6th grade science I mark as brainliest

What is the period of a sound wave having a frequency of 340 Hz

sertanlavr [38]

Answer:

lambda = 343 m/s divided by 340 Hz = 1.009 seconds

Hope it helps and have a wonderful day!

4 0

3 years ago

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you slide across home plate during baseball game. If you have a mass of 82 kg, and the coefficient of kinetic friction between y

Tju [1.3M]

m = mass of the person = 82 kg

g = acceleration due to gravity acting on the person = 9.8 m/s²

F = normal force by the surface on the person

f = kinetic frictional force acting on the person by the surface

μ = Coefficient of kinetic friction = 0.45

The normal force by the surface in upward direction balances the weight of the person in down direction , hence

F = mg eq-1

kinetic frictional force on the person acting is given as

f = μ F

using eq-1

f = μ mg

inserting the values

f = (0.45) (82) (9.8)

f = 361.6 N

8 0

3 years ago

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Which of the following is a characteristic of a mixture

BARSIC [14]

I believe you forgot to add the choices. I will tell you some of the characteristics of mixtures and I hope you find one of them in the choices you have.

A mixture is a physical combination between two or more elements. No chemical reaction is involved in the formation of mixtures.
The components of the mixture can be separated using physical methods such as filtration, boiling and condensation.
Examples of mixtures include mixture of sugar and water or mixture of salt and sugar.

5 0

3 years ago

A stone tumbles into a mine shaft strikes bottom after falling for 4.2 seconds. How deep is the mine shaft

ziro4ka [17]

$d = u \times t \: + \frac{1}{2} \times a \times {t}^{2}$
Since initial velocity is zero hence , u = 0

=> d = 1/2 * a * t2

$d = 0.5 \times 9.8 \times {4.2}^{2}$
on solving we get

d = 86.436 metres

Note ; Here Gravitational Acceleration is take as , g = 9.8 m/s2

3 0

3 years ago

Two taut strings of identical mass and length are stretched with their ends fixed, but the tension in one string is 1.10 times g

ollegr [7]

Answer:

The beat frequency when each string is vibrating at its fundamental frequency is 12.6 Hz

Explanation:

Given;

velocity of wave on the string with lower tension, v₁ = 35.2 m/s

the fundamental frequency of the string, F₁ = 258 Hz

velocity of wave on the string with greater tension;

$v_1 = \sqrt{\frac{T_1}{\mu }$

where;

v₁ is the velocity of wave on the string with lower tension

T₁ is tension on the string

μ is mass per unit length

$v_1 = \sqrt{\frac{T_1}{\mu} } \\\\v_1^2 = \frac{T_1}{\mu} \\\\\mu = \frac{T_1}{v_1^2} \\\\ \frac{T_1}{v_1^2} = \frac{T_2}{v_2^2}\\\\v_2^2 = \frac{T_2v_1^2}{T_1}$

Where;

T₁ lower tension

T₂ greater tension

v₁ velocity of wave in string with lower tension

v₂ velocity of wave in string with greater tension

From the given question;

T₂ = 1.1 T₁

$v_2^2 = \frac{T_2v_1^2}{T_1} \\\\v_2 = \sqrt{\frac{T_2v_1^2}{T_1}} \\\\v_2 = \sqrt{\frac{1.1T_1*(35.2)^2}{T_1}}\\\\v_2 = \sqrt{1.1(35.2)^2} = 36.92 \ m/s$

Fundamental frequency of wave on the string with greater tension;

$f = \frac{v}{2l} \\\\2l = \frac{v}{f} \\\\thus, \frac{v_1}{f_1} =\frac{v_2}{f_2} \\\\f_2 = \frac{f_1v_2}{v_1} \\\\f_2 =\frac{258*36.92}{35.2} \\\\f_2 = 270.6 \ Hz$

Beat frequency = F₂ - F₁

= 270.6 - 258

= 12.6 Hz

Therefore, the beat frequency when each string is vibrating at its fundamental frequency is 12.6 Hz

6 0

3 years ago