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3 years ago

What is the smallest frequency at which the string can be moved to produce any standing wave?

2 answers:

8 0

The first harmonic would be the smallest frequency for a string to produce a standing wave. In addition, the strings were fixed in a single attachment and have only limited motion. It is because standing waves require a specific medium for the sound to travel in it.

dem82 [27]3 years ago

4 0

Answer:

v/2l the first harmonic of the wave.

Explanation:

Since the first harmonic of standing wave is produced when the wavelength is double of the string length, so that crest and trough of the wave falls over one another to form a standing wave.

As v=fλ and λ=2L (L being the length of the string)

so, f=v/2L

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A 0.0400-g positive charged ball with charge q = 6.40 μC is resting on a flat, frictionless horizontal surface. For a time of t

Leto [7]

Answer:

The height is 0.1014 m

Explanation:

Given that,

Mass = 0.0400 g

Charge $q= 6.40\ \mu C$

Time t = 0.0420 s

Electric field $E=7.80\times10^{2}\ N/C$

We need to calculate the electric force on the particle

Using formula of electric force

$F=qE$

Put the value into the formula

$F_{e}=6.40\times10^{-6}\times7.80\times10^{2}$

$F_{e}=0.004992= 0.499\times10^{-2}\ N$

We need to calculate the gravitational force

Using formula of force

$F=mg$

Put the value into the formula

$F_{g}=0.0400\times10^{-3}\times9.8$

$F_{g}=0.000392 = 0.392\times10^{-3}\ N$

We need to calculate the net force

$F_{net}=F_{e}-F_{g}$

$F_{net}=0.499\times10^{-2}-0.392\times10^{-3}$

$F_{net}=0.004598= 0.4598\times10^{-2}\ N$

We need to calculate the acceleration

Using newton's law

$F = ma$

$a = \dfrac{F}{m}$

$a=\dfrac{0.4598\times10^{-2}}{0.0400\times10^{-3}}$

$a =114.95\ m/s^2$

We need to calculate the height

Using equation of motion

$s = ut+\dfrac{1}{2}at^2$

$s=0+\dfrac{1}{2}\times114.95\times(0.0420)^2$

$s=0.1014\ m$

Hence, The height is 0.1014 m

3 0

2 years ago

During an experiment, Ellie records a measurement of 0.0034 m. How would

Goshia [24]

Answer:

(A) She needs to move the decimal point by 3 places

8 0

3 years ago

Calculate the acceleration of a 1000 kg car if the motor provides a small thrust of 1000 N and the static and dynamic friction c

grin007 [14]

Explanation :

It is given that,

Mass of the car, m = 1000 kg

Force applied by the motor, $F_A=1000\ N$

The static and dynamic friction coefficient is, $\mu=0.5$

Let a is the acceleration of the car. Since, the car is in motion, the coefficient of sliding friction can be used. At equilibrium,

$F_A-\mu mg=ma$

$\dfrac{F_A-\mu mg}{m}=a$

$a=\dfrac{1000-0.5(1000)(9.81)}{1000}$

$a=-3.905\ m/s^2$

So, the acceleration of the car is $-3.905\ m/s^2$ . Hence, this is the required solution.

6 0

3 years ago

A constant magnetic field can be used to produce an electric current. True or False?

Umnica [9.8K]

this question is true

6 0

3 years ago

The radius of curvature of both sides of a converging lens is 18 cm. One side of the lens is coated withsilver so that the inner

Dahasolnce [82]

Answer:

n = 1.4

Explanation:

Given,

R1 = 18 cm, R2 = -18 cm

From lens makers formula

1/f = (n - 1)(1/18 + 1/18) = (n-1)/9

f = 9/(n-1)

Power, P = 1/f ( in m) = (n-1)/0.09

Now, this lens is in with conjunction with a concave mirror which then can be thought of as to be in conjunction with another thin lens

Power of concave mirror = P' = 1/f ( in m) = 2/R = 2/0.18 = 1/0.09

Net power of the combination = 2P + P' = 2(n-1)/0.09 + 1/0.09 = 1/0.05

n = 1.4

7 0

2 years ago