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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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Dominik [7]

Answer:

i Believe the correct answer is "An open circuit being changed into a closed circuit"

Explanation:

8 0
3 years ago
A 30kg boxed is pushed with a force of 20N. What is the boxes acceleration. Please show work
kozerog [31]

Answer:

<h3>The answer is 0.67 m/s²</h3>

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

a =  \frac{f}{m}  \\

f is the force

m is the mass

From the question we have

a =  \frac{20}{30}  =  \frac{2}{3}  \\  = 0.6666666...

We have the final answer as

<h3>0.67 m/s²</h3>

Hope this helps you

8 0
3 years ago
A reversible Carnot engine has an efficiency of 30% and it operates between a heat source maintained at T 1 , and a refrigerator
bixtya [17]

To solve this problem we will apply the concepts related to the thermal efficiency given in an engine of the Carnot cycle. Here we know that efficiency is given under the equation

\eta = 1 - \frac{T_2}{T_1}

Where,

T_2 = Temperature of Cold Body

T_1 =Temperature of Hot Body

\eta= Efficiency

According to the statement our values are:

\eta = 0.3

T_2 = 210K

Replacing we have that

0.3 = 1- \frac{210}{T_1}

\frac{210}{T_1} = 0.7

T_1 = \frac{210}{0.7}

T_1 = 300K

Therefore the temperature of the heat source is 300K

6 0
3 years ago
A 0.55 kg projectile is launched from the edge of a cliff with an initial kinetic energy of 1550 J and at its highest point is 1
SCORPION-xisa [38]

Answer:

a). 53.78 m/s

b) 52.38 m/s

c) -75.58 m

Explanation:

See attachment for calculation

In the c part, The negative distance is telling us that the project went below the lunch point.

5 0
2 years ago
A moving van travels 10km North, then 4 km east, drops off some furniture and then drives 8 km south. (a) Sketch the path of the
Juli2301 [7.4K]

Answer:

4.47 km

Explanation:

If we draw the path of the van then we get a shape with two exposed points A and D. If we draw a line from point D perpendicular to BA we get point E. This gives us a right angled triangle ADE.

From Pythagoras theorem

AD² = AE² + ED²

AD=\sqrt{AE^2+ED^2}\\\Rightarrow AD=\sqrt{2^2+4^2}\\\Rightarrow AD=\sqrt{20}\\\Rightarrow AD=4.47\ km

Hence, the van is 4.47 km from its initial point

3 0
2 years ago
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