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aliina [53]
3 years ago
12

Two violinists are trying to play in tune. However, whenever they play their A string at the same time they hear a beat frequenc

y of 5.0 Hz. One of the violinists measures with an electronic tuner and finds her A string is at 462 Hz. What are the possible frequencies for the A string of the other violinist?(multiple answers)
462 Hz
452 Hz
457 Hz
467 Hz
472 Hz
Physics
1 answer:
kaheart [24]3 years ago
4 0

Answer:

The possible frequencies for the A string of the other violinist is 457 Hz and 467 Hz.

(3) and (4) is correct option.

Explanation:

Given that,

Beat frequency f = 5.0 Hz

Frequency f'= 462 Hz

We need to calculate the possible frequencies for the A string of the other violinist

Using formula of frequency

f'=f_{1}-f...(I)

f'=f_{1}+f...(II)

Where, f= beat frequency

f₁ = frequency

Put the value in both equations

f'=462-5=457\ Hz

f'=462+5=467\ Hz

Hence, The possible frequencies for the A string of the other violinist is 467 Hz and 457 Hz.

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One horse is pulling a 755 kg sled straight ahead applying a force of 1988 N. If the acceleration of the sled is 1.36 m/s2, what
Inessa [10]

Answer:

The coefficient of kinetic friction is 0.13

Explanation:

Newton's second law states that the acceleration of an object is proportional to the net force on it, the factor of proportionality is the mass. So, we can express that law mathematically as:

\sum\overrightarrow{F}=m\overrightarrow{a} (1)

With F the net force, m the mass and a the acceleration of the object. In our case we're interested on what's happening to the sled, then we have to analyze the forces on it, those forces are the weight and the normal force on the vertical direction and the pulling force and frictional force in the horizontal direction. So, because (1) is a vector equation we can express that in their vertical (y) and horizontal (x) components:

F_y=ma_y (2)

F_x=ma_x (3)

On y we have that the acceleration is zero because the sled is not moving upward or downward, remember that the net force on y is the weight (W) pointing downward and the normal force pointing upward:

F_y=W+n=0

Following the convention that positive is upward and negative downward, W=mg=(755)(-9.81):

F_y=(755)(-9.81)+n=0

n=7406.55 N (4)

Now on the x direction we have the sum of the forces is the pulling force (T) and friction force (f)

F_x=F+f=ma_x

Choosing the direction where the horse is pulling F=1988N and the acceleration should be positive too, then:

1988+f=m(1.36)

f=(755)(1.36)-1988=-961.2 N

The negative sign means it's in the opposite direction the horse is pulling

The frictional force is related with the coefficient of kinetic friction in the next way:

|f|=\mu_k n

with μk the coefficient of kinetic friction, and n the normal force that we already found on (4), so we simply solve the last equation for μk:

\mu_k=\frac{|f|}{n}=\frac{961.2}{7406.55}=0.13

4 0
3 years ago
A common physics demonstration is to drop a small magnet down a long, vertical aluminum pipe. Describe the motion of the magnet
Rzqust [24]

Answer and Explanation:

This experiment is known as Lenz's tube.

The Lenz tube is an experiment that shows how you can brake a magnetic dipole that goes down a tube that conducts electric current. The magnet, when falling, along with its magnetic field, will generate variations in the magnetic field flux within the tube. These variations create an emf induced according to Faraday's Law:

\varepsilon =-\frac{d\phi_B}{dt}

This emf induced on the surface of the tube generates a current within it according to Ohm's Law:

V=IR

This emf and current oppose the flux change, therefore a field will be produced in such a direction that the magnet is repelled from below and is attracted from above. The magnitude of the flux at the bottom of the magnet increases from the point of view of the tube, and at the top it decreases. Therefore, two "magnets" are generated under and above the dipole, which repel it below and attract above. Finally, the dipole feels a force in the opposite direction to the direction of fall, therefore it falls with less speed.

7 0
3 years ago
Read 2 more answers
Young's Modulus refers to changes in the: a- Volume b- Length c- Body layers
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<u>Answer:</u> The correct answer is Option b.

<u>Explanation:</u>

Young's Modulus is defined as the ratio of stress acting on a substance to the amount of strain produced.

Stress is defined as force per unit area and strain is defined as proportional deformation in a material.

The equation representing Young's Modulus is:

Y=\frac{F/A}{\Delta l/l}=\frac{Fl}{A\Delta l}

where,

Y = Young's Modulus

F = force exerted by the weight

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\Delta l = change in length

Hence, the correct answer is Option b.

6 0
3 years ago
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Answer:

x = 6.94 m

Explanation:

For this exercise we can find the speed at the bottom of the ramp using energy conservation

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Final point. Lower

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            Em₀ = Em_{f}

            ½ m v₀² + m g h = ½ m v²

            v² = v₀² + 2 g h

             

Let's calculate

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Newton's second law

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               fr = μ N

               fr = μ m g

We replace

             μ m g x = ½ m v²

             x = v² / 2μ g

Let's calculate

            x = 5.89² / (2 0.255 9.8)

            x = 6.94 m

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Answer:

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3 years ago
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