.5j + 0 + 7.3i

√(22.5² + 7.3²)

the magnitude of the acceleration of the collar P just as it reaches B in ft/s²

a= 23.65 ft/s²

4 0
3 years ago
The electric field is measured for points at distances r from the center of a uniformly charged insulating sphere that has volum
RoseWind [281]

For the electric field is measured for points at distances r,Electric field  is mathematically given as

E=19.9*10^{-5}c/m^3

<h3>What is the Electric field?</h3>

Generally, the equation for the  electric field uniform charged is mathematically given as

E=\frac{lr}{3e0}

Therefore

E=lr/3E0=3Ee0/r

Therefore

E=3*6*10^4*8.85*10^{-12}

E=19.9*10^{-5}c/m^3

In conclusion, E=19.9*10^{-5}c/m^3

E=19.9*10^{-5}c/m^3

Read more about Electric field

brainly.com/question/9383604

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Complete question attached below

4 0
2 years ago
An ideal gas is confined within a closed cylinder at atmospheric pressure (1.013 * 105 Pa) by a piston. The piston moves until t
likoan [24]

Answer:

911700\ \text{Pa}

Explanation:

P_1 = Initial pressure = 1.013\times 10^5\ \text{Pa}

V_1 = Initial volume

V_2= Final volume = \dfrac{V_1}{9}\\\Rightarrow \dfrac{V_1}{V_2}=9

Temperature is the same in the initial and final state

From the ideal gas law we have

P_1V_1=P_2V_2\\\Rightarrow P_2=\dfrac{P_1V_1}{V_2}\\\Rightarrow P_2=P_1\times9\\\Rightarrow P_2=1.013\times 10^5\times 9\\\Rightarrow P_2=911700\ \text{Pa}

The final pressure of the system is 911700\ \text{Pa}.

5 0
3 years ago
A fountain of water and steam that erupts from the ground
Roman55 [17]

geyser.........................................



















3 0
3 years ago
Read 2 more answers
Steel train rails are laid in 12.0-m-long segments placed end to end. The rails are laid on a winter day when their temperature
Yanka [14]

Answer:

a) Space = 6.05 x 10^{-3} m = 0.605 cm

b) Stress= -100.8 x 10^{6} Pa

Explanation:

<em>1) Data Given</em>

L = 12 m , T_i = -9 C \degree, T_f = 33 C \degree

<em>2) Calculate the space using Linear thermal expansion formula</em>

We need to use Linear thermal expansion formula since the space created would be a change on 1 dimension, the increase of the temperature will increase the length of the steel.  The formula is given by:

\Delta L = L_i \alpha_{steel} \Delta T

We have everything except the \alpha_{steel} , so we look for this on a book and we find that \alpha_{steel} = 1.2 x 10^{-5} C^{-1}, so we can replace.

\Delta L = 12 m (1.2 x 10^{-5} C^{-1}) (33 C \degree -(-9 C \degree)) = 12 m (1.2 x 10^{-5} C^{-1}) 42 C \degree =6.048 x 10^{-3} m = 0.6048cm

<em>3) Calculate the stress of the steel </em>

The Stress is the ratio of applied force F to a cross section area - defined as

\sigma = \frac{F_n}{A}

Since we don't have the force and the Area, we need to look for another way to find the stress.

For this we can use the concept called Young's Modulus, defined as : "the mechanical property that measures the stiffness of a solid material", and the formula for this is given by:

Y =\frac{F L}{A \Delta L} (1)

Solving \frac{F}{A} from the previous formula we have this:

\frac{F}{A}  = (Y  Δ L)/L  (2)

From the <em>Linear thermal expansion formula</em> we can solve like this

\frac{\Delta L}{L} =  α  ΔT  (3)

And replacing equation (3) into equation (2) we have:

\frac{F}{A}  = Y α ΔT (4)

We have that the Young's Modulus for the steel is 20x10^{10} Pa, so replacing into equation (4)

\frac{F}{A} = 20x10^{10} Pa (1.2x10^-5 C^-1) (42C) = 100.8 *10^{6} Pa  

That represent the absolute value for the Stress, the sign on this case would be negative since there is a compression.

3 0
3 years ago
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