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Vedmedyk [2.9K]
2 years ago
10

A sinusoidal transverse wave travels along a long, stretched, string. the amplitude of this wave is 0.0885 m, it's frequency is

4.31 hz, and its wavelength is 1.21m.
(a) What is the transverse distance between a maximum and a minimum of the wave?
(b) How much time is required for 71.7 cycles of the wave to pass a stationary observer?
(c) Viewing the whole wave at any instant, how many cycles are there in a 30.7-m length of string?
Physics
1 answer:
timofeeve [1]2 years ago
6 0

Answer:

(a) 0.177 m

(b) 16.491 s

(c) 25 cycles

Explanation:

(a)

Distance between the maximum and the minimum of the  wave = 2A ............ Equation 1

Where A = amplitude of the wave.

Given: A = 0.0885 m,

Distance between the maximum and the minimum of the wave = (2×0.0885) m

Distance between the maximum and the minimum of the wave = 0.177 m.

(b)

T = 1/f ...................... Equation 2.

Where T = period, f = frequency.

Given: f = 4.31 Hz

T = 1/4.31

T = 0.23 s.

If 1 cycle pass through the stationary observer for 0.23 s.

Then, 71.7 cycles will pass through the stationary observer for (0.23×71.7) s.

= 16.491 s.

(c)

If  1.21 m contains  1 cycle,

Then, 30.7 m will contain (30.7×1)/1.21

= 25.37 cycles

Approximately 25 cycles.

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The distance between two particles is 2 centimeters. If the distance is increased to 4 centimeters, the force will be ?
postnew [5]

Answer:

The new force is 1/4 of the previous force.

Explanation:

Given

Initial\ Distance = 2cm ---- r_1

New\ Distance = 4cm --- r_2

Required

Determine the new force

Let the two particles be q1 and q2.

The initial force F1 is:

F_1 = \frac{kq_1q_2}{r_1^2} --- Coulomb's law

Substitute 2 for r1

F_1 = \frac{kq_1q_2}{2^2}

F_1 = \frac{kq_1q_2}{4}

The new force (F2) is

F_2 = \frac{kq_1q_2}{r_2^2}

Substitute 4 for r2

F_2 = \frac{kq_1q_2}{4^2}

F_2 = \frac{kq_1q_2}{4*4}

F_2 = \frac{1}{4}*\frac{kq_1q_2}{4}

Substitute F_1 = \frac{kq_1q_2}{4}

F_2 = \frac{1}{4}*F_1

F_2 = \frac{F_1}{4}

The new force is 1/4 of the previous force.

3 0
2 years ago
he Hobbits are building a watchtower so they can prepare to battle in case trolls decide to attack them. One Hobbit will always
joja [24]

Answer:

5m/8

Explanation:

Function T gives the time the Hobbits have to prepare for the attack, T(k), in minutes, as a function of troll's distance, k, in meters.

Function V gives visibility from the watchtower, V(m), in meters, as a function of the height of the watchtower, m, in meters.

Therefore, T(V(m)) will give the time the Hobbits have to prepare for the troll attack as a function of the height, m, of the watchtower.

We can input m into function V to obtain the visibility from watchtower, V(m), in meters. Since visibility indicates the distance you can see, this also gives the distance of the trolls. This can then be input into function T to obtain the time that the Hobbits have to prepare for a troll attack.

Let's find T(V(m)) by substituting the formula for V(m) into function T as shown below.

T(V(M))=T(50m)

=50m/80

We can simplify this as follows:

=50m/80

=5m/8

4 0
2 years ago
I am thinking of an unreactive nonmetal and Aisha suggested I use bromine. Is this a good choice? Explain your reasoning. If not
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3 years ago
Which options correctly describe the velocity of the object represented in the graph?
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2 years ago
An uncharged capacitor is connected to a 21-V battery until it is fully charged, after which it is disconnected from the battery
attashe74 [19]

Answer:

The voltage bewtween the plates will be 9.5V

Explanation:

Facts:

The capacitance of a parallel plate capacitor having plate area A and plate separation d is C=ϵ0A/d.  

Where ϵ0 is the permittivity of free space.  

A capacitor filled with dielectric slab of dielectric constant K, will have a new capacitance C1=ϵ0kA/d

        C1=K(ϵ0A/d)

        C1=KC   ----------- 1

Where C is the capacitance with no dielectric and C1 is the capacitance with dielectric.

The new capacitance is k times the capacitance of the capacitor without dielectric slab.  

This implies that the charge storing capacity of a capacitor increases k times that of the capacitor without dielectric slab.

Given points

• The terminal voltage of the battery to which the capacitor is connected to charge V=25V

• A dielectric slab of paraffin of dielectric constant K=2  is inserted in the space between the capacitor plates after the fully charged capacitor is disconnected

The charge stored in the original capacitor Q=CV

The charge stored in the original capacitor after inserting dielectric Q1=C1V1

The law of conservation of energy states that the energy stored is constant:

i.e Charge stored in the original capacitor is same as charge stored after the dielectric is inserted.

Q   =  Q1

CV = C1V1

  CV = C1V1  -------2

We derived C1=KC from equation 1. Inserting this into equation 2

   CV = KCV1

 V1 = (CV)/KC

         V/K

       = 21/2.2

      = 9.5

4 0
3 years ago
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