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yawa3891 [41]
3 years ago
9

When a tuning fork of frequency 201 Hz vi-brates beside a piano string, beats are heard.The string is tightened slightly and the

beatsgo away.What was the original frequency of thestring
Physics
1 answer:
steposvetlana [31]3 years ago
5 0

Answer:

the original frequency of the string is less than 201 Hz.

Explanation:

Given;

frequency of the tuning fork, F = 201 Hz

Decrease in length of the string by tightening it slightly increases the frequency of the string, thus the original frequency of the string should be less than 201 Hz to create a beat frequency with the tuning fork.

Once the string is tightened slightly, its frequency increases to be equal to 201 Hz, which cancels the beats heard.

Therefore, the original frequency of the string is less than 201 Hz.

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Kinetic and
beks73 [17]

Answer:

raise the board to a higher angle

Explanation:

Static friction is the force opposite to the applied force.

Static friction is dependent on the angle of inclination, it means as the angle of incline increases, the force of friction will increases as normal force will decrease.

So, if the board will be raised to a higher angle, it will increase the angle of incline and will overcome the static friction and block will be able slide.

Hence, the correct option is "raise the board to a higher angle".

5 0
3 years ago
An unknown charged particle passes without deflection through crossed electric and magnetic fields of strengths 187,500 V/m and
Kobotan [32]

Answer:

Explanation:

Given that

The electric fields of strengths E = 187,500 V/m and

and The magnetic  fields of strengths B = 0.1250 T

The diameter d is 25.05 cm which is converted to 0.2505m

The radius is (d/2)

= 0.2505m / 2 = 0.12525m

The given formula to find the magnetic force is F_{ma}=BqV---(i)

The given formula to find the electric force is F_{el}=qE---(ii)

The velocity of electric field and magnetic field is said to be perpendicular

Electric field is equal to magnectic field

Equate equation (i) and equation (ii)

Bqv=qE\\\\v=\frac{E}{B}

v=\frac{187500}{0.125} \\\\v=15\times10^5m/s

It is said that the particles moves in semi circle, so we are going to consider using centripetal force

F_{ce}=\frac{mv^2}{r}---(iii)

magnectic field is equal to centripetal force

Lets equate equation (i) and (iii)

Bqr=\frac{mv^2}{r} \\\\\frac{q}{m}=\frac{v}{Br}  \\\\\frac{q}{m} =\frac{15\times 10^5}{0.125\times0.12525} \\\\=\frac{15\times10^5}{0.015656} \\\\=95808383.23\\\\=958.1\times10^5C/kg

Therefore,  the particle's charge-to-mass ratio is 958.1\times10^5C/kg

b)

To identify the particle

Then 1/ 958.1 × 10⁵ C/kg

The charge to mass ratio is very close to that of a proton, which is about 1*10^8 C/kg

Therefore the particle is proton.

8 0
3 years ago
How does gravity affect the velocity of falling objects?
sladkih [1.3K]

When objects fall to the ground, gravity causes them to accelerate. Acceleration is a change in velocity, and velocity, in turn, is a measure of the speed and direction of motion. Gravity causes an object to fall toward the ground at a faster and faster velocity the longer the object falls

7 0
3 years ago
One of your summer lunar space camp activities is to launch a 1130 kg1130 kg rocket from the surface of the Moon. You are a seri
maxonik [38]

Answer:

∆U = 2.296×10^10Joules

Explanation:

Gravitational potential energy is defined as the energy possessed by an object under the influence of gravity due to its virtue of position.

Potential energy U = Fr where;

F is the force of attraction between the masses of the moon and the rocket.

r is the radius or height of the object.

From Newton's law of universal gravitation, F = GMm/r²

Potential energy U = (-GMm/r²)×r

Potential energy U = -GMm/r

The force is negative because the objects act upward.

M is the mass of the rocket

m is the mass of the moon

Gravitational potential energy possessed by the rocket

U1 = -GMm/r1

r1 is the altitude covered by the rocket

Gravitational potential energy possessed by the Moon

U2 = -GMm/(r2+r1)

r2 is the radius of the moon

Change in gravitational potential energy ∆U = U2-U1

∆U = -GMm/(r2+r1)-(-GMm/r1)

∆U = -GMm/(r2+r1) + GMm/r1

∆U = -GMm{1/(r2+r1)-1/r1}

Given

G = 6.67×10^-11m³/kgs²

M = 1130kg

m = 7.36×10²²kg

r1 = 215km = 215,000m

r2 = 1740km = 1,740,000m

∆U = -6.67×10^-11× 7.36×10²² × 1130{1/(215,000+1,740,000)-1/215000}

∆U= -55.47×10¹⁴{1/1955000-1/215000}

∆U = -55.47×10¹⁴{5.12×10^-7 - 4.65×10^-6}

∆U = -284×10^7 + 257.94×10^8

∆U = 22,954,000,000Joules

∆U = 2.296×10^10Joules

8 0
3 years ago
Institutions that anticipate and recognize significant external forces and modify their strategies and operations will prosper a
vesna_86 [32]

Answer:True

Explanation: External factors are factors outside of an organisation or an institution that has the capacity of either adversely or positively affect the institution or the Organisation.

For an institution to prosper and perform optimally in an economy.or a country, such Institution must put into cognisance the possible external threats that are capable of affecting it, when Organisations or Institution put strategies in place to control or mitigate such externalities,the institution or Organisation will sure Prosper.

5 0
3 years ago
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