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chubhunter [2.5K]
3 years ago
8

For a particular pipe in a pipe-organ, it has been determined that the frequencies 296 Hz and 370 Hz are two adjacent natural fr

equencies. Using 343 m/s as the speed of sound in air, determine the following. (a) fundamental frequency for this pipe ______ Hz
Physics
1 answer:
user100 [1]3 years ago
8 0

Answer:

fundamental frequency of pipe will be equal to 74 Hz

Explanation:

We have given for a particular organ pipe two adjacent frequency are 296 Hz and 370 Hz

Speed of the sound in air is 343 m/sec

We have to find the fundamental frequency for the pipe

Fundamental frequency will be equal to difference of the two adjacent frequency

So fundamental frequency = 370 - 296 = 74 Hz

So fundamental frequency of pipe will be equal to 74 Hz

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inna [77]

Answer:

Because the electricity flows through and creates static bonds around the metal case which creates a bond with other fields that protects it.

Explanation:

8 0
2 years ago
Two identical copper blocks are connected by a weightless, unstretchable cord through a frictionless pulley at the top of a thin
likoan [24]

Answer:

13.6 N

Explanation:

Since one side of the wedge is vertical and the wedge makes and angle of 33 with the horizontal, the angle between the weight of the copper block on the incline and the incline is thus 90 - 33 = 57.

Let M be the mass of the block that hangs, m be the mass of the block on the incline and T be the tension in the weightless unstretchable cord.

We assume the motion is downwards in the direction of the hanging block, M.

We now write equations of motion for each block.

So

Mg - T = Ma    (1) and T - mgcos57 - F = ma where F is the frictional force on the block on the incline and a is their acceleration.

Now, since both blocks do not move, a = 0.

So, Mg - T = M(0) = 0     and T - mgcos57 - F = m(0) = 0

Mg - T = 0    (3) and T - mgcos57 - F = 0 (4)

From (3), T = Mg

Substituting T into (4), we have

T - mgcos57 - F = 0

Mg - mgcos57 - F = 0

So, Mg - mgcos57 = F  

F = Mg - mgcos57

F = (M - mcos57)g

Since g = acceleration due to gravity = 9.8 m/s², and M = 2.94 kg and m = 2.85 kg.

We find F, thus

F = (2.94 kg - 2.85 kgcos57)9.8 m/s²

F = (2.94 kg - 2.85 kg × 0.5446)9.8 m/s²

F = (2.94 kg - 1.552 kg)9.8 m/s²

F = (1.388 kg)9.8 m/s²

F = 13.6024 kgm/s²

F ≅ 13.6 N

6 0
3 years ago
in the derivation of the time period of a pendulum in electric field when considering the fbd of bob to find the g effective why
Neko [114]

Answer:

we learned that an object that is vibrating is acted upon by a restoring force. The restoring force causes the vibrating object to slow down as it moves away from the equilibrium position and to speed up as it approaches the equilibrium position. It is this restoring force that is responsible for the vibration. So what forces act upon a pendulum bob? And what is the restoring force for a pendulum? There are two dominant forces acting upon a pendulum bob at all times during the course of its motion. There is the force of gravity that acts downward upon the bob. It results from the Earth's mass attracting the mass of the bob. And there is a tension force acting upward and towards the pivot point of the pendulum. The tension force results from the string pulling upon the bob of the pendulum. In our discussion, we will ignore the influence of air resistance - a third force that always opposes the motion of the bob as it swings to and fro. The air resistance force is relatively weak compared to the two dominant forces.

The gravity force is highly predictable; it is always in the same direction (down) and always of the same magnitude - mass*9.8 N/kg. The tension force is considerably less predictable. Both its direction and its magnitude change as the bob swings to and fro. The direction of the tension force is always towards the pivot point. So as the bob swings to the left of its equilibrium position, the tension force is at an angle - directed upwards and to the right. And as the bob swings to the right of its equilibrium position, the tension is directed upwards and to the left. The diagram below depicts the direction of these two forces at five different positions over the course of the pendulum's path.

that's what I know so far

8 0
3 years ago
A simple pendulum consisting of a bob of mass m attached to a string of length L swings with a period T. If the pendulum is take
Juli2301 [7.4K]

To solve this problem we will use the definition of the period in a simple pendulum, which warns that it is dependent on its length and gravity as follows:

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

Here,

L = Length

g = Acceleration due to gravity

We can realize that 2 \pi is a constant so it is proportional to the square root of its length over its gravity,

T \propto \sqrt{\frac{L}{g}}

Since the body is in constant free fall, that is, a point where gravity tends to be zero:

g \rightarrow 0 \Rightarrow T \rightarrow \infty

The value of the period will tend to infinity. This indicates that the pendulum will no longer oscillate because both the pendulum and the point to which it is attached are in free fall.

5 0
2 years ago
An astronomer is measuring the electromagnetic radiation emitted by two stars, both of which are assumed to be perfect blackbody
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There is not enough information to draw a conclusion about
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