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lesantik [10]
3 years ago
5

1. A pendulum has a period of 3 seconds. What's its frequency? 2. A pendulum has a frequency of 0.25 Hz. What's its period? 3. A

pendulum makes 10 back and forth swings in 20 seconds. What's its period AND frequency?
Physics
2 answers:
RideAnS [48]3 years ago
6 0

Answer:

(1) 0.333 Hz

(2) 4 sec

(3) 2 sec, 0.5 Hz

Explanation:

(1) We have given time period of pendulum is 3 sec

So T = 3 sec

Frequency will be equal to f=\frac{1}{T}=\frac{1}{3}=0.333Hz

(2) Frequency of the pendulum is given f = 0.25 Hz

Time period is equal to T=\frac{1}{f}=\frac{1}{0.25}=4sec

(3) It is given that a pendulum makes 10 back and forth swings in 20 seconds

So time taken to complete 1 back and forth swings = =\frac{20}{10}=2sec

So time period T = 2 sec

Frequency will be equal to f=\frac{1}{T}=\frac{1}{2}=0.5Hz

andre [41]3 years ago
6 0

Answer:

(1) 0.33 Hz

(2) 4 s

(3) 2 s, 0.5 Hz

Explanation:

The relation between frequency and time period is given as

f=\frac{1}{T}

(1) Given a pendulum has a period, T = 3 s.

So,  f=\frac{1}{3s}

f = 0.333 Hz.

Thus, frequency of pendulum is 0.33 Hz.

(2) Given a pendulum has a frequency, f = 0.25 Hz.

So, T=\frac{1}{0.25 Hz}

T = 4 s.

Thus , period of pendulum is 4 s.

(3) Frequency is measured in cycles/second.

Given a pendulum makes 10 back and forth swings in 20 seconds.

So,f=\frac{10 cycles}{20s}

f = 0.5 Hz

Time period,  

T=\frac{1}{0.5}

T = 2 s.

Thus, period and frequency of pendulum are 2 s and 0. 5 Hz respectively.  

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A person holding a lunch bag is moving upward in a hot air balloon at a constant speed of 7.3 m/s . When the balloon is 24 m abo
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Explanation:

Given that,

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3 years ago
One end of a thin rod is attached to a pivot, about which it can rotate without friction. Air resistance is absent. The rod has
Mars2501 [29]

Answer:

6.86 m/s

Explanation:

This problem can be solved by doing the total energy balance, i.e:

initial (KE + PE)  = final (KE + PE). { KE = Kinetic Energy and PE = Potential Energy}

Since the rod comes to a halt at the topmost position, the KE final is 0. Therefore, all the KE initial is changed to PE, i.e, ΔKE = ΔPE.

Now, at the initial position (the rod hanging vertically down), the bottom-most end is given a velocity of v0. The initial angular velocity(ω) of the rod is given by ω = v/r , where v is the velocity of a particle on the rod and r is the distance of this particle from the axis.

Now, taking v = v0 and r = length of the rod(L), we get ω = v0/ 0.8 rad/s

The rotational KE of the rod is given by KE = 0.5Iω², where I is the moment of inertia of the rod about the axis of rotation and this is given by I = 1/3mL², where L is the length of the rod. Therefore, KE = 1/2ω²1/3mL² = 1/6ω²mL². Also, ω = v0/L, hence KE = 1/6m(v0)²

This KE is equal to the change in PE of the rod. Since the rod is uniform, the center of mass of the rod is at its center and is therefore at a distane of L/2 from the axis of rotation in the downward direction and at the final position, it is at a distance of L/2 in the upward direction. Hence ΔPE = mgL/2 + mgL/2 = mgL. (g = 9.8 m/s²)

Now, 1/6m(v0)² = mgL ⇒ v0 = \sqrt{6gL}

Hence, v0 = 6.86 m/s

4 0
3 years ago
An office building has a 24-volt branch circuit installed for landscape lighting around the front of the building. The circuit w
Arturiano [62]

The circuit was installed in UF cable which requires a minimum burial depth of 6 inches for this circuit.

<h3>UF cable</h3>

UF cable is used as an underground feeder cable to distribute power from an existing building to outdoor equipment. UF cable can also be used as direct burial cable.

For the 24-volt branch circuit installed, the minimum burial depth will be 6 inches.

Thus, the circuit was installed in UF cable which requires a minimum burial depth of 6 inches for this circuit.

Learn more about UF cable here: brainly.com/question/8591560

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